# How Do You Calculate DC Collector Current In Output Transistors?

#### aac044210

Joined Nov 19, 2019
127
I have a class ab power amp circuit. I am hand-calculating all dc voltages and currents but I
can't figure out how I would find that the dc collector current for Q4 and Q5 is 25.5mA. I
have attached the sim file.

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#### crutschow

Joined Mar 14, 2008
25,242
Show us your calculations so far.

#### dl324

Joined Mar 30, 2015
11,223
Post a schematic for those of us who are using a device that can't render .asc files.

#### aac044210

Joined Nov 19, 2019
127
Show us your calculations so far.
 RTH1 = RB1 = RB11║RB12 3.590​ KΩ VTH1 = (VCC * RB12) / (RB11 + RB12) 10.769​ Volts IB1 = (VTH1 - VBE1) / (RTH1 + (βdc1 + 1) * RE1) 0.096​ mA IC1 = βdc1 * IB1 9.627​ mA IE1 = IB1 + IC1 9.723​ mA VE1 = RE1 * IE1 9.723​ Volts VB1 = VE1 + VBE1 10.423​ Volts VC1 = VCC - (IC1 + RC1) 20.373​ Volts VCE1 = VC1 - VE1 10.650​ Volts IBIAS = VB / RB2 1.861​ mA IRB11 = IBIAS1 + IB1 1.957​ mA IRB12 = IBIAS1 1.861​ mA IRC1 = IC1 9.627​ mA IRE1 = IE1 9.723​ mA IDC1 = IRB11 + IC1 11.584​ mA PRB11 = IRB11² * RB11 38.309​ mW PRB12 = IRB12² * RB12 19.395​ mW PRC1 = IRC1² * RC1 92.681​ mW PRE1= IRE1² * RE11 94.544​ mW PDQ1 = VCE1 * IC1 102.529​ mW PDC1 = VCC * IDC1 347.531​ mW RTH2 = RB2 = RB21║RB22 0.915​ KΩ VTH2 = (VCC * RB22) / (RB21 + RB22) 2.286​ Volts IB2 = (VTH2 - VBE2) / (RTH2 + (βdc2 + 1) * RE2) 0.144​ mA IC2 = βdc2 * IB2 14.399​ mA IE2 = IB2 + IC2 14.542​ mA VE2 = RE2 * IE2 1.454​ Volts VB2 = VE2 + VBE2 2.154​ Volts VC2 = VB5 14.238​ Volts VE3 = VC2 14.238​ Volts VB3 = VE3 + VBE3 14.938​ Volts VC3 = VB4 15.762​ Volts VRB3 = VCC - VC3 14.238​ Volts IRB3 = VRB3 / RB3 14.238​ mA VE4 = VCC * 0.5 15.000​ Volts VB4 = VE4 + VBE4 15.762​ Volts VC4 = VCC 30.000​ Volts VCE4 = VCC * 0.5 15.000​ Volts VE5 = VCC * 0.5 15.000​ Volts VB5 = VE5 - VBE5 14.238​ Volts VC5 0.000​ Volts VCE5 = VCC * 0.5 15.000​ Volts

#### aac044210

Joined Nov 19, 2019
127
Post a schematic for those of us who are using a device that can't render .asc files.
Schematic attached.

#### Jony130

Joined Feb 17, 2009
5,180
It is impossible to solve for Q4 and Q5 quiescent current without involving non-linear equations.

#### crutschow

Joined Mar 14, 2008
25,242
It is impossible to solve for Q4 and Q5 quiescent current without involving non-linear equations.
For an exact answer, that's true.
But you should be able to solve for reasonably approximate values by assuming all the transistor base-emitter drops are about 0.7V and the base currents are zero.

#### aac044210

Joined Nov 19, 2019
127
It is impossible to solve for Q4 and Q5 quiescent current without involving non-linear equations.
Hi Jony:

How would I do that?

aac

#### Jony130

Joined Feb 17, 2009
5,180
But you should be able to solve for reasonably approximate values by assuming all the transistor base-emitter drops are about 0.7V and the base currents are zero.
But TS mentions that the Icq = 25mA hence the voltage drop across emitter resistance is very low (25mV). Therefore the error of an approximated calculation will be very large, Isn't it?

#### Jony130

Joined Feb 17, 2009
5,180
How would I do that?
First you need to solve for Vbe multiplier (Q3) "output" voltage.
Can you do it?

#### aac044210

Joined Nov 19, 2019
127
First you need to solve for Vbe multiplier (Q3) "output" voltage.
Can you do it?
 VE3 = VC2 14.238 Volts VB3 = VE3 + VBE3 14.938 Volts VC3 = VB4 15.762 Volts

#### crutschow

Joined Mar 14, 2008
25,242
But TS mentions that the Icq = 25mA hence the voltage drop across emitter resistance is very low (25mV). Therefore the error of an approximated calculation will be very large, Isn't it?
Yes, that's a valid point, it could.
In that case you could use the data sheet nominal Vbe voltage of the specified output transistors, at their anticipated 25mA collector current, and see how close you come to that current with the calculations.
If not close, then the set point for the Q3 base pot might need to be tweaked and the bias point values recalculated.

#### MrAl

Joined Jun 17, 2014
7,748
I have a class ab power amp circuit. I am hand-calculating all dc voltages and currents but I
can't figure out how I would find that the dc collector current for Q4 and Q5 is 25.5mA. I
have attached the sim file.
Where did the sim file come from, were you given this or did you create it yourself?
I ask because if you intend to get all the calculations to match the sim file results you would have to use the spice models of the transistors involved in the calculations.
So a related question is, where did you get this 25.5ma figure from?

#### aac044210

Joined Nov 19, 2019
127
Where did the sim file come from, were you given this or did you create it yourself?
I ask because if you intend to get all the calculations to match the sim file results you would have to use the spice models of the transistors involved in the calculations.
So a related question is, where did you get this 25.5ma figure from?
Hi:

The circuit is one that I do got off the internet some time ago and just came across it the
other day and I created the sim file. Where the schematic came from I really can't say.

The 25.5 mA is the current flowing through the emitter resistors of the output transistors.

I used the same transistor parameters in my calculations as in LTS.

aac

#### MrAl

Joined Jun 17, 2014
7,748
Hi:

The circuit is one that I do got off the internet some time ago and just came across it the
other day and I created the sim file. Where the schematic came from I really can't say.

The 25.5 mA is the current flowing through the emitter resistors of the output transistors.

I used the same transistor parameters in my calculations as in LTS.

aac
Is this really homework then?
Ok then to start, see what you can determine about the subcircuit with Q3 in it. See if you can calculate the voltage across the collector to emitter as it relates to the Vbe of that transistor.
You know the resistor values associated with that transistor and if necessary assume the Vbe is 1 for now. Thus you can calculate the ratio Vce/Vbe without knowing Vbe explicitly just yet.

Since we have limited information, do you believe that we can assume that all the voltage drops for Vbe are the same for all transistors even though they are not all the same type?

Attached is a 'dark mode' rendering of the schematic.

Last edited:

#### Audioguru again

Joined Oct 21, 2019
1,721
it is a silly circuit. Why don't teachers use the circuit of a real amplifier?
1) No negative feedback so it distorts like crazy.
2) What is the trimpot used for? It is not used for calculations, instead it is used to adjust the idle current because the transistors have differences that mess up your calculations.

#### MrAl

Joined Jun 17, 2014
7,748
it is a silly circuit. Why don't teachers use the circuit of a real amplifier?
1) No negative feedback so it distorts like crazy.
2) What is the trimpot used for? It is not used for calculations, instead it is used to adjust the idle current because the transistors have differences that mess up your calculations.
It has theoretical significance.
What do you mean the trimpot is not used for calculations? If the trimpot adjusts the idle current then it must be included in the calculations because the idle current is calculated within the calculations. Why wouldnt it?

#### Audioguru again

Joined Oct 21, 2019
1,721
If the trimpot is simply calculated accurately then it would not need to be adjustable. But since transistors have a wide range of spec's then an accurate calculation cannot be made so the trimpot is needed to be adjusted for the desired current.

#### MrAl

Joined Jun 17, 2014
7,748
If the trimpot is simply calculated accurately then it would not need to be adjustable. But since transistors have a wide range of spec's then an accurate calculation cannot be made so the trimpot is needed to be adjusted for the desired current.
In this circuit in this case the trimpot setting is already determined. It's right on the schematic. Is it not set right though?

#### Jony130

Joined Feb 17, 2009
5,180
 VE3 = VC2 14.238 Volts VB3 = VE3 + VBE3 14.938 Volts VC3 = VB4 15.762 Volts
The "output" voltage of a Vbe multiplier (Q3) is a Vce3 voltage.