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I need to power a Thermoelectric module with 10A maximum load current. I'm designing a card with a minimal design in order to change the current or the voltage across the load, given a standard voltage source.

I just need a minimal regulation of the load, which is, just be able to decrease the power of the thermoelectric module from 100% to maybe 80-90%. Thus I was thinking to control the module by voltage, since precise regulation is not needed and voltage control is usually much easier than current control.

I am choosing among a linear voltage or a switching regulator solution. The point is that the output voltage should be absolutely flat, with Maximum 5% ripple. A potentiometer should be used to "regulate" the power of the system. That's why to start with I was looking into linear voltage regulators. The point is that, to be on the safe side, I would require at least a 15V max output voltage and 20A output current model. Thus I was thinking in designing several regulators in parallel, but I am aware that this is a bad idea. How could I design a safe parallel regulator circuit?

If the linear regulator is a bad idea, is there any off-the-shelf step down switching regulator which does not require any external components (no coils or caps) more than a few resistors?

As said I'm not concerned about the accuracy of the system but just about reliability, design robustness and cost.

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    \$\begingroup\$ Switch mode devices can easily achieve <5% ripple. \$\endgroup\$ – Peter Smith Dec 21 '16 at 13:19
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    \$\begingroup\$ Voltage and current control are about as complicated as each other – you can build a nice const-current source out of nearly every supply architecture by making the feedback voltage a function of the current flowing, e.g. by replacing the voltage over a voltage divider by a voltage over a shunt resistor. \$\endgroup\$ – Marcus Müller Dec 21 '16 at 13:21
  • \$\begingroup\$ Anyway, your ripple requirement isn't that hard – what's more problematic is that your output voltage * output current = 300 W! That's quite a lot. You would not want to burn a couple of volts at 20A over a linear regulator (which, by the way, makes no sense – that will convert the energy that you don't put inot your thermoelectric module to thermal energy, which is kind of what you wanted to regulate in the first place....) \$\endgroup\$ – Marcus Müller Dec 21 '16 at 13:23
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    \$\begingroup\$ is there any off-the-shelf step down switching regulator which does not require any external components (no coils or caps) That's a simple: No unless you buy a complete module. A switching regulator by definition needs caps and coils to work, these are too large to be integrated on-chip. At least not at the power levels you're looking at. \$\endgroup\$ – Bimpelrekkie Dec 21 '16 at 13:24
  • \$\begingroup\$ what now, absolutely flat or <5%? Also just calculate the losses in a linear version, thats insane. \$\endgroup\$ – PlasmaHH Dec 21 '16 at 13:29
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Your ripple requirement isn't that hard – what's more problematic is that your output voltage * output current = 300 W!

That's quite a lot. You would not want to burn a couple of volts at 20A over a linear regulator (which, by the way, makes no sense – that will convert the energy that you don't put inot your thermoelectric module to thermal energy, which is kind of what you wanted to regulate in the first place....).

Linear regulators simply work by having an "adjusting" internal resistance that simply drops the voltage difference that's between your in- and output and converts that to heat. So, if the in-output difference is just 2V, at 20A, your linear regulator would dissipate 2 V * 20 A = 40 W of power. That's a terrible thing to cool.

If the linear regulator is a bad idea, is there any off-the-shelf step down switching regulator which does not require any external components (no coils or caps) more than a few resistors?

Terminology: When talking about regulators, it's not perfectly sure whether you're only referring to the thing that regulates the currents flowing, or mean the complete system including all the necessary energy-storage components. Usually, we'd use the former meaning. For the other thing (controller + switch + power storage (coil)), we'd say supply, or at least module.

You can of course buy readily made power supplies. Every laptop has one, and they even exist for the currents you need. Getting one that is adjustable might be a little harder, but you might want to think about just using PWM on the output to reduce the average power going into your module. Of course, that'll technically absolutely break the "5% ripple" requirement (PWM is actually 100% ripple, if you want to consider it that way), but I'm not sure where that requirement came from in the first place. Maybe you'd want to also specify the acceptable/unacceptable frequencies for deviations from the intended current/voltage point, and explain why you'd need so strict regulatory limits for something as slow as a thermal element. **Update:* nope, not PWM then, according to your comment :)

You can also buy adjustable 300W supplies – but these tend to be a little more costly.

Regarding modules: The module we're talking about will most probably be sold as "open|closed frame power supply".

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  • \$\begingroup\$ How would using PWM not cause the same ripple problems OP said s/he wants to avoid? \$\endgroup\$ – The Photon Dec 21 '16 at 15:47
  • \$\begingroup\$ @ThePhoton Well, I don't really know why OP "needs" the "absolutely flat voltage" – sounds a bit doubtful for a thermal system, as those are typically "slow as hell". \$\endgroup\$ – Marcus Müller Dec 21 '16 at 15:50
  • \$\begingroup\$ OP clarified "absolutely flat" means "maximum 5% ripple". I agree it sounds like an overzealous spec, but I think your answer should explain that before suggesting alternatives that don't achieve it. \$\endgroup\$ – The Photon Dec 21 '16 at 15:51
  • \$\begingroup\$ @ThePhoton very fair point! Specified. \$\endgroup\$ – Marcus Müller Dec 21 '16 at 16:07
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    \$\begingroup\$ Thermoelectric devices using in cooling applications perform very bad when controlled by pulses. This is because of thermal losses, induced by the junctions that build a thermoelectric module. Besides, pulses have a bad effect over long time reliability of thermoelectric modules \$\endgroup\$ – Francesco Dec 21 '16 at 17:02
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Just ask your favorite supplier or search engine for "Adjustable Power Resistors" and use one of those together with your load as a voltage divider.

Let's assume that your load is strictly ohmic. Your output specs require 300 W of output power, and you want to regulate (burn away) 10-20% of that. That's 30-60 Watts, which is easily possible for power resistors.

Just be careful that your load has a positive temperature coefficient (resistance increases with temperature) or else you may experience thermal runaway and your load will start to glow soon. ;-)


By the way, a linear regulator would also burn those 30-60 W as heat, so it would need a big heatsink and probably a fan.

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If you want design simplicity, why not just toggle the power to the thermoelectric module on and off at very slow PWM rate using an N-channel MOSFET driven by a timer circuit?

Remember, your home fridge works in similar way, except even slower running every 10 minutes or so.

You could use say a 1 second pulse rate, so a 10% reduction would be 0.1 second off, 0.9 second on. Since cooling is very slow and your module can withstand the full supply voltage, then 1 second rate would be just fine and stress the electronics less, etc.

You would need to check the load regulation specs of your power supply to see what kind of ripple you might get with switching a 10 Amp load. For example, a load regulation spec of 1% at full load would mean the output will drop 1% in voltage between no load and full load.

An N-channel MOSFET would be very easy to direct drive with a microcontroller, a 555 timer circuit, discrete or opamp oscillator or whatever. And you can make the timer circuit use a pot to change the duty cycle from 80% to 100%.

Choose an N-channel MOSFET with very low Rds-on that can handle very high currents. Use the Rds-on * load current to calculate the power dissipation of the MOSFET. Choose a low enough Rds-on and you will not need much (if any) heatsinking for the MOSFET. The gate capacitance will increase with lower Rds-on resulting in slower switching times, but at 1 second rate, it won't matter much.

If this approach sounds good, I can help with the slow PWM timer circuit. Hope that helps, -Vince

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