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Bits
02-22-2012, 02:04 AM
I am designing a circuit that controls current very accurately. My first design restriction is no PWM so it’s all analog for this gal. Also its only DC so that makes things a tad bit easier.

Check it out…
…I am using the following rudimentary circuit to get things started and all on a bread-
Questions:
How do I go about calculating the gate resistor R1 in the picture? Keep in mind switching speed are not a factor here.

If the mosfet I have selected has a Gate threshold voltage of say 2-4 volts. How much deviation can I expect, other than temperature related changes? In other words, how much variation can I expect given that the temperature remains constant?

Bits
02-22-2012, 02:48 AM
Moving forward I have biased the mosfet using an adder circuit. I need to still use some type of feedback system (replacing Gate_TH_voltage). Perhaps a differentiation op-amp circuit below R3 and above R3 respectively.

What do

tonyp12
02-22-2012, 03:48 AM
Your power supply says 10Amp, at these high currents you gone need need to chopper circuit pwm.
http://www.stepperworld.com/Tutorials/chopwave.gif

http://www.stepperworld.com/Tutorials/pgCurrentControl.htm

Tracy Allen
02-22-2012, 04:17 AM
Leave aside for the moment the power implied by the notation of 10A beside the 5V source and the 1Ω resistor. That is definitely a concern.

If you do want to get an accurate current using analog and linear, you are on the right track by including an op-amp in the circuit. However, the mosfet will have to be included in the feedback loop along with a current sensing resistor. The following is the basic idea. The resistor R1 is in series in the loop with the power source, the load, and the mosfet. The voltage across that resistor feeds back to the input of the op-amp and is compared with Vin, and the output of the op-amp drives the mosfet gate one way or the other until the voltage across R1 is exactly equal to Vin. That makes the current in the power loop IL = Vin / R1. You get to choose the value of R1. But also take into consideration the other limits such as power dissipation.

89786

To make the picture bigger, double click on it once you see it in your edit window. Size options will pop up.

Bits
02-22-2012, 04:45 AM
Tracy Allen (http://forums.parallax.com/member.php?40745-Tracy-Allen)

You my friend are reading my mind.:)

I just finished drawing this sipping a warm beer. Take a look below.

My only alteration from yours is the R1. That would have to be a fairly large resistor and I don't have the board space let alone heat evacuation

89791

Bits
02-22-2012, 04:49 AM
tonyp12

Ordanairly I would agree with you but, this is a time where my skills are going to be pushed in a different direction. I can’t PWM client insist and for good reason.

tonyp12
02-22-2012, 02:52 PM
You did not say what current range you want or what you are powering.
Or if this is just a learning experiment and that there are IC's made to do this is not what you want.

Bits
02-22-2012, 03:06 PM
Tonyp

That is a great link, thanks, Ill read it over more today.

This is in fact going to be a product if I can make it work. Here are 3 requirements that I have to meet basically

Tracy Allen
02-22-2012, 03:17 PM
That new circuit idea looks fine to sense the voltage and thus the current across the load resistor directly.

On first glance, the sense of the feedback is wrong, positive instead of negative. R7 could I believe connect to the inverting summing junction between R4 and R5. If R8 is made 20k instead of 10k, then the U1A gain will be the same thru from both the feedback and from the DAC.

In the basic circuit in post #4, the sensing resistor can be quite small, provided that the op-amp has high accuracy. Suppose a 0.04Ω shunt resistor, it develops 0.2 V at 5 amps.

Bits
02-22-2012, 03:43 PM
Again anyone know how to calculate the gate resistor for a mosfet? Is it not so important since switching speeds are low?

Tracy,

Thanks for the input. I will continue looking into a feed back method, but I have to take baby steps :) I would like to someway detect the Gate threshold voltage and add that to R7. This way if the mosfet generates heat and the gate threshold voltage goes up, then the adder circuit can accommodate for it.

Tracy Allen
02-22-2012, 04:04 PM
There is no reason I can see to go to extra work to detect the gate threshold voltage. The op-amp circuit automatically compensates for that. The feedback comes directly from the current sensing resistor and the gate voltage is adjusted to whatever it takes to make that match the input from the DAC.

The resistor R1 from the op-amp output to the mosfet gate should probably be something more like 1kΩ, and 50k is too high. The reason to have a resistor there at all is to isolate the output of the op-amp from the input capacitance of the mosfet, which can be considerable, on the order of 0.01 µF. My inclination would be to dispense with C1, and to make the pulldown resistor R2 1M.

Is the load going to be constant, like a 1Ω resistor as you have it shown? You say switching speeds are going to be slow.

tonyp12
02-22-2012, 04:08 PM
"The MOSFET's advantages in digital circuits do not translate into supremacy in all analog circuits. The two types of circuit draw upon different features of transistor behavior."

I guess you have to look at the VGS chart of your mosfet to find what voltage range you want to be in a slightly open/closed state.

Bits
02-22-2012, 05:09 PM
"The MOSFET's advantages in digital circuits do not translate into supremacy in all analog circuits. The two types of circuit draw upon different features of transistor behavior."

I guess you have to look at the VGS chart of your mosfet to find what voltage range you want to be in a slightly open/closed state.

I agree and will do so.

For the record I chose a mosfet vs NPN because of the availability and cost. NPNs are more difficult to find when we are talking about the higher current requirements.

I did try a smaller value gate resistor R1 but it shorted out my power supply. This is why I went with the larger value gate resistor. R2 is just to keep the input from floating and I could remove C1 I suppose.

Bits
02-22-2012, 06:00 PM
Ahh I apologize, I am now using a smaller R1 (1K) and its not shorting the PS. Can this be due to the (10M) R2? Perhaps I made an error in my testing.

Leon
02-22-2012, 07:46 PM
What is the reason for not using PWM?

davejames
02-22-2012, 08:28 PM
For the record I chose a mosfet vs NPN because of the availability and cost. NPNs are more difficult to find when we are talking about the higher current requirements.

FYI...2N3055 - the work horse since the 60's: 15A, 60V NPN that tops out around 115W (w/heat sink)...and they cost \$2-ish.

tonyp12
02-22-2012, 10:11 PM
All of these can drive a N-mosfet with current control that is set with an external resistor.
Most a highside with charge pump
I'm not sure if they all control curent or just shot it off.
http://www.mouser.com/Semiconductors/Power-Management-ICs/Hot-Swap-Power-Distribution/_/N-wnwuZscv7?P=1z0xtxm

This one even have the mosfet built-in (-2 version =retry)
http://www.mouser.com/ProductDetail/Texas-Instruments/TPS2421-1DDA/?qs=sGAEpiMZZMuUQJU%252b7yHKCnF98N99pHTC

Bits
02-22-2012, 10:38 PM
What is the reason for not using PWM?

Well I think its for a military application and that is specifically what they requested - no PWM.

FYI...2N3055 - the work horse since the 60's: 15A, 60V NPN that tops out around 115W (w/heat sink)...and they cost \$2-ish.

This is good to know. I may end up using a NPN transistor.

All of these can drive a N-mosfet with current control that is set with an external resistor.
Most a highside with charge pump
I'm not sure if they all control curent or just shot it off.
http://www.mouser.com/Semiconductors/Power-Management-ICs/Hot-Swap-Power-Distribution/_/N-wnwuZscv7?P=1z0xtxm

I will check the part/s out. I still think that I wont need all this but I am not counting it out completely.

Tracy Allen
02-22-2012, 10:44 PM
What do you mean that the load will be dynamic? There is a lack of information that limits the help here to less than two cents worth!

The circuit in post #5 (the 2nd circuit, delete the first) shows a 1 ohm resistor which I was assuming would be the constant load, maybe a heater? The circuit you have there will regulate the voltage across it and only secondarily the current. If the resistance changes (dynamically or because you switched in another resistor), then the proportionality changes.

The circuit could work equally well with either a mosfet or bipolar transistor, and either way as a linear circuit it would burn a lot of power. As shown, the maximum power point occurs when 2.5V is across the transistor and 2.5V is across the resistor, the current is 2.5A, and the total power is 12.5W, half in the resistor and half in the mosfet. The maximum current is 5A when the mosfet is all the way turned on, and all 25W is dissipated in the 1Ω resistor and only a residual in the mosfet. If you were doing this with PWM, the mosfet would be either all the way off or all the way on, in either case only residual power would be dissipated in the mosfet. Do you understand those power equations?

tonyp12
02-22-2012, 11:16 PM
You are planning to rebuild something complex like this. (does not even show the inner working of the constant power engine)
If you want the in-rush to be the constant current you could tie timer-pin to gnd ,this one have constant max 6amp and it fast trip at 1.6x the set limit.

tonyp12
02-22-2012, 11:56 PM
89823
You are planning to rebuild something complex like this. (does not even show the inner working of the constant power engine)
If you don't want power reset you could override the timer (or use a very long time)

kwinn
02-23-2012, 12:38 AM
@Bits,

You can't control voltage and current independently across a load. They are related by E=I x R(Z) and its derivative equations. You can control the voltage and let the load resistance/impedance determine the current, you can control the current and let the load resistance/impedance determine the voltage, or you can control the power (E x I) and let the load resistance/impedance determine both. It sounds like you may be trying to avoid the noise that comes with PWM by going to a linear regulator. If that is the case and the current and power dissipation is very high you may want to consider using PWM for a pre-regulator and the linear circuit as a final regulator.

Bits
02-23-2012, 01:22 AM
What do you mean that the load will be dynamic? There is a lack of information that limits the help here to less than two cents worth!

I am so sorry. Let me clear things up a bit. The load is dynamic in the sense that as temperature rises the resistance will change some and yes its a heating device. It may be that I am thinking too deeply and need not consider the small details. The mosfet Gate threshold voltage will also change with temperature, please look here (http://www.fairchildsemi.com/ds/FQ/FQP33N10.pdf).I believe figure 2 in the data sheet will illustrate this change. I am not sure if these things need to be considered and so that is why I say the load is dynamic.

The circuit in post #5 (the 2nd circuit, delete the first) shows a 1 ohm resistor which I was assuming would be the constant load, maybe a heater? The circuit you have there will regulate the voltage across it and only secondarily the current. If the resistance changes (dynamically or because you switched in another resistor), then the proportionality changes.

The circuit is working on paper but, I am not convinced that the differentiator circuit is going to function as I expect it to (Post #5). I wanted to someway subtract the gate threshold voltage from V1. My belief is that the "Gate threshold voltage" will change with respect to temperature thus throwing my DAC resolution off. I plan on using a 16 or even 20 bit DAC to control the mosfet gate voltage.

The circuit could work equally well with either a mosfet or bipolar transistor, and either way as a linear circuit it would burn a lot of power. As shown, the maximum power point occurs when 2.5V is across the transistor and 2.5V is across the resistor, the current is 2.5A, and the total power is 12.5W, half in the resistor and half in the mosfet. The maximum current is 5A when the mosfet is all the way turned on, and all 25W is dissipated in the 1Ω resistor and only a residual in the mosfet. If you were doing this with PWM, the mosfet would be either all the way off or all the way on, in either case only residual power would be dissipated in the mosfet. Do you understand those power equations?

I totally agree that the circuit will work with all sorts of transistors, mosfets too. And I can agree that the power consumed by the mosfet will be significant. I calculated it as P = I^2 R, inserting the mosfet RDson at 52 mOhms then I can say that if I was to run 8 amps through the mosfet I would get P = 8.5a * 52 mOhms = 3.757 Watts. The heat that is created from the mosfet will be managed with a active heat-sink.

@Bits,

You can't control voltage and current independently across a load. They are related by E=I x R(Z) and its derivative equations. You can control the voltage and let the load resistance/impedance determine the current, you can control the current and let the load resistance/impedance determine the voltage, or you can control the power (E x I) and let the load resistance/impedance determine both. It sounds like you may be trying to avoid the noise that comes with PWM by going to a linear regulator. If that is the case and the current and power dissipation is very high you may want to consider using PWM for a pre-regulator and the linear circuit as a final regulator.

I understand this but, I feel that the mosfet has plenty of room to handle the power as a linear device.

This is my thinking. Total power 5 * 8.5 = 43 Watts. Mosfet takes ~ 4 Watts and the load takes 39 watts.giving a efficiency of 91%. This is really not that bad.

tonyp12
02-23-2012, 01:30 AM
>RDson at 52 mOhms
But are you not adjusting gatevoltage to change this RDson to control current?
Maybe your rdson is 1ohm at low gate volatges and more heat.

Bits
02-23-2012, 01:37 AM
>RDson at 52 mOhms
But are you not adjusting gatevoltage to change this RDson to control current?
Maybe you rdson is 1ohm at low gate volatges and more heat.

I am using the DAC in conjunction with a Gate threshold voltage source. I am trying to finish up a feedback system or ref voltage source, at this point its being developed as depicted in post #5.

The RDsOn is constant as far as I am considered since I am operating the mosfet under 20 amps it remains flat.

What are you thinking?

Phil Pilgrim (PhiPi)
02-23-2012, 01:44 AM
Do not worry about the gate threshold voltage. The op-amp will take care of that automatically, as Tracy said. You do need to be concerned about the changing resistance of the load with temperature, though, as that will affect the current, if you're regulating voltage. Are you sure you don't want to regulate temperature instead of current or voltage?

-Phil

Bits
02-23-2012, 02:07 AM
Do not worry about the gate threshold voltage. The op-amp will take care of that automatically, as Tracy said. You do need to be concerned about the changing resistance of the load with temperature, though, as that will affect the current, if you're regulating voltage. Are you sure you don't want to regulate temperature instead of current or voltage?
-Phil

What, why?

I cant believe I made a circuit and cant fully understand it, yet everyone else does. That circuit was made for me to conceptualize things and you guys say it works already?!! OkayI believe you guys but why?

Bits logic...

If the feedback circuit subtracts V1 from the low side of R3 then how is that enough voltage to feed into the adder circuit?
Moreover, if the voltage measures 0 on the low side of R3 (load has max current, voltage is ~0) then the adder circuit is going to be off, right?

Phil Pilgrim (PhiPi)
02-23-2012, 04:52 AM
I'm not claiming to have examined your circuit well enough to understand all of it -- only to point out that the op-amp will do whatever it can to attain stasis, based upon the feedback it receives. If the MOSFET's operating point changes due to temperature, the op-amp will adjust for that. You do not need to make special accommodation for that aspect of the circuit.

-Phil

Tracy Allen
02-23-2012, 05:58 AM
The first circuits you posted in posts #1 and #2 did not have the transistor in the feedback loop, and they would indeed be sensitive to the temperature dependence of the gate-source voltage.

You referred to your #5 circuit as a "differentiator". That is the wrong term. The circuit comprised of U1B is a classic differential amplifier circuit with a gain of x1, and it functions to bring whatever voltage is across the 1Ω heater resistor on the high side of the fet over to a matching voltage referenced to ground. So, whatever voltage there is across the power resistor R1 appears at the left end of R7 wrt ground. 1 volt differential in ==> 1 volt single ended out. In your post #5 circuit, that resistor R7 is connected to the non-inverting input of the other op-amp, U1A, and when the voltage there increases, the output of U1A will increase, and that will cause the current in the mosfet and the power resistor to increase, and the voltage across the resistor increases more, which makes the transistor turn on more, and soon the transistor is all the way on. That's the end of it. The circuit has what is called positive feedback and the result of positive feedback is usually a snap action to one rail or the other, not fine level control that I think you want.

What to do about it? Connect the right end of R7 to the inverting input of U1A instead of to the non-inverting input. Also remove R4 as it complicates the analysis. Leave in R5. Now U1A is a comparator, and there is negative feedback around the loop as a whole. Suppose your DAC produces 2V. There is now 1V at the non-inverting input of the U1A. Reason as follows: Suppose the current through the 1Ω heater happens to be a little less than 2A,. The voltage on the inverting input of U1A will be a little less than 1V, compared to the 1V on the non-inverting input, and that will cause the voltage output of the op-amp/comparator to go up, more positive on the mosfet gate, increasing the current in the mosfet and heater. On the other hand, if the heater current happens to be a little more than 2A, then the inverting input of U1A will be a little above the 1V from the DAC, and that will make the op-amp output go down, decrease the voltage on the mosfet gate, and decrease the current in the heater. The negative feedback will always drive the output toward the point where the voltage across the heater is the same as the voltage from the DAC.

Tracy Allen
02-23-2012, 06:10 AM
Another thing, you said, "I calculated it as P = I^2 R, inserting the mosfet RDson at 52 mOhms then I can say that if I was to run 8 amps through the mosfet I would get P = 8.5a * 52 mOhms = 3.757 Watts. The heat that is created from the mosfet will be managed with a active heat-sink."

That is true, but I was not sure that you realize that the mosfet resistance is only that low and consumes that minimum power when it is fully turned on.

"The RDsOn is constant as far as I am considered since I am operating the mosfet under 20 amps it remains flat."

No no no! In a linear circuit like this, the transistor is essentially serving as a variable resistor and in the middle range its effective resistance will be much higher. I gave an example before. Another, with 8 amps flowing through a 1Ω heater and a 12V power supply, the voltage across the transistor has to be 4V, and the power in the transistor is 4V*8A = 32 watts, while the power in the resistor is 64 watts. You could get the same division of voltage and power by substituting a 1/2Ω resistor in place of the mosfet, so it is as if the mosfet has 1/2Ω of resistance at that operating point.

Bits
02-24-2012, 01:03 AM
I stand corrected. Thanks for clearing this up. Ill be back with more questions once I get some rest. Its been a hell of a week so far and a few more to come.

Bits
02-24-2012, 01:05 AM
While bench testing this circuit I think I have discovered that op-amps hate capacitive loads. I am experiencing an unwanted oscillation.

Mark_T
02-24-2012, 01:40 AM
While bench testing this circuit I think I have discovered that op-amps hate capacitive loads. I am experiencing an unwanted oscillation.

Yes - you'll probably have to find a high current op-amp. Ideally you need to analyse how the phase-shifts associated with the transistor affect the stability, but thats complex stuff... You can add a little high-frequency feedback directly from the opamp output to its inverting input (small capacitor) to get the loop stable (this impacts the bandwidth at which the load follows the DAC input though)

Phil Pilgrim (PhiPi)
02-24-2012, 01:46 AM
Some op-amps are better than others at tolerating capacitive loads. I've had good luck with an Analog Devices OP113 (http://www.analog.com/en/all-operational-amplifiers-op-amps/operational-amplifiers-op-amps/op113/products/product.html) driving a cable, which has similar requirements to driving a MOSFET.

-Phil

Circuitsoft
02-24-2012, 01:53 AM
Just realized you have a positive feedback loop from U1B/R7. Move R7 from U1A+ to U1A-.

An Op-amp tries to balance the input. So, as you increase the + input, the output goes up turning Q1 on further. More current flows, more voltage across R3, more voltage out of U1B. Feed that into U1A-, and it'll balance out.

How significantly does the load resistance change with temperature?

Lawson
02-24-2012, 03:38 AM
You can also calm down unwanted oscillations in a circuit like this by making sure the op-amp circuit doesn't respond too much faster than the Fet + load + load wiring combination. (stray inductance in the load wiring is a real pain for circuits like this) The simplest way to slow this op-amp circuit is to add a Capacitor in place of (or parallel with) R4 . (in the dual op-amp circuit from post #5) Try several values of capacitor, the capacitor size will control the balance between speed and oscillation damping.

Lawson

Bits
02-24-2012, 12:42 PM
Could leaving some of the pins on the LM258 be causing this oscillation? I am using one half of the IC and left the remaining pins floating.

Ill read, research, and try what you guys are suggesting, wish me luck today.

idbruce
02-24-2012, 02:14 PM
I wish you luck Bits :)

Somehow you have managed to attract some of the very sharpest minds in the forum to your thread and you are definitely in good hands. You seem to be very knowledgable and intelligent, so I am sure that you really don't need any luck, but since you asked for it, I wish you the best of luck. Additionally welcome to the forum.

Is that Liberty Lake, IL? Near Wauconda?

Bruce

davejames
02-24-2012, 03:36 PM
Could leaving some of the pins on the LM258 be causing this oscillation? I am using one half of the IC and left the remaining pins floating.

Nowhere near the group of sharpest minds on the Forum, but I do know that leaving opamp inputs floating is asking for trouble (oscillations, output railing, general yuckiness).

Tie the (-) input to the output, and ground the (+) input to avoid the above.

Bits
02-25-2012, 12:50 AM
Apparently I became one of "them". I only read about these nightmares.

So I made a few PCBs to test the design and it works, well some do and some dont!. In fact only one of my boards is/are {{cant remember if its IS or ARE}} working. In fact Its working better than I even imagined. The others I cant speak about because I get sick. The don't work well. :innocent:

This is where it gets strange. The board that works has a definitive oscillations and its something I did not intend on having designed in to the system. The other boards have no oscillation and as mentioned before they are not preforming at all the way I expected them to. I can say that the preform well just not good enough.

The oscillation is, in my opinion, rooted in the feedback circuit and its not consistent in that it will oscillate for a few seconds then subside for a few more only to return again.

Any ideas on how I can troubleshoot this discrepancy?

Bits
02-26-2012, 05:17 PM
Select a mosfet.

Here is what I am thinking. My current mosfet data sheet shows that the Drain to Source voltage with respect to Drain to Source current. I noticed that its, for a lack of better words, is sloppy.

Keep in mind I am not operating this device beyond 10 amps and the Drain to source voltage will never be greater than 12V. That said could I get better resolution using a mosfet with less current to gate voltage ratio? I feel that my circuit is pulsing because I am operating the mosfet in no mans land for a short time then operating the mosfet in a know region for a short time.

Current mosfet selected plot. notice the slope.
90025

Deciding if this will preform better, the slope is greater.

90026

It seems to me that the lesser one will allow me to control the current better. Am I headed in the right direction?

Leon
02-26-2012, 05:35 PM
It's best to drive MOSFETs really hard with a suitable driver. I use the Micrel MIC2246, it delivers 1.5 A into the gate.

Phil Pilgrim (PhiPi)
02-26-2012, 05:39 PM
It's best to drive MOSFETs really hard with a suitable driver.
... unless, of course, your intent is to operate them in their linear region, which I believe is still the objective here.

-Phil

Leon
02-26-2012, 05:41 PM
I should have checked. :(

Bits
02-26-2012, 07:27 PM
Another question -

I read here (http://www.play-hookey.com/analog/non-inverting_amplifier.html) the following:

Resistor Rz has no effect on the gain of the circuit. However, to balance out variations caused by the small input current to the amplifier, Rz should be made equal to the parallel combination of Rf and Rin.

Since I left out the input resistor Rz could this contribute to my oscillation in preceding amplification circuits? Where can I find more information on Rz issues?
90036

Tracy Allen
02-26-2012, 07:32 PM
Bits, those curves are marginally relevant to operation in the linear mode. The main thing they tell you is that either of those transistors can supply the maximum current required, within the Vgs range that the op-amp can supply. It also tells you how much overhead the transistor is going to require. Take that 10 amp figure. The first transistor has a minimum internal resistance of 0.1Ω (the slope of the line, V/I). So when 10 amps passes through it, there will be 1 volt left across the transistor. You can read that right off the graph: 10 amps, 1 volt. That level of current can be achieved with a gate-source voltage of 4.5V, well within what the op-amp/comparator can supply.

If the load is a 1Ω heater, you have to have a power supply of at least 11V to allow for the 1V across the mosfet + 10V across the load. Theoretically, with 10A flowing, there will be 10V across the heater, 1V across the transistor, and the transistor will be burning 10W and the heater 100W. Better a 12V supply though. Those mosfet curves are typical, not best case or worst case. With a 12V supply, the transistor will have to burn more power. With 10A flowing, the transistor has to drop 2V instead of 1V and that means 20 watts for the transistor and 100W for the heater.

With the second transistor, under the same conditions, its internal resistance is closer to 0.01Ω, so you could possibly get by with less power supply overhead, but overall that won't make much difference at all, because you will be operating it in the linear mode, not full on. Those curves do not tell you anything about that.

You might also see a curve like the following:

90035

This shows you the range of gate-source voltages that the op-amp would have to supply as Vgs to the mosfet gate in order to control the current through the power loop. This is an IRF520, which has a much higher threshold than the two transistors in your graphs, but the idea is the same. The local slope of the line is called the transconductance gain (in units of mhos, amps/volts). This graph shows a transconductance gain in the mid-range of about 8A/2V = 4 amps per volt. That gain might be part of the stability equation. I'm not clear on what circuit you are working with now. There were problems with the one in post#5.

Bits
02-26-2012, 08:29 PM
Thanks for the reply Tracy. Ill trash that notion that I have poorly selected a mosfet and move on to the design.

I basically have 4 blocks that deal with controlling the load.

Cir A. Takes the 20 bit DAC output (range of 0 - 5 volts) and multiples it by a factor of 2. Possible issue is that there is no input resistor on pin 3.

Cir B. This is the feedback / voltage offset amp circuit. Takes the overall power (5volts) and subtracts it from the load. As far as I can understand it this circuit will drive the output until the inputs are matched.

Cir C. Takes Cir A and Cir B and combines them together to drive the Gate of the mosfet.

Cir D. Mosfet used as common source configuration. R10 creates a voltage divider with respect to R9 however the large difference of resistance between the two makes looking at this circuit as simple as R10 a pull down resistor. This keeps the gate from floating. My problem area I believe resides in Cir B. Somehow there is an oscillation on pin 6.

Are my equations correct?
90041

Lawson
02-26-2012, 10:05 PM
The listed LM258 op amp has a 45nA input bias current. When this current flows through the parallel 15K resistors of circuit A it will create about a 0.3mV offset. That's far less than the specified 2mV input offset voltage, so it's safe to ignore.

My problem area I believe resides in Cir B. Somehow there is an oscillation on pin 6.
As pin 6 of the LM258 is an input, oscillation at that pin comes from somewhere else. I.e. look at the other side of R4. BTW, op-amps only hold the inputs at equal voltages if they are in a stable circuit with negative feedback. If the op-amp circuit is unstable, op-amps react identically to comparators.

To stabilize the circuit I'd remove R9 (or replace it with 10-100 ohms) and put a 1uF capacitor in parallel with R7. The circuit should now be stable but SLOW with a lot of offset error. Next I'd remove R7 to eliminate the offset error and try successively smaller capacitors till the circuit oscillates again. Finally I'd operate with the smallest capacitor that stabilized the circuit in testing.

Lawson

Bits
02-26-2012, 10:46 PM
As pin 6 of the LM258 is an input, oscillation at that pin comes from somewhere else. I.e. look at the other side of R4. BTW, op-amps only hold the inputs at equal voltages if they are in a stable circuit with negative feedback. If the op-amp circuit is unstable, op-amps react identically to comparators.
Lawson

I was leaning towards the feedback into pin 6 as the oscillation. Since the feedback stems from the mosfet its as if it creates a loop where one feeds the other.

To stabilize the circuit I'd remove R9 (or replace it with 10-100 ohms) and put a 1uF capacitor in parallel with R7. The circuit should now be stable but SLOW with a lot of offset error. Next I'd remove R7 to eliminate the offset error and try successively smaller capacitors till the circuit oscillates again. Finally I'd operate with the smallest capacitor that stabilized the circuit in testing.
Lawson

I will give this a try.
Thanks for the help everyone so far I have learned an overwhelming amount of "stuff" I feel gluttony-ish.

Tracy Allen
02-27-2012, 08:07 AM
First off, is it really the voltage across the heater that needs to be controlled, not the temperature or something like that?

As for the circuits, your math for A and B is fine, but C is not quite right, and D needs more consideration to close the loop. For circuit C, the equation for the mosfet gate-to-source voltage should be
Vgs = (DAC * 2) * (R7 / R8 + 1) - Vref * R7 / R8
The main difference is that there is an inverting gain factor for Vref. It is possible to work that out from basic principles, but when you become practiced at looking at op-amp circuits, it is just something you come up with "by inspection". With the specific resistor values you chose, that simplifies to,
Vgs = (DAC * 2) * 6 - Vref * 5.
(I used 50k for R7 instead of 47k, just to make neat numbers) The (DAC * 2) gain factor comes from your circuit number "A". That does not really tell you how to find Vgs, because it is part of a larger feedback loop, closed in circuit D. What you call Vref is equal to the voltage across the heater, which is IL * RL. So you can plug that in to find,
Vgs = (DAC * 2) * 6 - IL * RL * 5.

So what does that tell you? Still not much. One equation in at least two unknowns. Okay. You know you want a certain stable relation between input and output, but how do you get rid of Vgs?

Enter here the transconductance gain curve for the IRL530 from a data sheet, plotted on linear coordinates. (Sometimes it is done on semilog, so that you can discern more detail as Vgs decreases.) The trick is to plot the above equation on the same graph. Where the two lines intersect (assuming that they do so) is called the operating point. A little algebra can solve the above equation for IL versus Vgs. Here it is (if I got it right), with the further assumption that RL = 1Ω.
IL = (DAC * 12 - Vgs) / 5.
The dark blue line on the graph is with DAC = 1 V, and the light blue line is with DAC = 2.5 V.

90083
One thing to notice is that if and when the transfer characteristics of the mosfet shift or drift a little left or right, the operating point also moves a little up or down. Bummer. Mosfet transfer characteristics vary not only from one to the next of the same part #, but also, as you know, with temperature. So the circuit is left with the dependence you have been trying to avoid.

Something funny happens when you get rid of resistor R7. The math becomes simpler, and the curve becomes like the horizontal red line, and mosfet transfer characteristic become pretty much irrelevant.
The math is,
Vgs = ((DAC * 2) - IL * RL) * Aol
where Aol is the open loop gain of the op-amp (now acting as a comparator), usually a number around 100_000. Rearranging, and again assuming RL = 1Ω
IL = (DAC * 2) / 1Ω - (Vgs / (Aol * 1Ω))
= (DAC * 2) / 1Ω
I dropped the term that has the huge number in the denominator. Violà--No more Vgs. The red line on the graph corresponds to DAC = 2.5V. Solid 5A no matter what Vgs does. The accuracy of the circuit depends on a high gain through the loop.

Stability is another question. It would really be a lot simpler if you could connect the heater between the mosfet source and ground, and operate it as a source follower. However, again, is the goal really to regulate the voltage across the heater?

Circuitsoft
02-27-2012, 09:29 PM
You can simplify the circuit and limit oscillation by replacing R7 with a capacitor, such as 0.18 uF, which will give you a cutoff frequency of ~88Hz.

The question stands, do you really want to regulate heater voltage?

A simpler constant current option may be:http://forums.parallax.com/attachment.php?attachmentid=90098&d=1330381710

This circuit should be stable, and will give you 10 amps output with 5V input, linearly.

Circuitsoft
02-27-2012, 09:29 PM
Oh, you will want to put a resistor in series with the gate of the mosfet, maybe 10k or so, or it could oscillate on it's own.

tonyp12
02-27-2012, 10:22 PM
Pretty high watt rating for that 0.03ohm needed.
P = I^2 * R

= 10*10 * 0.03 = 3watt
So at least a 5 watt just to be safe.
http://www.mouser.com/Passive-Components/Resistors/Current-Sense-Resistors/_/N-7fjcfZscv7?P=1z0vjvmZ1z0vo1zZ1z0vl80Z1z0vnpi&FS=True&Ns=Pricing|0

Bits
02-28-2012, 01:38 AM
Okay sleeves are rolled up, time to dig in my high heels and debate a bit (friendly of course) ...

So quite a few people have mentioned or, at least as I interpret it to be, that I am trying to vary the voltage. This circuit can’t do that and I don’t want to vary the voltage. I want to vary the current.

I declare that the voltage on the load is constant. Take a look at my post #47 and in particular Cir D.

The voltage before the load is 5V and that can not change no matter what the mosfet “resistance” becomes or what the load resistance is for that matter. It will still maintain 5 volts.

So why are people asking me this?

You can simplify the circuit and limit oscillation by replacing R7 with a capacitor, such as 0.18 uF, which will give you a cutoff frequency of ~88Hz.

The question stands, do you really want to regulate heater voltage?

A simpler constant current option may be:90098

This circuit should be stable, and will give you 10 amps output with 5V input, linearly.

This is a sweet option but I don't like the feedback resistor.

The main difference is that there is an inverting gain factor for Vref. It is possible to work that out from basic principles, but when you become practiced at looking at op-amp circuits, it is just something you come up with "by inspection".

Is it that obvious? :smile: Thanks for correcting me. Its a bit difficult for me to see this in my head but I am starting to see through the fog. Please dont take that the wrong way I am thankful for your help and the help of everyone else.

Tracy Allen
02-28-2012, 04:37 AM
So quite a few people have mentioned or, at least as I interpret it to be, that I am trying to vary the voltage. This circuit can’t do that and I don’t want to vary the voltage. I want to vary the current.
I declare that the voltage on the load is constant. Take a look at my post #47 and in particular Cir D.
The voltage before the load is 5V and that can not change no matter what the mosfet “resistance” becomes or what the load resistance is for that matter. It will still maintain 5 volts.

It seems you are equating what we are asking about voltage with the power supply voltage. No. It is nice that the power supply is so stiff and stable, but that is irrelevant so long as the it is adequate. We were talking about the voltage across the load.. Across means the voltage you would measure between the power terminal and the node you have called "feedback". The voltage across the resistor/heater/load. The mosfet absorbs the rest of the voltage and power, and the sum across the whole circuit is the total power supply voltage.

So, you really do want to regulate via the DAC current through the load, not the voltage across the load, nor the power dissipated by the load, nor the temperature of the load? We are just asking. There may be good reasons for doing any of those things.

If it is current through, you will definitely need a way to sample that current and use it for feedback. Like it or not (why do you say not?), the circuit that Circuitsoft posted just above is a superbly sweet option!

Tracy Allen
02-28-2012, 04:50 AM
but when you become practiced at looking at op-amp circuits, it is just something you come up with "by inspection". Is it that obvious? :smile: Thanks for correcting me. Its a bit difficult for me to see this in my head but I am starting to see through the fog. Please dont take that the wrong way I am thankful for your help and the help of everyone else.

No, it is not obvious. I salute you for searching your way through the fog! If you want, I could show how it is derived, or you can find it in many books. After you've been through it a few times, you will begin to see the patterns. I didn't want to get into those details, because I wanted to get into rest of the loop.

In Horowitz and Hill, Art of Electronics, they end many chapters with a collection of bad circuits, things that might look good on the surface but have a serious flaw. It is a good exercise to work through circuits even if they don't work as expected. By the way, if you don't have a copy of that book, put it on your list.

Circuitsoft
02-28-2012, 06:26 AM
Horowitz and Hill, Art of Electronics, ... By the way, if you don't have a copy of that book, put it on your list.Definitely second that.

For my circuit, you can use a smaller resistor between the source of the MOSFET and ground, and just adjust the input voltage divider to match.

Basically, pick a resistor for the current sense. I chose .03 ohms because it made math easy, but smaller, such as .01 ohms, would make more sense. Once you know what the voltage across that resistor will be at full scale (current * R = voltage), then you want full scale input (5) equal to that voltage times ((r2+r1)/r1) for the divider.

r1 = input resistor, currently 4.7K
r2 = input shunt, currently 300
r3 = source resistor, currently 30m
vin = maximum input voltage from DAC (I'm assuming 5)
Iout = maximum output current to heater (I'm assuming 10)

vin = ((r1+r2)/r2) * Iout * r3

Power dissipation of r3 = Iout*Iout*r3

Circuitsoft
02-28-2012, 06:29 AM
Oh, and if that circuit oscillates, just make the capacitor bigger until it stops oscillating.

Bits
10-24-2012, 02:21 AM
I need one more detail explained to me if someone can.

Circitsoft suggested in post #51 a circuit for me to try. It works great but there are some personal nagging concerns I have. I now know that the feedback resistor cant be to low or the performance degrades below my needs. Initially I implemented a resistor of .01 Ohms just to keep the power dissipation low on this part of the circuit board. Nevertheless I am not sure what to call this; too low of a feedback resistor problem but, it must be some type of spec on the op-amp I am reaching. Perhaps when a feedback resistor is too low the op-amp will in turn not notice it as much and thus the output will reflect this. Is it gain? I found that if I use any feedback resistor below .03 Ohms the circuit wont operate the way I need it to. Not a big deal because a .05 Ohm resistor is perfect. Perhaps an analog guru could explain why this is.

Then my other question is why wont the output of the op-amp go below the gate turn on voltage? When the + op-amp input is 0 V the op-amp output is always right at the gate turn on voltage. Is this just simply the way feedback is designed to do and if not what is this process called?

Thank you Circuitsoft for helping me on this as well.:smile:

Tracy Allen
10-24-2012, 03:16 AM
Hi Bits, good to see you around.

"I found that if I use any feedback resistor below .03 Ohms the circuit wont operate the way I need it to. "

Can you say a bit more about what happened? Did you also make the changes at the input voltage divider that circuitsoft suggested?

"why wont the output of the op-amp go below the gate turn on voltage?"

That may be because the op amp is still maintaining a small current through the load. Did you measure the voltage across the load, when the input voltage is zero? You'd expect zero there too, right?

Op-amps however have what is called an input offset voltage, and that acts like a small signal that is always present, and it adds to your input signal. You can always find it in the data sheet for an op-amp. Here it is for the LM258. (Is that still the op-amp you are using?)

96484

What that means is that there is typically an effective 2 millivolt input signal even when your input from outside is ostensibly at zero. That would make the circuit try to sink a current of 2mV/0.03Ω = 67 mA through the heater. The op-amp holds the mosfet gate near threshold to make that happen.

If you apply a few millivolts NEGATIVE at your input, the op-amp will try its best to generate a negative current through the heater, however, the best it can do is shut the mosfet off completely, and to do so it will drive the gate all the way down to ground potential.

Bits
10-24-2012, 04:01 PM
Hi Bits, good to see you around.

"I found that if I use any feedback resistor below .03 Ohms the circuit wont operate the way I need it to. "

Can you say a bit more about what happened? Did you also make the changes at the input voltage divider that circuitsoft suggested?

It basically wont control the heater as well. It begins to oscillate. Now I did change the capacitor on the gate but this approach did not affect the oscillations.

Circuitsoft
10-24-2012, 04:17 PM
What frequency is it oscillating at? A few Hz, or several MHz? You may need to increase the capacitor significantly more to make a difference if it's oscillating slowly. If it's fast, then put a resistor (1k or so) between the op-amp and the MOSFET.

Bits
10-24-2012, 04:32 PM
Its a few HZ and I have tried larger Capacitors 100uF as an example. I really think that the feedback resistor is the issue as too small of a value creates an unwanted oscillation. Then again I could be wrong.

Circuitsoft
10-24-2012, 04:58 PM
That could be it. The MOSFET probably has a pretty high gain near it's cutoff voltage, so when the resistor gets that low, it acts a lot more like a switch. A larger current-sense resistor actually lowers the effective gain of the MOSFET reducing the frequency response of the system as a whole and bringing any poles closer to the 0 provided by the cap in the negative feedback of the op-amp.

Tracy Allen
10-24-2012, 05:02 PM
"I did change the capacitor on the gate but this approach did not affect the oscillations. "

Could you draw out the circuit values as you have them? Here is the circuit from post #51. When you have 0.01Ω for the shunt, what values do you have for the other components?

I think I would try connecting the right-hand side of the capacitor to the source of the mosfet instead of to the gate. (That is, the node where the source meets the 10kΩ, 30mΩ resistor.) Again, experiment with the C value.

http://forums.parallax.com/attachment.php?attachmentid=90098&d=1330381710

"I really think that the feedback resistor is the issue as too small of a value creates an unwanted oscillation."

A resistor is a passive element and won't create an oscillation by itself. The oscillation comes from delays, phase shifts, in the whole feedback loop.

Tracy Allen
10-24-2012, 05:14 PM
"The MOSFET probably has a pretty high gain near it's cutoff voltage"

Circuitsoft, but the mosfet in this circuit is a source follower, so the gain in any case is just a hair's breadth below unity. What do you think about connecting the capacitor to the source instead of the gate? There shouldn't be much of a phase difference, unless...

I'm wondering if Bits put a resistor in between the op-amp output and the gate. I'd go with more like 100Ω there, otherwise it will cause a big phase delay with the input capacitance of the large mosfet, and a compensating phase lead directly from the op-amp output would certainly be necessary.

Circuitsoft
10-24-2012, 05:21 PM
Also, try swapping out your MOSFET for a pair of paralleled 2SK1056 transistors. They're very linear devices made for audio use.

tonyp12
10-24-2012, 05:36 PM
>Here is the circuit from post #51
For newbie to analog type circuits (like me), I will try explain what it is doing.

The 4.7k and 300 ohm on the left creates a voltage divider (300/4700*5v =0.31915volt)

The two 10k in series resistors, limits the currents and smooth's out any spikes.

The 30mO (0.03 Ohm) before gnd allows you to measure voltage drop vs current and the amp will compare it to 0.31915volt.

The output from the amp will be in the mosfets half open/half closed state region (does create heat)
the more current flowing the more is try choke itself, but that could induce oscillation and capacitor placed at right place will help.

Circuitsoft
10-24-2012, 05:50 PM
Almost.

300/(4700+300)*5v = 0.3v

The 10k resistors are there to equalize voltage differences that could arise from any input current of the op-amp.

Bits
10-24-2012, 07:45 PM
From my testing the voltage divider is not even needed in the circuit and it all boils down to this simple equation.

Current through heater element = Voltage at + op-amp / power resistor or...

I = (+v) / (.05 Ohms) as I am using a .05 Ohm resistor at the moment.

That simple! I swapped out all the resistors with assorted values and conclude that the equation stays the same no matter the resistor sizes.

Bits
10-24-2012, 07:53 PM
Also, try swapping out your MOSFET for a pair of paralleled 2SK1056 transistors. They're very linear devices made for audio use.

I have this on the back burner but I did not consider using 2 transistors. I went with one power transistor BDW46G and BD910. Playing with this config next week.

jmg
10-24-2012, 08:29 PM
Its a few HZ and I have tried larger Capacitors 100uF as an example. I really think that the feedback resistor is the issue as too small of a value creates an unwanted oscillation. Then again I could be wrong.
A few Hz is very low, so I would suspect something else.
Try building the circuit in #65, with a simple trim pot source, and check your current control range.

Also note that the figures give around 10 Amps, and depending on the heater resistance, your MOSFET will generate a lot of heat.
Even at ~ 1ohm of heater, the 6A drive level, will give 36 Watts (!) in the FET.

Most heater driver designs switch the power, and use thermal inertia to smooth the ripple.

Some use Dual heaters, to lower the ripple effect, as often you have two requirements : Fast heat-up, and then fine-control,and the fine-control only needs balance the losses.

Or is this application thermal-ripple paranoid ? (like a Oven controlled crystal, or similar ? )

Circuitsoft
10-24-2012, 09:13 PM
Based on other posts Bits has made, I'm guessing she's trying to drive a heater that's a ways down a cable and minimize any EMI generated by the heater wires.

tonyp12
10-24-2012, 09:24 PM
Having a 70-90% duty cycle to finetune heat on a 1hz digital signal (fully on/off) to mosfet should not create any EMI?

Or have two parallel circuits,
one that is always on that reach 90% off the heat needed
and a digital 1hz that add the 0-15% needed, should create less spikes.

Bits
10-24-2012, 09:40 PM
Based on other posts Bits has made, I'm guessing she's trying to drive a heater that's a ways down a cable and minimize any EMI generated by the heater wires.
You are correct :)

Others, again I am not using PWM because of the 3 meter cable between the heater and mosfet. And well PWM 10 amps causes problems unless you want to use big inductors etc. For the record the mosfet is coupled to a large active heat-sink inside a wind tunnel for maximum cooling.

The circuit works perfect as I stated. I just wanted some answers to my nagging thoughts. No big deal ill keep testing until I get the eureka moment.

jmg
10-24-2012, 09:44 PM
Based on other posts Bits has made, I'm guessing she's trying to drive a heater that's a ways down a cable and minimize any EMI generated by the heater wires.

Ok, then I'd go for a trapezoid control - not really PWM, and not high dissipation linear either.
{ I think this was also measuring a bridge sensor down other wires ?}

Use a circuit like #65, and slow the slew to 10ms or slower, and generate control that avoids watts in the MOSFET.
At 'edge of the curve' powers, eg < 10%, > 90%, then work linear/fully saturated.
For worst case area like ~50% power, switch between those levels, with slow current modulation. Peak FET W is still 36W, but only briefly.
A more manageable thermal average of a couple of watts would be the target.

If a sensor is involved, those modulation changes could also be reading-synchronized. (even tho a 10ms slew is already very low EMC )

jmg
10-24-2012, 09:47 PM
For the record the mosfet is coupled to a large active heat-sink inside a wind tunnel for maximum cooling.

None of that sounds cheap, but 'a large active heat-sink inside a wind tunnel' does sound unexpected and impressive :)

Tracy Allen
10-24-2012, 10:05 PM
From my testing the voltage divider is not even needed in the circuit and it all boils down to this simple equation.

Current through heater element = Voltage at + op-amp / power resistor or...

I= (+v) / (.05 Ohms) as I am using a .05 Ohm resistor at the moment.

That simple! I swapped out all the resistors with assorted values and conclude that the equation stays the same no matter the resistor sizes.

That's the fundamental principle of negative feedback with an op-amp. The op-amp if possible will make the voltage at the two inputs (+) and (-) equal to one another. All the other parts in the circuit are there to make a given output from a given input possible and stable within the allowable range of the op-amp inputs.

Bits
10-24-2012, 10:52 PM
That's the fundamental principle of negative feedback with an op-amp. The op-amp if possible will make the voltage at the two inputs (+) and (-) equal to one another. All the other parts in the circuit are there to make a given output from a given input possible and stable within the allowable range of the op-amp inputs.

True.

But what I was tough a long time ago is not working today. I still cant visualize why I place 0 volts on the + and the output is 2-4 volts depending on mosfet selected. Even when I run this on my SPICE program it behaves the same. So I am missing something or being daft.

Jmg,
Heatsink is \$1.20 and the tunnel is part of the package this PCB mounts into. Oh a TO-220 isolation kit runs me about .50 cents so all in all its a cheaper than any other circuit I can come up with to do the same thing.

jmg
10-24-2012, 10:58 PM
I still cant visualize why I place 0 volts on the + and the output is 2-4 volts depending on mosfet selected.

Do you mean 2-4 V on the Opamp output pin ?
That is 100% expected, as the Opmap output moves to make the + and - pins equal.
That 2-4V on the Opamp output is the linear operating range. The Source voltage is what it cares about.

If you happen to hit an opamp with a high offset voltage, such that the -ve pin cannot go negative enough, then the Opamp output could go to ~ 0V, and the circuit would be outside the linear region. As the +ve ip increases, you would see a change to linear operation, and 2-4V on the Op/Gate again.

Phil Pilgrim (PhiPi)
10-24-2012, 10:59 PM
I still cant visualize why I place 0 volts on the + and the output is 2-4 volts depending on mosfet selected.
It's because the feedback is coming from the MOSFET's source pin, not the output from the op amp. The voltage on the output pin will be whatever the MOSFET's gate threshold voltage is, IOW the voltage that it takes to just begin making the MOSFET conduct and provide a non-zero feedback voltage to the op amp's negative input.

-Phil

Bits
10-24-2012, 11:02 PM
Here is the thing I can even cut the trace above the mosfet and it still reads over 0 volts on the op-amp output. How is this possible if both pins are 0 then the output should be 0 right?

96499

Phil is it that simple? I am looking too deep into this I think.

jmg
10-24-2012, 11:19 PM
Here is the thing I can even cut the trace above the mosfet and it still reads over 0 volts on the op-amp output. How is this possible if both pins are 0 then the output should be 0 right?

Only in an ideal opamp, with truly zero offset voltage (but still finite gain) will the output be 0v.

In the real world, the open-loop-DC voltage will depend on the offset voltage and the Gain.

With the Drain open, the opamp could be expected to be well above the FET threshold, for those positive offset cases.

An LM258 specs |Vos| of 2.9mV typ and 5mV max, and Ios of 3nA typ and 30nA max (Ios adds 300uV over 10k).
The Gain is Min 50V/mV and Typ 100V/mV
With a 12V supply, only for voltages within 120uV of the |Vos| will the output not be saturated, open loop.

If you are really lucky, and have a +20uV nett-circuit-offset part, shorting the pins on that would give an average DC Vo of ~2V.
{ the LM258 data does not even mention the noise voltage, but your 180nF does help limit the noise BW }

Phil Pilgrim (PhiPi)
10-24-2012, 11:27 PM
Even with an ideal op amp having zero offset, there could still be a non-zero output with "0 volts" on the positive input. Any noise picked up by the positive pin will cause the op amp's output to make an excursion to the MOSFET's threshold voltage.

-Phil

jmg
10-24-2012, 11:47 PM
Jmg,
Heatsink is \$1.20 and the tunnel is part of the package this PCB mounts into. Oh a TO-220 isolation kit runs me about .50 cents so all in all its a cheaper than any other circuit I can come up with to do the same thing.

36W is pushing practical limits with a TO-220, if you do intend to operate at that peak, you could consider spreading over two FETs - essentially just double the circuit.

Bits
10-25-2012, 08:39 PM
No the MOSFET is rated well over 100 watts.

Phil Pilgrim (PhiPi)
10-25-2012, 09:17 PM
... you could consider spreading over two FETs - essentially just double the circuit.
MOSFETs load-share just fine in switching circuits, where they're driven to saturation. I doubt that the loads would be shared anywhere near equally in the linear region, due to variations between them in source/drain resistance vs. gate voltage.

-Phil

jmg
10-25-2012, 09:32 PM
MOSFETs load-share just fine in switching circuits, where they're driven to saturation. I doubt that the loads would be shared anywhere near equally in the linear region, due to variations between them in source/drain resistance vs. gate voltage.

Yes, which is why is said essentially just double the circuit. - most opamps come as Duals, and the +ve IP could be paralleled, but the Source feedback need to be duplicated per-fet for best linear matching.

Duane C. Johnson
10-25-2012, 09:34 PM
Yes, it can be rated for lots of power dissipation.

However, this power dissipation is done while the MOSFET is heavily turned on.

It is generally not recommended to dissipate high power while in the linear region.
I would derate the maximum power specification to around 25% or so.

Here is the problem:
MOSFETs are not constructed of one single homogenous device. It is actually composed of maybe 100,000 individual MOSFET regions on the die. Since they do not all have exactly the same gate threshold voltage some regions will conduct heavily before other regions.

The problem is some of these regions may be in over current and heating excessively and other regions relatively cool.
When in conventional turned on mode all regions tend to have close to even current flow so there is much less problem with hot spots.

International Rectifier had a writeup on this one time.

Duane J

jmg
10-25-2012, 09:58 PM
No the MOSFET is rated well over 100 watts.

That's in an ideal world of infinite heatsinks, and no mounting insulators, and 25'C controlled environments, and maximum thermal cycling on the Tj.

I prefer to look at the TO220F package, which has package-insulation included.

Those packages indicate 41W @ 25'C and 20W @ 100'C Tcase, for Tj @ MAX.

If this is a hi-rel/military/long life design, I would design some distance off maximum thermal cycling for margin.

Also note Clip mounting is more consistent than nut+bolt, and much more reliable.

See
http://www.nxp.com/documents/application_note/AN11172.pdf

and if you need insulation, the best solution is a Clip plus heatsink compound and 0.25 mm maximum alumina insulator

Bits
11-16-2012, 04:06 PM
Okay I think this is solved.

All I ended up doing is increasing the feedback resistor to a 10M instead of the 10K. Naturally I also increased the input impedance matching resistor to the same value.

Thanks to everyone that helped me along the way on this one. Circuitsoft big hugs out to you thank you for all the help.