bjtpower amplifierpush-pull
My textbook says that biasing current through Q8 and Q9 is about 20mA and through Q10 and Q11 about 100mA. How can I analyze the circuit to get these values with back-of-the-envelope calculations?
So far I only calculated that biasing voltage provided by Q5, R8 and R7 is 2V, but cannot get any further.
There are two problems:
1) The first stage can only work properly at base voltages below zero. When the collector-base junction is reverse-biased, i.e. in normal operation mode, the the total voltage drop on R67 and R62 equals that on R56 (there is a small difference, which we can ignore here). This value can't be more than about 11 V, as the total voltage drop on all of these resistors, diodes and the transistor is 24 V. When when the base voltage is anywhere from slightly below zero to positive, the collector doesn't have anywhere to go, so to speak, it cuts off.
2) floating input, and we don't know anything about your source.
You may want to use an emitter follower with a diode bias, or two emitter followers in series, npn and pnp.
You are using the wrong topology for what you are trying to achieve. Here is a better circuit:
First, we figure out R2. This is what sets the LED current. You want 700 mA. D1 drops 1.4 V, and let's say the transistor saturates to 300 mV. That leaves 3.3 V across the resistor. (3.3 V)/(700 mA) = 4.7 Ω. Note that this resistor will dissipate (3.3 V)(700 mA) = 2.3 W.
Now for the base resistor. You need 700 mA collector current, so you need at least that divided by the transistor gain as base current. Let's say the minimum guaranteed gain of the transistor at 700 mA is 50. (700 mA)/50 = 14 mA minimum required base current. Figure the B-E drop will be 750 mV, so that leaves 4.25 V across R1. (4.25 V)/(14 mA) = 304 Ω. So the largest common value that guarantees everything within spec is 270 Ω.
Best Answer
The truth is that it is impossible to tell/predict what current will flow without making a "big" error. There are far too many unknowns.
The wild guess for the drive stage will be \$ \frac{V_{BEQ10}}{R_{11}} \approx 20mA \$
I used Bob's Cordell models in LTspice and get these results: \$I_{CQ8} = 18.1mA\$ and the output stage current is \$20.5mA\$ and \$\textrm{Vbe}\$ multiplier (Q5) bias voltage equal to \$2.4V\$.
Of course, when you build the circuit in real life you will get completely different results. Because the output transistors will get hot and after some time they will reach the thermal equilibrium and the bias current will settle on the new value.
This is why in practice we adjust on the benchtop the \$R_7\$ resistor value to set the output stage bias current to the desired value after warming-up the amplifier.