# Analysis of power amplifier bias circuit: does it need to be thermally bonded?

I'm in the process of building a medium power amplifier for the 20m band (14MHz) and I'm experimenting with bias circuits. The amplifier is a classic BJT push-pull arrangement with a transformer input having a centre tapped secondary into which the bias is fed.

I've come across the bias circuit shown below and I've been using it with mixed results.

simulate this circuit – Schematic created using CircuitLab

One thing I haven't done is thermally bonded the controlling transistor (Q2 above) to the power transistor heatsink (I should clarify that this is the heatsink for the push-pull power transistors and not Q1 shown above) and I wanted to generate an expression that would indicate how the output (Vout) changes with the temperature of Q2. I always find these coupled transistor circuits difficult to analyze, but I've arrived at an expression for VE that I find interesting.

$$V_{b1}=V_{cc} - I_c R_c$$

$$V_{E} = V_{b1} - V_{be1} = V_{cc} - I_c R_c - V_{be1}$$

$$I_{c} = I_{b2}\beta_2$$

$$I_{b2} = (V_{E} - V_{be2})/R_{b} - V_{be2}/R_{\text{adj}}$$

If I combine these expressions into an equation for VE, I get:

$$V_E = \frac{V_{cc}+V_{be2}(\beta_2 R_c R_t)-V_{be1}}{\beta_2 R_c R_{\text{adj}} + 1}$$ where $$\R_t = R_b + R_{\text{adj}}\$$

If we assume the Rt is equal to Radj and neglect the + 1 in the denominator, then as Q2's beta rises due to temperature, the value of VE would seem to drive towards Vbe2. Do I have that right?

But, Vbe2 is decreasing as a result of the rise in temperature. Does this suggest that Q2 should be thermally bonded?

• But Q2 is not used for the bias purpose but as short circuit protection (overcurrent protection).
– G36
Nov 23, 2021 at 16:37
• @G36 That makes it sound like Q2 is off until it's needed in some over-current mode. But, the circuit acts more like a regulator: Q1 controls Q2, which controls Q1, etc. I guess what I'm asking is if bolting Q2 to the main power transistor heatsink will help avoid thermal runaway in the PA (additional to its inherent regulation). Nov 23, 2021 at 16:53
• Can you show me a circuit diagram? Because I never saw this type of biasing in the push-pull stage. How is it supposed to work in push-pull? The emitter follower is not a constant current source. So, how can the output voltage increase, if Q2 will limit the output current to bias level?
– G36
Nov 23, 2021 at 17:14
• LTspice says that variations of VE(out) are caused by temperature of Q1(2N3904) and very, very little by temperature of Q2(TIP31). Nov 23, 2021 at 17:30
• Sorry for confusion: Yes, Q2(2N3904) should be not just on the heatsink - I'd try to thermally bond it to the metal tab of the classAB transistor, or as near as you can. There may be thermal gradients that don't have time to settle to steady-state during short-term testing for run-away. I think there is little chance of over-compensating if thermal coupling is too close. Nov 23, 2021 at 18:19

Don't depend on beta changing with temperature; it's too unpredictable. Best to assume beta is infinite initially and then analyze the effects of (negligible) base current.

If VE is considered an output, its value is VBE(Q2)*(RADJ+RB)/RADJ. If you want this to track some other circuits, then Q2 needs to be thermally coupled to them. Q1 is not particularly critical in the circuit.

Base current of Q2 (approximately VC/RC/beta) might be about 13.7/820/50 = 300 uA. 300 uA in RB (100 Ω) is 30 mV -- this variation is negligible in your circuit.

• I think your 2nd paragraph addresses my main concern. Intuitively speaking, if the Ic of Q2 rises due to an increase in temperature (external to Q2), then the voltage on the base of Q1 falls: which reduces the current flowing from VE (important for a bias circuit). I just wanted to make sure that by bolting Q2 to the heatsink I don't inadvertently create a positive feedback system!! Nov 23, 2021 at 17:16