Been a while since college,
but every way I tried to solve the circuit with an emitter resistor instead of a voltage divider didn't work out.
Am I solving that wrong?
Hmm. Well, time to dust off the college cobwebs.
Looking at the top half of the bridge, the resistance into the base of the 2N3904 is h
FE*R
E ≈ 100*100 ≈ 10kΩ. Assuming a 0.65V V
BE drop, 4.35V/10kΩ = 0.43mA. I
RE depends on the V
BE drop, which is narrow range, and 1/h
FE which is small since h
FE is big, so the tolerance of I
C is easily within a few percent when using a high-h
FE transistor like the 2N3904. In reality it might have an h
FE around 250-300, so the tolerance of R
E will dominate, but 100 is a safe minimum.
The current across the collector resistor (680Ω) above depends on I
C in the 2N3904, which matches I
RE within 1/h
FE (≈ 1%) as outlined above. 45mA*680Ω = 30.5V, so V
C can dip that much below +Batt. This V
C becomes the input of the class B driver pair (2N4401, 2N4403). The gate voltage will be within one V
CE(sat) each of +Batt and +Batt-12V (the common produced by the ≈ 12V 1N5242B zener). The 1kΩ resistor to the far left sets the zener current.
The bottom side of course just mirrors the top, for the other half of the bridge.
A problem is the circuit is dependent on a clean logic input signal, if it's not clean the MOSFETs can end up sitting in the ohmic region, which in turn could kill them if they can't dissipate enough heat. Maybe stick an inverter in front of it if there's also a connector, wiring, etc and it's not clear what the origin of the logic input is at all times.
A 30V swing seems a bit excessive to me also... that's like 1.3W over the 680Ω resistor. I can't see any reason for such a large swing; it just generates heat.
I'd either reduce the current (change the 100Ω RE on the 2N3904, to say 330Ω), or reduce the swing to ≈ 12V by changing 680Ω to 270Ω. That makes the dissipation 0.54W. Drop the dissipation to under half a watt, say 0.3W, but reducing the current to 0.3W/12V = 25mA. This puts R
E on the 2N3904 at 4.35V/25mA = 174Ω. Say 180Ω to err on the low side for I
C.
Oh, wait, it's not a bridge - it's a basic push-pull. So two of these circuits will make a bridge, driven at opposing phase...