# Class AB Audio Amplifier With Class A Driver - Questions

Discussion in 'Analog & Mixed-Signal Design' started by elec_dude, Mar 26, 2018.

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1. ### elec_dude Thread Starter New Member

Mar 31, 2017
20
1
I built this amplifier circuit and it works but I have some questions.
Component values are as follows:
R1 = 100 K ohm; R2 = 15K ohm; R3 = 1K ohm; R4 = 100 ohm
C1 = 1 uF; C2 = 1000 uF
D1 = D2 = 1N4148
Q1 = 2N3904; Q2 = TIP31C; Q3 = TIP32C
Supply Voltage = 12 V

VCQ1 = 6.90V; VBQ1 = 1.10V; VEQ1 = 0.40V; VCEQ1 = 6.57V
VCQ2 = 11.94V; VBQ2 = 8.11V; VEQ2 = 7.50V; VCEQ2 = 4.48V
VCQ3 = 0.00V; VBQ3 = 6.90V; VEQ3 = 7.50V; VCEQ3 = 7.51V
IcQ1 = 0.11mA; IcQ2 = IcQ3 = 6.75mA

Questions:
1) Should VEQ2 = VEQ3 not be 6.00 Volts?
2) According to my calculations, I expected VCQ1 = 5.30V, VBQ1 = 1.60V
and VEQ1 = 0.90V. I checked the resistor values and they are pretty
close to what they should be. I must be missing something.

• ###### BJTPA - Class AB Complementary-Symmetry - Single Supply - Diode Bias - Common-Emitter Driver.png
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Last edited: Mar 26, 2018
2. ### MrChips Moderator

Oct 2, 2009
18,696
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The proof of the pudding is in a real test, not a simulation.

Adjust R1 and/or R2 until you get even distortion on both positive and negative excursions at maximum output levels.

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,868
1,364
Well in perfect amp yes. But keep in mind that you do not know the exact value of a beta and Vbe. Also resistors tolerance.
Tto get Ve2 = Ve3 = 6V you need around 5.4V at Q1 collector (VBQ3).
Have you include the base Q1 base current in your calculations?

Ib1 ≈ (Vth - Vbe)/(Rth + (β+1)*R4)

Vth = Vcc * R2/(R1 + R2) = 1.56V

Rth = R1||R2 ≈ 13kΩ

Ib1 ≈ (1.56 - 0.66V)/(13kΩ+ (120+1)*100Ω) ≈ 36μA

and

Vb1 = 36μA*(120+1)*100Ω + 0.66V = 1.1V

And Ie1 ≈ (1.1V - 0.66V)/100Ω ≈ 4.4mA

4. ### crutschow Expert

Mar 14, 2008
22,217
6,474
What are the values of R5 and R6?

5. ### elec_dude Thread Starter New Member

Mar 31, 2017
20
1
Actually, I removed these from the real circuit temporarily. Sorry, forgot to mention that.

6. ### crutschow Expert

Mar 14, 2008
22,217
6,474
So were they out when you made the measurements?

7. ### elec_dude Thread Starter New Member

Mar 31, 2017
20
1
Yes, they were out when I made the measurements.

8. ### elec_dude Thread Starter New Member

Mar 31, 2017
20
1
If I disconnect the driver circuit from the rest of the circuit, the voltage
on Q1's base is 1.60V as expected. However, when the rest of the circuit
is connected, the voltage drops to 1.10V. Why would this base voltage
drop when the voltage divider (R1 & R2) produces the correct voltage?

Mar 14, 2008
22,217
6,474
10. ### Audioguru Expert

Dec 20, 2007
11,089
1,289
Audio amplifiers almost always bias your Q1 transistor from the center point of the output transistors like I show R1 so that the DC negative feedback adjusts the center point voltage.
You do not need R4 because then my new placement of R1 does the AC negative feedback that R4 did before along with the source impedance of the signal source.
Lookup "bootstrapping" then your R3 will be two series resistors with a capacitor to the center point to their junction.

• ###### bootstrapping2.png
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11. ### ebp Well-Known Member

Feb 8, 2018
2,332
811
'"The collector current for Q1 and Q2 is very high in the simulation (.23A) because of no R5 and R6."

To elaborate on crutchow's comment:
Without R5 and R6 the no-signal current though the output transistors is set by the voltage across D1 and D2 in a logarithmic relationship. Neither Vbe nor Vf of the diodes is a well-known quantity and varies from unit to unit. They are are all close to each other because they are all silicon PN junctions, but they not perfect matches. If Vbe of the transistors happens to be on the low side and Vf of the diodes on the high side, the quiescent current through the transistors can be quite high. If the relationship is the opposite way, it could be quite low. R5 and R6 provide a form of negative feedback, sometimes called emitter degeneration, that helps overcome the mismatch. R4 serves the same function for Q3.

If you have LTspice you could take crutchow's circuit and add a voltage source in series with D1 and D2. If you play with the value of that voltage with different values of R5 and R6 you will get a feel for the effect.

crutschow likes this.

Mar 31, 2017
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13. ### sghioto Well-Known Member

Dec 31, 2017
1,039
157
This is the same thread you started back in January, are we still doing homework? Apr 5, 2008
19,550
3,984
Hello,

@elec_dude ,Watch your language or you won't be long here.

Bertus

15. ### elec_dude Thread Starter New Member

Mar 31, 2017
20
1
Don't you dare threaten me. You tell that creep to stop harassing me all the time.
I am not a student and this is not homework and he has been told that before. His big mouth is going to get him into a whole lot of trouble. Kick him out and you owe me an apology!

Apr 5, 2008
19,550
3,984
Hello,

Then You will be banned for one day in a moment to cool down.

Bertus