With diode biasing how does the the input signal get to the npn transistor's base.


Moderator edit: Rotated image and enhanced contrast.

 

The base of Q2 is 2-diode voltages above that of Q3, i.e. about 1.3V higher.

Whatever appears on Q3 base also appears on Q2 base but 1.3V higher.

 

Note that R5 provides bias voltage from +Vcc to both Q1 and Q2.
The diodes basically act as a 1.3V fixed voltage between the two.

 

And supply just enough voltage to keep current running through both transistors (when the resistor to the positive supply is the right value), which minimizes distortion that would occur when one transistor turns off and the other turns on.

The distortion is referred to as crossover distortion.

 

D.C. has the explanation here: the diodes are indeed forward biased and so with current flowing their voltage drop is fairly constant and thus the effective resistance is low. This easily allows the signal portion of the current to vary the base current of Q2.
Keep in mind that at all times and under all conditions the current in any element is the sum of the currents from all sources. So for this circuit the first part of analysis is to understand the static (no signal) conditions.

 

Thanks guys , I was stuck on the diodes being one way.

 

Thanks guys , I was stuck on the diodes being one way.

The two diodes are there not to pass current or signal but to act as a constant 1.3V voltage reference (like a 1.3V zener diode, which they don't make because of the physics).

 

So the only purpose of the diodes is to eliminate cross over distortion and the varying current at the base of Q3 effects the base current of Q2

 

So the only purpose of the diodes is to eliminate cross over distortion and the varying current at the base of Q3 effects the base current of Q2

The purpose of the diodes is to allow both of the transistors base-emitter voltages to be set for linear operation. The result is a large reduction in crossover distortion because the transistors are not in the cutoff mode for part of the cycle. In some amplifiers the diodes are used to adjust the bias as the transistors heat and the base control characteristics change a bit.

 

So the only purpose of the diodes is to eliminate cross over distortion and the varying current at the base of Q3 effects the base current of Q2

Correct.

If you replace the two diodes with a resistor, and analyze the current through that resistor at a positive signal peak, at a zero-crossing, and at a negative signal peak, you will see that the while those are three different current values, they all have the same polarity/direction. While the input current is positive, zero, and negative, the interstage current is positive, less positive, and less less positive.

ak

 

There are three classes of linear amplifiers, Class A, AB, B.

Class A - the transistor is biased on the linear region and is wasting electrical power as heat. There is minimal cross-over distortion.

Class B - one of the two push-pull transistor pair is always in the cut-off region. There is severe cross-over distortion.



Class AB - as you can guess, is a sweet spot between Class A and Class B, i.e. a compromise between the two while taking advantage of both classes.

If you want to experiment with Class AB amplifiers, add a third diode or a resistor, shifting the amplifier towards a Class A amplifier.