# Power dissipation BJT transistor as voltage follower

Following the discussion from this thread I am finalizing my calculations. I am not sure that I am calculating the power dissipation correctly. Given the following circuit

I need to operate the opamp between 0-2.4V in 100mV steps, sometimes holding the values for 30min. With V_cc=3V and V_Bas=1V, V_Emit=0.5V. The collectorcurrent reads I_c=0.13mA and I_emit=0.043mA. At this conditions the Power loss calculates as

P_loss = (V_cc-V_Emit)*(I_c-I_emit)

I draw the power loss with the above formula over the range of V_in

Can I choose the right transistor based on this calculation (max ~430uW)?

• Without getting bogged down in details, are you asking if finding a transistor that that dissipate a maximum of $430\:\mu\text{W}$ is hard to find? If so, the answer is "no, they are easy to find -- as the worst case is probably a TO-92 with about 200 C per watt or an increase of less than a tenth of one degree Celsius. It will be very hard to find a device with a problem with this dissipation -- and I'd love to hear about it, if you find one. – jonk Jan 28 '20 at 7:44
• My main concern is if my calculation is correct – v3xX Jan 28 '20 at 7:46
• Oh. Well power dissipation will be $\approx V_\text{CE}\cdot I_\text{C}$ when not saturated. Can you work out those details? – jonk Jan 28 '20 at 7:53
• With Vce=Vcc-Ve? – v3xX Jan 28 '20 at 7:55
• No, $V_\text{CE}=V_\text{C}-V_\text{E}$. But since (I think) $V_\text{C}=V_\text{CC}$ in this case (as I quickly see it) then perhaps so. – jonk Jan 28 '20 at 7:56