# DC Analysis Of This Circuit

Joined Jun 16, 2023
128
Hi:

How would you start a DC analysis of this circuit?

#### dl324

Joined Mar 30, 2015
16,127
How would you start a DC analysis of this circuit?
What circuit?

EDIT: even though the OP's post doesn't reflect that it was edited, it was edited to add the circuit in question after my post...

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• drjohsmith

Joined Jun 16, 2023
128
What circuit?
Hi Dennis
LOL. The circuit that is missing of course. Brain fart.
RS

• drjohsmith

#### dl324

Joined Mar 30, 2015
16,127
Are you simulating this circuit?

Joined Jun 16, 2023
128
Are you simulating this circuit?
Yes but I would like to do a manual dc analysis so that I can come up with a set of design
steps for such a circuit. I attempted the analysis but went in circles and got no where fast.

#### MisterBill2

Joined Jan 23, 2018
16,534
If it is the circuit shown in post #1, I would start the analysis with out the feedback resistor "Rf" Also remove the input section and the output connections
The open loop DC analysis will be interesting if all of the forward diode and base-emitter voltages are assumed to be 0.7000 volts, because with no signal it would be a class B amplifier.

#### LvW

Joined Jun 13, 2013
1,683
If it is the circuit shown in post #1, I would start the analysis with out the feedback resistor "Rf"
The resistor Rf defines the DC bias point for Q1.
So - it is mandatory for a DC analysis of the circuit.

#### dl324

Joined Mar 30, 2015
16,127
I attempted the analysis but went in circles and got no where fast.
It will take an iterative process to determine DC bias.

#### LvW

Joined Jun 13, 2013
1,683
My approach (assuming 0.7 volts across each pn-junction):
* Base of Q2: V_B2=Vout+1.4 volts
* Base of Q4: V_B4=V_B2-2.1 volts (3 diode drops)
* Base of Q4: V_B4=Vout-0.7 volts

Now we have three equations for three unknown voltages (Vout, V_B2 and V_B4)
This should be enough for finding all other DC voltages and currents,

• BobTPH

#### crutschow

Joined Mar 14, 2008
33,346
Because of negative feedback through Rf and the bias point depending upon the beta gain of the transistors, that is not an easy circuit to analyze, or a very good circuit in practice.

Note that the 10uF output coupling capacitor to the 8 ohm load, means the low frequency cutoff of the amp is about 2kHz.

• Ian0

#### dl324

Joined Mar 30, 2015
16,127
I'd take a simpler approach.
1. Find the base current of Q1 by calculating the current through RB1 and the BE junctions of Q2, Q3, Q1, and RB2. $$I_{RB1} = \frac{V_{Sup} - 3*V_{BE}}{RB1+RF+RB2} = \frac{27.9V}{151.22kΩ}=0.18mA$$
2. Q1 is now on and beta Q1 (I used 100 for simplicity) times the current calculated into the base of Q1 (18mA) is now flowing in RB1. Giving about 12V at the anode of D1.
3. D1, D2, and D3 are now conducting, biasing Q2-5 on and you can calculate the voltage at the output (10.6V).
EDIT: corrected arithmetic error for output voltage, was 10.4V.

Last edited:

Joined Jun 16, 2023
128
Because of negative feedback through Rf and the bias point depending upon the beta gain of the transistors, that is not an easy circuit to analyze, or a very good circuit in practice.

Note that the 10uF output coupling capacitor to the 8 ohm load, means the low frequency cutoff of the amp is about 2kHz.
This circuit is used by many schools of engineering. Why would they use it if is a poor circuit?

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#### crutschow

Joined Mar 14, 2008
33,346
This circuit is used by many schools of engineering.
How do you know that?
Why would they use it if is a poor circuit?
Probably because the professors picking the circuits have had little experience designing real circuit.

Joined Jun 16, 2023
128
I'd take a simpler approach.
1. Find the base current of Q1 by calculating the current through RB1 and the BE junctions of Q2, Q3, Q1, and RB2. $$I_{RB1} = \frac{V_{Sup} - 3*V_{BE}}{RB1+RF+RB2} = \frac{27.9V}{151.22kΩ}=0.18mA$$
2. Q1 is now on and beta Q1 (I used 100 for simplicity) times the current calculated into the base of Q1 (18mA) is now flowing in RB1. Giving about 12V at the anode of D1.
3. D1, D2, and D3 are now conducting, biasing Q2-5 on and you can calculate the voltage at the output (10.6V).
EDIT: corrected arithmetic error for output voltage, was 10.4V.
Hi Dennis:
The sim is giving dc Vout = 12.25 V

#### crutschow

Joined Mar 14, 2008
33,346
The sim is giving dc Vout = 12.25 V
Those two values are as close as would be expected between a hand calculation and a simulation.
The simulator uses more accurate models so likely would be more accurate.

#### dl324

Joined Mar 30, 2015
16,127
Hi Dennis:
The sim is giving dc Vout = 12.25 V
I said it would take an iterative process to determine DC bias. I only gave 2 iterations.

If you want a "better" answer, you need to consult datasheets for diode and junction drops at the approximate current I gave and refine the currents and voltage drops. You should get a more converged answer after a couple more iterations.

But, the datasheets I have for 2N3904 and TIP31A don't give me enough information to refine the calculations...

#### Ian0

Joined Aug 7, 2020
8,942
Hi Dennis:
The sim is giving dc Vout = 12.25 V
What does it give if the hfe of Q1 is its minimum value? Or its maximum value?

Joined Jun 16, 2023
128
What does it give if the hfe of Q1 is its minimum value? Or its maximum value?
Here are some data sheets that I have on file.

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Joined Jun 16, 2023
128
Dennis / Ian:

If Q1 β = 40 then IC(Q1) = 6.35 mA and if β = 100 then IC(Q1) = 10.25 mA.

RS

#### Ian0

Joined Aug 7, 2020
8,942
Dennis / Ian:

If Q1 β = 40 then IC(Q1) = 6.35 mA and if β = 100 then IC(Q1) = 10.25 mA.

RS
So you now understand why it is such a bad circuit?