How Do You Calculate DC Collector Current In Output Transistors?

Thread Starter

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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Thread Starter

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
 

crutschow

Joined Mar 14, 2008
24,318
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.
 

Jony130

Joined Feb 17, 2009
5,145
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?
 

crutschow

Joined Mar 14, 2008
24,318
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
7,120
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?
 

Thread Starter

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
7,120
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.
 

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Last edited:

Audioguru again

Joined Oct 21, 2019
858
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
7,120
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
858
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
7,120
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?
 
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