BJT Temperature Compensation 2.0

174

02:00:15

The two circuits are identical except that the right circuit is hotter, which I simulate by doubling the gain of the transistors. As can be seen, the currents through the two lamps are nearly identical. This is because as the top transistor’s gain increases the current through the lamp increases, but the gain of the lower transistor also increases thereby taking voltage away from the top transistor thereby keeping the current through the lamp the same.

I use a resistor in series with a low resistance potentiometer because the circuit is very sensitive to the exact resistance so by tuning the potentiometer I can make the temperature compensation compensate exactly. For a temperature that doubles the gain 30.4% is the right value for the potentiometer. For a temperature that increases the gain only slightly a value around or slightly above 10% is ideal. With increasing temperature and a pot value of 10% the current first stays the same then decreases. At 30% the current first rises, then plateaus, then decreases and meets the original current at exactly twice the gain, but continues to drop further with increasing temperature.

published 5 years ago

CrazyEce91tblt

5 years ago

Nice and useful circuit . Just to note that this is applicable only for Bipolar transistors as in the case of Mosfets the current and thus the gain is inversely proportional to temperature.

jason9

5 years ago

So MOSFETs have increasing resistance with temperature? Interesting.

CrazyEce91tblt

5 years ago

Not exactly. What happens is that mobility as well as threshold voltage drops as temperature increases . That's why additional circuitry is needed in a IC ensuring that the gm of the amplifier will remain constant over the temperature. Search PTAT current for more info.

jason9

5 years ago

I see. Some more googling reveals that at voltages just above the threshold the current increases with temperature as the threshold drops and that for voltages significantly above the threshold the current decreases with temperature as the carrier mobility drops. From this, I can see that for something where the MOSFET is used as a solid-state relay, I would expect the resistance to increase with temperature. If that’s the case, then why is my friend observing a few MOSFETs getting much hotter than all the others when using them in parallel? This seems indicative of thermal runaway caused by decreasing resistance with temperature.

jason9

5 years ago

Here’s where you can find out more about my friend’s issue (we’re discussing it in the comments section): http://everycircuit.com/circuit/6374866377506816

CrazyEce91tblt

5 years ago

Yes it does not surprise me that this happens. You have to bare in mind that electrical devices are not perfect. In reality even same model devices will have slightly different parameters (mobility , treshold voltage etc) due to process variations . In plain English this means that when you put some mosfet transistors in parallel, they will have different resistances and thus different current consumption. This phenomenon is called current hogging and leads to uneven heat distribution among the devices. Is it allowed in California to have a Tesla coil in your garage or backyard 😛?

jason9

5 years ago

I see. I suppose it could even be handled by tuning potentiometers, one for each MOSFET.

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