High current/voltage load

Ronnie B · 2009-01-31 23:23

Hi, I have been spending the whole day searching for a solution of my problem but so far I haven't succeeded.
So now I make a new post hoping some of you can help me.·

·
I want to make a temperature control with a sensor and a heater. The heater is 24V and 60W so there will be quite high current.
I could use a relay but I'd prefer to have a PWM solution for better accuracy. I also want my circuit to supply the positive voltage to the heater. The negative side of the heater will be connected to a common ground. In all the examples the load is supplied the voltage and then the transistor cuts the negative side. This is not a common solution in European Industries where I'm working that’s why I need the other way around.
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Also I don't want the 24V to be supplied out from the circuit if the output from the Propeller Chip is low (i.e. the circuit is off). For personal and machine safety I want to have a active high signal from the Propeller to supply the 24V 2.5A.
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Also, it's a low-budget project so I prefer low-cost components.
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Hope you can help me here.
·
Ronnie

mctrivia · 2009-01-31 23:29

sounds like you are trying to build a ac line cliper circuit. If so you need an optoisloater and a Triac.

Leon · 2009-01-31 23:37

Do you have a suitable DC supply to power it?

Leon

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Ronnie B · 2009-01-31 23:44

Hi, thanks for the reply.

A Opto and a triac would be something like the lightbulb in
http://www.parallax.com/Portals/0/Downloads/docs/cols/nv/vol8/col/nv146.pdf

But it's no really how I like to do. I try to sketch my needs.

·······

VCS-|················· |- +24VDC
VSS-|················· |
··· ·· |·· CIRCUIT··· |----HEATER----|
··· ·· |················· |----HEATER----|
··· ·· |················· |····················---
··· ···

····················· -· 0VDC

Here is a exmple how I like to supply·2 heaters with 24V. The 24V is connected to the circuit.
Then only one wire per heater from the circuit should be needed. The 0VDC will be connected directly from the powersupply to the heaters.

Hope I don't make things harder to understand.

Leon:
Yes, I have a 24V 10Amp powersupply so thats not a problem.

Ronnie

mctrivia · 2009-01-31 23:54

then you need heavy duty transisters to switch the 24V dc power. you may need to cascade a few transisters but it should not be hard

Phil Pilgrim (PhiPi) · 2009-02-01 00:24

For this job, I'd consider a Micrel MIC5021 high-side MOSFET driver and an appropriate n-channel MOSFET.

-Phil

Peter Jakacki · 2009-02-01 02:20

Triacs are not suitable for DC voltages, NPN darlingtons will drop 1.2V in and out and thus will need to be driven with at least 24V already plus they will dissipate a few watts of heat. The best solution is to use an NPN to buffer and invert the signal from the Prop and use a PNP or P-channel MOSFET to drive the load. The part numbers are representative of what is required. 3 amps is not really high current and the heater load is resistive making the circuit fairly simple. The PWM should be very low frequency though to minimize the switching losses and utilizing the thermal inertia of the heater to even it out. Parts cost should be around a buck.

*Peter*

P.S. I have included R3 to reduce the Vgs to less than the 20V or so breakdown.

Post Edited (Peter Jakacki) : 2/1/2009 2:38:39 AM GMT

mctrivia · 2009-02-01 02:36

will that circuit work for running leds also if 3.3V is used for source?

kwinn · 2009-02-01 02:53

What you need is an optoisolator with a transistor (npn) or darlington transistor output and a PNP power transistor. The transistor in the optoisolator only conducts when the propeller pin outputs a high to the LED (through a current limiting resistor), the base is connected to ground and the collector of the transistor is connected (through a resistor) to the base of the PNP transistor. The emitter of the PNP is connected to the +24V, and the collector goes to the heater (which has the other end connected to ground).

PS If you do not already have a 24VDC supply it would be simpler to use a 24V transformer and a solid state relay to drive the heater with 24V AC. Vary the ratio of cycles on/off each time period and you achieve the same result. Works good, lasts a long time.

Peter Jakacki · 2009-02-01 03:36

@kwinn, the CTR of an opto is normally very low in the range of 10 to 50% which means that the opto has to be pumped with at least 100ma (spec = 25ma max) as the pnp requires at least 50ma base current to achieve saturation. Not a workable solution if you do your maths. Sure it will work if you use a MOSFET instead of the PNP but why bother with an opto anyway?

Your triac solution is certainly a simpler and practical if AC is available as you stated.

@mctrivia, this circuit provides +24V drive from a logic signal. In the case of LEDs it is more of a matter of how they are hooked-up, certainly if there are several in series with the current limit resistor and multiple parallel strands then of course it will work. If you are driving a single led then this is a most inefficient circuit.

*Peter*

mctrivia · 2009-02-01 03:59

i am driving a 60x14 grid of leds but only 1 is lit at a time. currently they are being driven direct by SN74HC14N ic(well 30ohm resister in series) for the positive side and CD74HC154E for the negative side. works but not that bright.

kwinn · 2009-02-01 05:17

Peter, you are correct about the CTR of an optoisolator with a transistor output. I should have suggested only the darlington output opto. As for the transistor you are correct again. I have a drawer full of TIP125's that I use for such applications and I sometimes forget that it also is a darlington transistor. On the whole though I still feel that an optoisolator such as the 4n33 and a PNP darlington transistor such as the TIP125 (or a mosfet) is a safe and economical solution for such an application. I suggest them because I have used them often with extremely few problems or failures. You can't really argue with "works good, lasts a long time, and is cheap'.

Ronnie B · 2009-02-01 10:34

Hi everybody, thanks for your efforts in this matter.

@kwinn, I agree a relay would be very easy to use but I like to make a "standard" output I will use for most of my controls. So it might include pulse signals to other equipments including servo/stepper motor drives. That's why I can't use the relay.

I also feel the TIP125 is a good solution so maybe we can continue with that one?
My provider in Sweden don't have the TIP125 in stock, so I'm thinking of using a BD680 instead. Do you think that would be OK?
Data https://www1.elfa.se/data1/wwwroot/webroot/Z_DATA/07104433.pdf

So then how do I interface the transistor in a safe way so I don't overload the Propeller?
Can I use a small signal transistor as Peter made a suggestion above?

Usually I just wire up and test, mostly it's working and sometimes it smoking.

I just don't want my Propeller to let the smoke out. So I hope someone can help me with a little circuit example.

kwinn · 2009-02-01 17:56

I will check the BD680 specs and get back to you with a schematic later today. Have to run out now.

kwinn · 2009-02-01 20:17

Hi Ronnie,
I have checked the data sheet for the BD680 and it should work. Very similar specs to the TIP125. If you are going to use PWM keep the frequency below 10KHz. The turn on/turn off time of the opto isolator is around 25uSec and the transistor will be dissipating power during that transition, so the higher the frequency the higher the transistor power dissipation. Most of the heater circuits I work with run at 60 or 120Hz (power line frequency) so the transistor barely gets warm.

You will note that there is no direct connection from the prop to the heater circuit. That means there is virtually no risk to the prop. It may seem a bit paranoid, but I have never had the control circuitry damaged by a failure on the high voltage side, and I have used this type of circuit on many instruments for over 20 years. Worst case you end up replacing the transistor and opto isolator.

Ronnie B · 2009-02-01 20:55

@kwinn.
I'm very thankful for all your help. I will order the parts tonight so I can start working with the control next week.
But I have one last question, my supplier they dont have the 4n33 but they do have the 4n35, 4n36 and 4n37 can I use any of them instead?

Thanks
Ronnie

kwinn · 2009-02-02 00:31

Ronnie, those may work, but they are all single transistor outputs not darlington transistors. Try to see if you can get one of these instead ( 4N29M, 4N30M, 4N32M, 4N33M, H11B1M, TIL113M ). If not you can try one of the 4n35/36/37. The problem may be that they will not provide enough current to the base of the PNP transistor to turn it fully on. If that happens you will have less than 24V on the heater and the transistor will be dissipating a lot more power. You can try making the resistor on the input to the optoisolator 180 ohms instead of 220 ohms or adding a small transistor such as a 2n3904 or 2n2222 to the output of the optoisolator.

Ronnie B · 2009-02-09 21:53

@kwinn: I got the parts TIP125 and opto 4N33-M last week and during the weekend I tried the setup. But I don't get it to work really good sad to say.

I'm using a PCF8574 to drive the opto. If I let the PCF supply the input on the opto I will get a current of 0,13mA but if I sink through the PCF I get a current of 20mA through the opto.

But no matter I dont get more then 25mA in the output from the opto to the TIP125, making the TIP terrible hot when loading it even at 50%(2.5A).

Any suggestions what might be wrong?

I guess it shouldn't be that hot? If the transistor is really open, will it be hot if I load it with 5A so I will need a heat-sink?

/Ronnie
·

Ronnie B · 2009-02-09 22:18

I tried to replace the 1k resistor to a 440 Ohm and then I get a current of litle more than 50mA and while loading the TIP with 2.5 Amp it doesn't get hot. Now what's get hot is the 440 Ohm resistor. Infact I'm using the 2 of te 220 I bought already. They are 0.75W and they get so hot you can't touch them. What size is recommended?

If I take the current = 50mA and the R=440 Ohm I get a power of 1.1 Watts...

/Ronnie

kwinn · 2009-02-10 06:42

That is strange. The TIP125 has a minimum gain spec of 1000, so 25mA from the output of the optoisolator should drive it into saturation. Even 5 mA should be enough.

The CTR of the 4N33 is 500%. That is for 1mA in you should get 5mA out as long as the collector resistor does not limit the current.

With a 24 volt power supply and a 1K resistor you should get a little less than 25mA through the optoisolator/resistor/TIP base (~24-1.5/1000=22.5mA). That should be more than enough current to drive the TIP125 into saturation (22.5mAx1000=22.5A). The power dissipation for the 1K resistor would be .0225A x 22.5V = .506W. The power dissipation of the 440 ohm resistor would be .050A x 22.5V = 1.1W as you stated.

2 problems. First, the TIP125 has an absolute maximum current rating of 5A. If you try to run it near the maximum for any length of time it will fail. I selected the TIP125 based on the original post where you said 24V 60W and 24V 2.5A. If you need 5A then we need to select another transistor.

The second problem seems to be that the gain of the TIP125 is much lower than what is in the data sheet. Can you send me a schematic of how you hooked things up along with current and voltage measurements for both 440 and 1K resistors. Also measure voltage across the heater or dummy load.

kwinn · 2009-02-10 06:47

PS - Can you measure the resistance of the heater when it is cold and when it is hot. Measure the voltage across a small value (2 ohms or less) series resistor as it warms up.

kwinn · 2009-02-10 07:01

PPS- I did goof on the resistor value (1K). That is what I normally used on circuits operating from a 12V supply so 1/4 watt was more than adequate. For 24V the 1K should be 1W and the 440ohm 2W to give them a bit of margin. I hope that will not be necessary since the TIP125 should provide well over 2.5A at a base current of less than 5mA which would be around 0.12W.

dMajo · 2009-02-10 08:26

Have a look at this modules.

Ronnie B · 2009-02-10 08:47

@kwinn: I will try again and measure all the data. I will also put in all new components to make sure there are no defects on any of them. Then I will measure again. I will write down all the data. Thanks for the inputs so far.

@dMajo: Thanks for the input, however this is a low-budget hobby project. So far I bought components for 25 Channels, 5Amp and that cost just little more than 1 Crydom output. But yes, the Crydom is surely a nice solution which I might have chosen if I had that kind of budget.

/Ronnie
·

Ronnie B · 2009-02-11 11:50

Now I have documented my different wirings.

Connection 1, original design.
I(R1) = 0.13mA but also the volt out from the PCF is only 0.99V.
I(R11) = 0.15mA.
I(load) = 2.5 A, but the TIP125 is getting very hot. A wet finger will be sizzling after only 5 seconds load.

Connection 2,·Alt design #1.
Opto 4N33-M is suppied 5V and grounded through the PCF. This will make a higher current.
I(R1) = 17mA but also the volt out from the PCF is only 0.99V.
I(R11) = 25mA.
I(load) = 2.5 A, but the TIP125 is getting very hot. A wet finger will be sizzling after only 5 seconds load.

Connection 3,·Alt design #2.
Same connection as Connection 2, but the R11 is adjustet from 1k to 440 Ohm.
I(R1) = 17mA but also the volt out from the PCF is only 0.99V.
I(R11) = 53 mA, but the resistor is getting·very hot.
I(load) = 2.5 A, but the TIP125 is NOT getting very hot.

What do you suggest? Any ideas? I could change the R11 to a bigger resistor but I don't like the heat, even if a take a 2W resistor to avoid getting hot, I still have the 1W energy to ventilate out.
Would it be a good idea to use a signal-transistor to control the TIP125? The OPTO controls the small transistor which controls the TIP125? A few more components, but then I'm sure the TIP is in saturation.

I haven't measured the resistance of the heater, but I do measure the current load. When the heater is cold the current will be 3.2 A then when the heater is getting hot (2-3 seconds) the current will drop to 2-2.5 A.

/Ronnie

Edit: Corrected the schematic that I was drawing wrongly.

Post Edited (Ronnie : 2/11/2009 3:22:17 PM GMT

PJAllen · 2009-02-11 12:24

Use a relay with appropriately rated contacts (5A).· You won't have to spec high-current semiconductors and you can dispense with further angst, hand-wringing, &c.· Low-cost, low-budget, simple.

Ronnie B · 2009-02-11 13:02

@PJ Allen: Thanks for the relay design, I might use it in the future for a Relay output card. However in this project I need a transistor since I'm making a "standard output" card even suitable for PWM loads, Stepper Motor, etc. That's why I want a High Power transistor solution.

Hum, hum. Since I have some relays at home (6V) I think I will wire them up according to your sketch. It's always good to have a 8 Ch, relay output card ready to use.

/Ronnie

Leon · 2009-02-11 13:19

Try connecting the collector of the photo-transistor to +24 V and the emitter to the driver base. I think it will work a lot better.

Leon

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Post Edited (Leon) : 2/11/2009 1:27:19 PM GMT

PJAllen · 2009-02-11 13:29

Ronnie B said...
I'm making a "standard output" card even suitable for PWM loads, Stepper Motor, etc. That's why I want a High Power transistor solution.

Then you should use Jakacki's circuit.

Otherwise, to turn on/off a heater, I'd use my circuit.· (Not much to be gained from PWM'ing a heater coil, IMO.)

dMajo · 2009-02-11 18:00

Ronnie B said...

Now I have documented my different wirings.

Connection 1, original design.
I(R1) = 0.13mA but also the volt out from the PCF is only 0.99V.
I(R11) = 0.15mA.
I(load) = 2.5 A, but the TIP125 is getting very hot. A wet finger will be sizzling after only 5 seconds load.

Connection 2,·Alt design #1.
Opto 4N33-M is suppied 5V and grounded through the PCF. This will make a higher current.
I(R1) = 17mA but also the volt out from the PCF is only 0.99V.
I(R11) = 25mA.
I(load) = 2.5 A, but the TIP125 is getting very hot. A wet finger will be sizzling after only 5 seconds load.

Connection 3,·Alt design #2.
Same connection as Connection 2, but the R11 is adjustet from 1k to 440 Ohm.
I(R1) = 17mA but also the volt out from the PCF is only 0.99V.
I(R11) = 53 mA, but the resistor is getting·very hot.
I(load) = 2.5 A, but the TIP125 is NOT getting very hot.

What do you suggest? Any ideas? I could change the R11 to a bigger resistor but I don't like the heat, even if a take a 2W resistor to avoid getting hot, I still have the 1W energy to ventilate out.
Would it be a good idea to use a signal-transistor to control the TIP125? The OPTO controls the small transistor which controls the TIP125? A few more components, but then I'm sure the TIP is in saturation.

I haven't measured the resistance of the heater, but I do measure the current load. When the heater is cold the current will be 3.2 A then when the heater is getting hot (2-3 seconds) the current will drop to 2-2.5 A.

/Ronnie

Edit: Corrected the schematic that I was drawing wrongly.

Try with this Connection#2 (mod) R12=150

Edit: Because of R1=220 I suppose that PCF Vcc=5V
······ I assume also you have a heatsink on TIP as at 3A the datasheet states V_CEsat=2V and that is 6W. The datasheet states a P_C of 2W at ambient T=25°C

Post Edited (dMajo) : 2/11/2009 6:28:50 PM GMT

Ronnie B · 2009-02-11 20:01

@dMajo: OK, thanks for confirming the heatsink. I suspected that was what I needed since I'm above 50% load, I just didn't remember how to calculate the need. So then I will design the PCB and add a heatsink to the design. Thanks everybody for all your assistance!

/Ronnie

High current/voltage load

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