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.
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.
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.
"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.
"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.
"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.
>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.
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.
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 ? )
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.
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.
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.
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 )
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.
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.
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.
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.
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?
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 }
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.
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.
... 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.
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.
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.
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.
Comments
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.
"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.
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.
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.
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.
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.
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.
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 ? )
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.
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.
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 )
None of that sounds cheap, but 'a large active heat-sink inside a wind tunnel' does sound unexpected and impressive
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.
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
Attachment not found.
Phil is it that simple? I am looking too deep into this I think.
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
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.
-Phil
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.
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
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
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.