How Do You Calculate DC Collector Current In Output Transistors?

MrAl

Joined Jun 17, 2014
11,389
The "output" voltage of a Vbe multiplier (Q3) is a Vce3 voltage.
That's a nice illustration of how the output stage crossover distortion is removed.
The analysis of the "Vbe multiplier" is not exactly accurate though as there is an effect due to the upper power supply resistor also which could contribute significantly to the transistor CE drop. A complete analysis shows this effect clearly as an equation such as:
Vce=K1*Vbe+K2*Vcc
where K1 is the "multiplier" and where K2 is small but significant for finding the real bias point of the output stage.
The reason for the second term is probably because the Beta of the transistor is finite and not really super high, but even a Beta of 100 shows this effect is significant.
 

Jony130

Joined Feb 17, 2009
5,487
That's a nice illustration of how the output stage crossover distortion is removed.
The analysis of the "Vbe multiplier" is not exactly accurate though as there is an effect due to the upper power supply resistor also which could contribute significantly to the transistor CE drop. A complete analysis shows this effect clearly as an equation such as:
Vce=K1*Vbe+K2*Vcc
where K1 is the "multiplier" and where K2 is small but significant for finding the real bias point of the output stage.
The reason for the second term is probably because the Beta of the transistor is finite and not really super high, but even a Beta of 100 shows this effect is significant.
Yes, I know about that. But we can add a small Rc resistance to overcome this problem.
See my answer here
https://electronics.stackexchange.c...emitter-resistance-cancellation/450134#450134
Or read Jonk's more detail analysis.
 

MrAl

Joined Jun 17, 2014
11,389
Yes, I know about that. But we can add a small Rc resistance to overcome this problem.
See my answer here
https://electronics.stackexchange.c...emitter-resistance-cancellation/450134#450134
Or read Jonk's more detail analysis.
Yes i was just pointing out that there was a little something missing that's all.

But the analysis i did for one such circuit came out to this:
Vshunt=(3100*E2*R4)/1021031-(1000*E1*R4)/1021031+(2110000*E2)/1021031+(11021*E1)/1021031

where E2 is the Vbe and E1 is the power supply voltage and there is a upper series resistor that appears in the actual amplifier circuit,

and now setting E2=0 and Vshunt=0 we can solve for R4 (the "comp" resistor) and then when we apply that resistance to R4 in the equation we get exactly:
Vshunt=2.1*E2
which was the target voltage.
Of course this is sort of theoretical in nature.
With E2=0.7 we get 1.47 so if each power transistor Vbe is 0.7v also then we get 35ma quiescent current in the output section.

I did notice that for a given resistor set the "comp" resistor value varies a bit. We'd have to figure out how to get around that if we wanted to keep it perfectly adjustable without too much trouble.

Nice idea though. Another idea is to make the two set point resistors lower in value so that they swamp out effects from the upper series resistor.

The other thing that kinda bothered me about these circuits is that the output power transistor Vbe is probably neer the same exactly as the smaller transistor Vbe. Maybe use an op amp tracking regulator to set the Vbe multiplier circuit voltage (he he) :)

Actually i just call it a shunt regulator, which acts like a zener.
 

Audioguru again

Joined Oct 21, 2019
6,672
In this circuit in this case the trimpot setting is already determined. It's right on the schematic. Is it not set right though?
The trimpot setting is already determined so that a student can do the calculations. But in real life the spec's for transistors vary widely so the current must be measured as the trimpot is adjusted, with NO calculations.
 

MrAl

Joined Jun 17, 2014
11,389
The trimpot setting is already determined so that a student can do the calculations. But in real life the spec's for transistors vary widely so the current must be measured as the trimpot is adjusted, with NO calculations.
Oh yeah sure, this is a theoretical exercise i presume. For these we also usually assume a Beta like 100 for example, or a range of Beta from maybe 50 to 200 for example.

The full analysis though would try to find the min and max to see if the design is feasible to begin with. That's all pure calculation too.
 

MrAl

Joined Jun 17, 2014
11,389
Do you know how to analyze transistor circuits by assuming the transistor is a current controlled current source?
The transistor CE becomes a CCCS and the base emitter becomes a constant voltage diode where the base current and collector current both flow though that diode.
If you want to model the diode using the non linear diode equation then you end up with a non linear equation that usually has to be solved numerically so that's only done when needed.
We can look at a simpler circuit to get you started if you like.
 
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Thread Starter

aac044210

Joined Nov 19, 2019
178
If this is the case we can now use a KVL and solve for the output stage collector current.

Vce3 = Vbe4+Vbe5+Icq*(Re31 + Re32) (1)

Icq = (Vce3 - 2Vbe)/(Re31 + Re32) (2)
using equation (2) above

ICQ = (1.524 - 1.4) / (1 + 1) = 62 mA ≠ 25.5 mA per sim did I misunderstand?
 
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Jony130

Joined Feb 17, 2009
5,487
The calc is 62mA and the sim is 25.5 mA. That isn't even in the same universe.
You could always use a nonlinear equation and solve it for Ic current.
For simplicity we can assume that Is = 1E-14 and that Vbe4+Ie*Re31 = 1.524V/2 = 0.762V

And we can use this oversimplified circuit and solve for Ie current this way:

3 (3).PNG


Ie = (Vin - Vbe)/Re = (0.762V - 0.7V)/1Ω = 62mA (initial guess)

Vbe = Vt* ln(Ie/Is) = 26mV * ln(62mA/1E-14) = 0.7658V

New Vbe value (0.7V + 0.7658V)/2 = 0.7329V

Ie = (0.762V - 0.7658V)/1Ω = 29.1mA

Vbe = 26mV * ln(29.1mA/1E-14) = 0.74618V


New Vbe value (0.7329V + 0.74618V)/2 = 0.73954V

Ie = (0.762V - 0.73954 V)/1Ω = 22.46mA

Vbe = 26mV * ln(22.46mA/1E-14) = 0.73944V

(0.73954V+ 0.73944V)/2 = 0.73949V

Ie = (0.762V - 0.73949V )/1Ω = 22.51mA


I will end here and assume that Ie = 22.5mA is the solution.

But if I use Vt = 25.8mV :

Ie = 25.23mA; Vbe = 0.7367V.


But what is the point of doing such calculations?
 

Thread Starter

aac044210

Joined Nov 19, 2019
178
You could always use a nonlinear equation and solve it for Ic current.
For simplicity we can assume that Is = 1E-14 and that Vbe4+Ie*Re31 = 1.524V/2 = 0.762V

And we can use this oversimplified circuit and solve for Ie current this way:

View attachment 198289


Ie = (Vin - Vbe)/Re = (0.762V - 0.7V)/1Ω = 62mA (initial guess)

Vbe = Vt* ln(Ie/Is) = 26mV * ln(62mA/1E-14) = 0.7658V

New Vbe value (0.7V + 0.7658V)/2 = 0.7329V

Ie = (0.762V - 0.7658V)/1Ω = 29.1mA

Vbe = 26mV * ln(29.1mA/1E-14) = 0.74618V


New Vbe value (0.7329V + 0.74618V)/2 = 0.73954V

Ie = (0.762V - 0.73954 V)/1Ω = 22.46mA

Vbe = 26mV * ln(22.46mA/1E-14) = 0.73944V

(0.73954V+ 0.73944V)/2 = 0.73949V

Ie = (0.762V - 0.73949V )/1Ω = 22.51mA


I will end here and assume that Ie = 22.5mA is the solution.

But if I use Vt = 25.8mV :

Ie = 25.23mA; Vbe = 0.7367V.

But what is the point of doing such calculations?
Thanks Jony.

If you are designing an amplifier, you need to estimate the current that the power supply
will need to produce. If you know (within reason) the collector current then you know the
current that the circuit will draw (with no load). That is my logic anyway. Thinking out loud.
I realize that with a load, a lot more current will be required. I haven't gone that far yet.
 

MrAl

Joined Jun 17, 2014
11,389
Make each transistor Q2, Q3, Q4, Q5 a current controlled current source collector to emitter, and use a diode for the base emitter.
The current source goes from collector to anode of diode, the base goes to anode, the emitter comes from the cathode of the diode. The collector current is then the base current times the Beta. You can approximate the diode as a voltage source of 0.7 volts for now.

You do that for each of the four transistors then analyze the whole thing. You can then calculate the current through the two 1 Ohm resistors.

Here is a quick drawing showing a transistor like this which also has an emitter resistor, the resistor R12 is an internal base resistor you can make simply 1 Ohm for now.

Note the polarity of the emitter diode given by the voltage source, and the current source in the collector points down, and the collector current is iB*Beta.
This transistor happens to have an emitter resistor but you dont need that for all the transistors unless of course they have an actual emitter resistor.
 

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MrAl

Joined Jun 17, 2014
11,389
Hi,

Here is a full example of a single transistor amplifier.
Note iC=iB*Beta and in the equations B=Beta.
The schematic is also labeled with more common names for the components.
The nodes are labeled 1 through 5 and the voltages in the equations are v1 through v5 respectively.
Solving the equations shown leads to the solution for each and every node voltage in terms of the components.
 

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MrAl

Joined Jun 17, 2014
11,389
Here is an example of the "Voltage Multiplier" we had been talking about.
Note the components used for the transistor are shown to the right of the circuit.
I1 is the collector current current source, E2 is the base emitter voltage drop.
R9 is an optional base resistor.
Note that I1=iB*B where B is the Beta, and since iB is (v3-v2)/R9 the collector current is:
iC=B*(v3-v2)/R9

and you can spot the right side of that in the equations for the circuit shown near the bottom.
 

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