Input coupling to Class AB amp with diode bias. One capacitor or two?

Question

When AC coupling the input signal to a Class AB (Push-Pull / Complementary Pair) which is diode biased I see two different approaches:

1. Signal connected between biasing diodes with single decoupling capacitor:

2. Signal connected directly to each transistors base with separate capacitors:

What is the practical difference between these two approaches? Is one better than the other?

Here is an editable circuit showing the basic idea of the 2nd approach (NB: values are not that realistic):

^{simulate this circuit – Schematic created using CircuitLab}

Here's another simulation of a the first circuit (courtesy of Tony Stewart).

Answer 1

The purpose of the diodes is to set a bias voltage between the transistors' bases, which sets a small idle current through the push-pull. This makes it work in class-AB and lowers crossover distortion. However the diodes should be thermally coupled to the transistors, to prevent thermal runaway. Also, emitter resistors should be used for this reason.

Anyway.

As long as both diodes conduct, say a few mA current through the diodes, their dynamic impedance will be rather small, like 10-20 ohms, so the transistors will be driven from a low impedance. What matters here is that this bias current is generated by resistors R1 and R2.

So, when we want high positive output voltage (and presumably high output current) voltage on R1 will be low as TR1 is driven to a voltage close to the positive power rail. Since TR1's base current comes only from R1, this is a problem: for a high enough output current, TR1's base current will suck out all the current R1 can provide, so D1 will turn off and it no longer works.

The second configuration will work better if the two input caps are large enough to have low impedance at the frequency of interest: in this case, AC base current is provided from the signal source through the caps, and R1/R2 only set the DC operating point.

Thus the second configuration is a better choice, if the extra performance is required. It would also allow higher values for R1/R2 since it solves the problem of the resistors having to be small enough to let enough current through for the base current required for maximum output current.

Answer 2

It's a bit more complicated when you drive high currents because of the choice of every component affects the results of output impedance, quiescent current of drivers, harmonic distortion, dampening ratio which affects voltage from back EMF on low frequencies and thus "muddy bass".

Naturally from Shockley effects on Vbe vs Tjcn and same for diode, even if thermally matched may cause problems if the diodes have too small or too large a power rating and thus ESR with changes in Vbe from bias R's affecting output idle current greatly.

In order to determine the optimum Cap configuration, you need to understand that this amplifier is less than unity gain. So why is there loss and where is it? and why is that important to minimize the voltage attenuation for good low frequency response, but it will come at cost in idle power dissipation and bigger output C values rated for ripple current or load current in this case.

The question is simply comparing the capacitor impedance at some f vs the source and input impedance to see if the cap is impedance is significant. The differences in these two choices are minor compared to the other factors in R ratio design and Pd ratio selection for the Transistor and the diode so that they bias the output stage at the desired current to achieve low output impedance , which is essentially the source impedance driving the base /hFE.

Do you want to know more?

Then you need to define more specs.

Including: Pmax, Vmax, Load min, f min, THD max, min dampening factor ( usually 10 is cheap designs, 100 is better) Source impedance..

The lower your speaker impedance, like 4 Ohms , the more critical is thermal runaway settings and hFE matching between PNP and NPN, yet with +/5V you can easily generate 5W. A better design capable of 0.3W into 60 Ohm headphones or a few 8 Ohm speakers. Using 1N400x diodes instead of small signal 1N4148 must use a pot between diode string gives lower Vf changes but adding a 50 or 100 Ohm pot between them must be tuned for the speaker load and desired output power and mismatch of hFe. ( want them within 20%)

tinyurl.com/y9pdw3uv is an example my this in my latest simulation. Note RMS power in speaker, you can change R value and RMS power from each supply (-ve) ought to be 30% efficient at best or 60% from both supplies. Note how pot affects each signal and DC minimum current. This gives very good dampening factors and DC response on output. You can DC couple input if source is 0Vdc.

• unknown hFE power transistors can created issues if not matched.
  • these S8050/S8550 are graded for hFE, take note by suffix.

Answer 3

Nobody makes an amplifier output like that. Both circuits were used many years ago in school. Very important DC coupling should be used and negative feedback is missing.