Well, I have a serious problem with heat in my circuit. I was trying to build a voltage controlled current source. Everything went fine in the building process. But when I turn ON the power supply the TIP122 gets very warm and so do the IRF820. I know this guys tend to get hot when dissipating some considerable amount of energy, however, my circuit just intent to deliver some hundreds of mA. And the main problem: the TDA2822 is getting very hot. Hotter than all the transistors together. Although all that dissipated thermal energy, the circuits works pretty well TBH.


Any help is appreciated! Cheers.

  • 1
    \$\begingroup\$ The TDA2822 is a power amplifier, not an operational amplifier. Are you sure that it is suitable for this type of connection? And regarding Q1, at the worst case it would be dissipating 7 x 0.5 = 3.5W, not a low ammount of power... \$\endgroup\$ – Claudio Avi Chami Sep 13 '17 at 3:01
  • \$\begingroup\$ Oops, I really missed it for the Q1, thanks for that. And about the amplifier, I thought that this amplifier could work, since it's "high" power. ratings. Am I wrong? \$\endgroup\$ – KawaungaXDG Sep 13 '17 at 3:06

The IRF820 is poorly suited for this application- choose a logic-level MOSFET with a low Rds(on). Your output current will depend on the value of Rds(on) + R1 so if you stay with this circuit you would probably want Rds(on) to be extremely low compared to 10 ohms.

Since the IRF820 may not turn on fully it may get hot.

The TIP122 will get hot when the circuit is dropping a lot of voltage at high current- that is just the math of the situation. Worst case is 5V in (0.5A if the MOSFET was very low resistance) into a short, or 3.5W, which will require a heatsink of some kind.

Finally, the amplifier output current may be large- if the load goes open- which will cause the op-amp to heat- add 1K in series with the base.

The resistor will also dissipate another 2.5W worst case, and that will tend to heat nearby components (the total 2.5W+3.5W = 6W which corresponds to 0.5A and 12V.

Rather than adding the power MOSFET as you did, I would tend to consider switching the input to a more precise single-supply op-amp using a CMOS analog switch. Then you have only the 10 ohm resistor to ground. The amplifier you are using is not really specified for DC applications.

  • \$\begingroup\$ Thanks for the answer! It really helped. And about the MOSFET substitution, can a solid state relay be a good option? \$\endgroup\$ – KawaungaXDG Sep 13 '17 at 9:21
  • \$\begingroup\$ You don't need the isolation an SSR provides, I don't see any particular advantage over a properly specified MOSFET. The voltage drop is the important thing. \$\endgroup\$ – Spehro Pefhany Sep 13 '17 at 12:35
  • \$\begingroup\$ Well, I was using the MOSFET to take down the current source, so I could measure the voltage of any batteries that I will charge with this circuit. So, it's basically for isolation. But I will find a better MOSFET for this project, as you suggested. \$\endgroup\$ – KawaungaXDG Sep 13 '17 at 13:19
  • \$\begingroup\$ Come to think of it, with an SSR you could put the switch on the other side of the current sink (eg. in the collector or in series with the 12V supply). The base resistor would then become even more important. \$\endgroup\$ – Spehro Pefhany Sep 13 '17 at 14:11
  • 1
    \$\begingroup\$ Yeah, I really need that resistor, as you pointed. Thanks for your attention! You really saved my circuit from melting down :) \$\endgroup\$ – KawaungaXDG Sep 13 '17 at 14:19

1) The power amp is prone to high frequency oscillations which can be improved with the 4.7 ohm 0.1 uF snubber load as they show on most audio power amps including this. (which you neglected)

2) The MOSFET is a miss-fit for this application. It is rated for low current and 3 Ohms RdsOn at Vgs=10 but if you only supply a couple volts less than Vcc=5 V voltage so it could be >> 10 ohms and not have a low Vds when enabled. Remember this. If the MOSFET Vgs "threshold" is given use at least 3x the max Vgs(th) threshold. not 3~4V drive for Vgs(th)=2 to 4V range.

3) TIP122 Darlington has pretty high gain which drops when saturated but Vce(sat) starts at 1V and rises to 1.2 at a few 00' mA. and then only needs <1mA base current so you can drive this without a Power Amp (U1) just using 12V on an Op Amp and 12V on the MOSFET switch.

That being said , the FET must be much lower than 10 ohms for R sense, if you want precision current control. Such as 100 mohms. Then all of the lost power from voltage drop in the FET will be add to voltage drop in the TIP122 which naturally increases its temperature unless you designed a heatsink . Basic Ohm's Law Pd=VI from the drop voltage and current selected.

You can easily eliminate the FET with inserting a large series R> 1K from pot wiper then switch Gnd to Vin(-) to disable current.


I forgot to mention in a standard VC-CC sink or active load, you can use a current sense R of only 1/2W (derated by 50%) by choosing a Vdrop at max current of I^2*R = 250mW. Thus if max current is say 0.5A then R=0.25W/0.5^2= 1 ohm .

Thus the max Vref on your pot is now only 0.5V to match 0.5 A * 1 ohm. So you add a series fixed R to the pot to V+ to reduce the control range. Now you eliminate the FET and disable as above using a switch to ground on Vin(+) to ground.

IC's use this approach but use a standard Vref of 75 mV to 100mV for even less power dissipation instead of a 500mV drop. Then all the voltage drop is in your low RdsOn FET or Darlington.

Then if you are choosing large currents like many Amps, then you use a dummy load like a halogen bulb of much larger power rating to dissipate most of the heat, as these tend to be constant current sinks in the 25'C to 500'C range with a large PTC coefficient but still use the above to fine tune and regulate CC with an emitter or drain sense resistor in milliohms.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.