Class AB Voltage Gain And Output Impedance

Thread Starter

aac044210

Joined Nov 19, 2019
127
I have simulated the attached class AB amplifier and it has a voltage gain of 588.

My calculations (attached) come up with a voltage gain of 279. I need to
determine where my numbers have gone wrong. I also would like to
determine the output impedance of each stage.

Thanks very much.

aac
 

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Audioguru again

Joined Oct 21, 2019
1,272
The RC2 is not 168 ohms, instead it is a little higher than 450 ohms. So AV2 is about 4.5 times and the gain of AV1 is about 160 times so the total gain is about 720 times.

I simulated the gain of the 1st transistor and got a gain of 160 times:
 

Attachments

Jony130

Joined Feb 17, 2009
5,155
Show us the 2N3904JP model.

Av2 = (RB3||Zin3)/(re2 + RE2) ≈ (1kΩ||900Ω)/101Ω ≈ 470Ω/101Ω ≈ 4.6V/V

Zin3 ≈ 101*(8Ω + 1Ω) ≈ 900Ω

And

In class A

AV3 ≈ (RL + 0.5*R31)/(0.5*re4 + (RL+R31)) * RL/(R31 + RL) ≈ (8Ω + 0.5Ω)/(0.5Ω + 8.5Ω) * 8Ω/9Ω ≈ 0.88V/V

Av ≈ 172*4.6*0.88 ≈ 696V/V
The LTspice resoult
av: v(vout)/v(vin)=(688.806,11.7811°) at 1000
 

Audioguru again

Joined Oct 21, 2019
1,272
The beta and Vbe of a transistor can be minimum, typical or maximum in a fairly large range. The circuit must be deigned so that any passing transistor works properly.

This circuit is designed only for a transistor with 2N3904JP spec's that might not even be available if you buy thousands and test them all. I simulated the first transistor using a transistor with typical spec's and it was biased wrong.
 

Thread Starter

aac044210

Joined Nov 19, 2019
127
Show us the 2N3904JP model.

Av2 = (RB3||Zin3)/(re2 + RE2) ≈ (1kΩ||900Ω)/101Ω ≈ 470Ω/101Ω ≈ 4.6V/V

Zin3 ≈ 101*(8Ω + 1Ω) ≈ 900Ω

And

In class A

AV3 ≈ (RL + 0.5*R31)/(0.5*re4 + (RL+R31)) * RL/(R31 + RL) ≈ (8Ω + 0.5Ω)/(0.5Ω + 8.5Ω) * 8Ω/9Ω ≈ 0.88V/V

Av ≈ 172*4.6*0.88 ≈ 696V/V
The LTspice resoult
Here is the model.

.model 2N3904JP NPN(IS=1E-14 VAF=100 Bf=100 IKF=0.4 XTB=1.5 BR=4 CJC=4E-12 CJE=8E-12 RB=20 RC=0.1 RE=0.1 TR=250E-9 TF=350E-12 ITF=1 VTF=2 XTF=3 Vceo=40 Icrating=200m mfg=NXP)
 

Jony130

Joined Feb 17, 2009
5,155
Here is the model.

.model 2N3904JP NPN(IS=1E-14 VAF=100 Bf=100 IKF=0.4 XTB=1.5 BR=4 CJC=4E-12 CJE=8E-12 RB=20 RC=0.1 RE=0.1 TR=250E-9 TF=350E-12 ITF=1 VTF=2 XTF=3 Vceo=40 Icrating=200m mfg=NXP)
Well, in that case if you want the exact solution you need to include the Early effect and add ro resistance into equations.

ro = (VAF + Vce)/Ic
 

Thread Starter

aac044210

Joined Nov 19, 2019
127
All this means that the Early effect can be model as a ro resistor connected between the collector and the emitter of a “perfect” transistor in your small signal analysis.
My sim gives overall voltage gain of 588. Are we not using the same transistor models?
My TIP32C has a warning that there is an unrecognized parameter "c"
 
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Thread Starter

aac044210

Joined Nov 19, 2019
127
Yes, this is mainly due to ro effect on the first stage. The gain of a first stage is 144V/V
So, Av = 144 * 4.6 * 0.88 = 583 V/V


Now we are using the same model.
I get the same Av2 = 4.6

I get 1, not 0.88 for Av3 using your formula.

I still get an Av1 = 172 using

Av1 = rc1 / r'e1 = RC1║Zin2 / r'e1 = 0.458 KΩ║0.845 KΩ / 2.66 Ω = 172
 
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Thread Starter

aac044210

Joined Nov 19, 2019
127
The beta and Vbe of a transistor can be minimum, typical or maximum in a fairly large range. The circuit must be deigned so that any passing transistor works properly.

This circuit is designed only for a transistor with 2N3904JP spec's that might not even be available if you buy thousands and test them all. I simulated the first transistor using a transistor with typical spec's and it was biased wrong.
I have 2 questions:
1) How would you design the circuit to work with any passing transistor?
2) This is a circuit from a book that I found on line so I don't know what is wrong with the biasing (although I
did wonder why RC1 and RE1 were both 1K resistors).

I am trying to learn this stuff by reading and researching on my own so I am counting on these
internet sources to be correct.

aac
 
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Jony130

Joined Feb 17, 2009
5,155
I get 1, not 0.88 for Av3 using your formula.
AV3 ≈ (RL + 0.5*R31)/(0.5*re4 + (RL+0.5*R31)) * RL/(0.5*R31 + RL) ≈ (8Ω + 0.5Ω)/(0.5Ω + 8.5Ω) * 8Ω/8.5Ω ≈ 0.88V/V

I still get an Av1 = 172 using

Av1 = rc1 / r'e1 = RC1║Zin2 / r'e1 = 0.458 KΩ║0.845 KΩ / 2.66 Ω = 172
It is not a surprise, you simply did not include all resistances.

re1 = Vt/Ie = 25.86mV/9.73mA = 2.66Ω

And from your BJT model
.model 2N3904JP NPN(IS=1E-14 VAF=100 Bf=100 IKF=0.4 XTB=1.5 BR=4 CJC=4E-12 CJE=8E-12 RB=20 RC=0.1 RE=0.1 TR=250E-9 TF=350E-12 ITF=1 VTF=2 XTF=3 Vceo=40 Icrating=200m mfg=NXP)
RE = 0.1Ω and RB = 20Ω

therefore

re1 = 2.66Ω + 0.1Ω + 20Ω/100 = 2.96Ω

ro1 = (VAF + Vce)/Ic = (100V + 10.62V)/9.64mA ≈11kΩ


And finally, we have:

Av1 = (Rc1||RB21||RB22||Zin2||ro)/re ≈ 437Ω/2.96Ω ≈ 147V/V


I have 2 questions:
Add an overall negative feedback loop.
 

Thread Starter

aac044210

Joined Nov 19, 2019
127
AV3 ≈ (RL + 0.5*R31)/(0.5*re4 + (RL+0.5*R31)) * RL/(0.5*R31 + RL) ≈ (8Ω + 0.5Ω)/(0.5Ω + 8.5Ω) * 8Ω/8.5Ω ≈ 0.88V/V



It is not a surprise, you simply did not include all resistances.

re1 = Vt/Ie = 25.86mV/9.73mA = 2.66Ω

And from your BJT model

RE = 0.1Ω and RB = 20Ω

therefore

re1 = 2.66Ω + 0.1Ω + 20Ω/100 = 2.96Ω

ro1 = (VAF + Vce)/Ic = (100V + 10.62V)/9.64mA ≈11kΩ


And finally, we have:

Av1 = (Rc1||RB21||RB22||Zin2||ro)/re ≈ 437Ω/2.96Ω ≈ 147V/V




Add an overall negative feedback loop.
I now get Av3 = 0.888 as well. Not sure why we use the 0.5 factor.

Questions:
1) Is the 100 in the re1 formula the beta of 100?
2) Can I have more detail for the Av1 calculation please. It looks like Zin2 would have to
be about 10KΩ in order to get 147V/V and it is only ≈ 840Ω as far as I can tell.
 
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Audioguru again

Joined Oct 21, 2019
1,272
Adding an emitter resistor or biasing the transistor from its collector is a method of dc negative feedback that reduces the effects of Vbe change and beta change. Overall negative feedback is also necessary.
Some articles on the web are written by kids who are only 10 years old and they don't know much.
 

Jony130

Joined Feb 17, 2009
5,155
I now get Av3 = 0.888 as well. Not sure why we use the 0.5 factor.
Because you set the output stage quiescent current at 26mA. So, the LTspice in AC analysis will calculate the gain for the case when both output transistors are ON.
Therefore from the AC signal point of view, we have two BJT's content in parallel.
And this is why we have R31||R32 = 0.5*R31 and re4||re5 = 0.5*re4

1) Is the 100 in the re1 formula the beta of 100?
Yes, you are right.

2) Can I have more detail for the Av1 calculation please.
The AC equivalent circuit will look like this:

12 (3).PNG

Where

RB and RE are defined in your 2N3904JP model statement.



It looks like Zin2 would have to
be about 10KΩ in order to get 147V/V and it is only ≈ 840Ω as far as I can tell.
No, Zin2 = (β +1)*(re2 + RE2) = 101*(1.8Ω + 100Ω) ≈ 10.2kΩ
 

Thread Starter

aac044210

Joined Nov 19, 2019
127
I separate RB21 and RB22 form Zin2

This is why I have (Rc1||RB21||RB22||Zin2||ro) in my Av1 gain equation.
Ok. So Zin2 in the equation is really Zin2(Base). Got it.

The calculated Overall Gain is 613 while the sim is 588. Do we need to include
the voltage division factor between stage 1 and stage 2 and then between stage
2 and stage 3? I was hoping to get an estimate that was closer.
 
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