Transistor Thermal Dissipation Question

Thread Starter


Joined Feb 8, 2011
I am implementing the circuit shown in Figure 15 of the Application Note at

In my application, I have a 1Ohm TEC that will draw at most 3A (at 3V). I am trying to determine the amount of power dissipated in each of the two transistors for a given scenario (i.e., Q1 and Q3 on or Q2 and Q4 on). I believe that I can just put a 1Ohm resistor in for the load (where it says TEC in the circuit) and ignore the 0.02Ohm resistor (for this rough order of magnitude calculation). However, I am having a difficult time calculating Vce for the two transistors (abstracted -- I can add the actual characteristics of the transistors later).

My ultimate input signal can vary between 0 and 5V, so I believe that my Vb (referred to ground) can also vary between 0 and 5V. However, I don't know how to get Ve of Q1.

Any advice would be appreciated.


Joined Feb 19, 2009
Why bother with a TEC for heat sink? You'll need a larger heat sink to remove the heat from the TEC compared to properly heat sinking the transistors.

When operating, what is the Vce of the transistor, and at what current is that Vce measured? Multiply them for roughly the amount of power generated, then get appropriately sized heatsink and mount transistors to heat sink.

This sounds like an H-Bridge application, which would lead me to believe that MOSFETs would be a better choice, please post schematic of circuit, and what the issues are.

Thread Starter


Joined Feb 8, 2011
I'm not "bothering" with a TEC for a heatsink, it is part of the overall design goal (i.e., provide a circuit to power this TEC). Finding the Vce of the transistor is what I am having difficulty doing. Once I have that, I can easily compute the dissipated power. The schematic is on page 7 of the document referenced by the link in the first post -- Figure 15.


Joined Sep 9, 2010
At full on, the best an NPN transistor can do is to drop ~0.65V across itself. I believe the PNPs drop even more.

At low current the TEC may not behave as a 1 ohm resistor but under use it will behave more-or-less as a predictable load.

Be aware that when cooling, 3A won't do a lot compared to the heat it generates on the hot side. I just mean you'll be consuming 10W for every 1 watt you move from the cold to the hot side. When heating, all 11 watts will appear on the hot side. So you get 1 watt of cooling or 11 watts of heating. A bit unbalanced. It's also a bad idea to switch from one mode to the other, due to thermal shock. The design you're looking at doesn't seem to prevent that, but maybe I just missed it.