I’m trying to simulate that amplifier from Proteus!

enter image description here

First I found quiescent current for Q1, Q2 transistors: IcQ = (12 – 0.7)/R1 = 35mA

But to set that current through R1-D1-D2-CE(Q3)-R5 I need the proper biasing of BC184. How I can calculate the R2 and R6 for that proper bias? I have the BC184 datasheet: http://pdf1.alldatasheet.com/datasheet-pdf/view/50719/FAIRCHILD/BC184.html And know that UR1 = 11,2V and UcQ3 = 11,2V From datasheet I may found the hfe of BC184 but there is no minimal hfe for Uce = 11,2V what value I need to choose?

That schema may have a very big output DC voltage and the very little changes in base voltage divider resistors changes output DC dramatically. What I must do to tune the output voltage correctly?


First, please put proper DC supplies on your simulation.

Since the push/pull output stage has a unity gain, the DC level on Q3's collector, \$V_{C_{Q3}}\$, will be seen on VOUT node as \$V_{OUT}=V_{C_{Q3}}+V_{BE_{Q2}}\$ -neglecting the drop on 0R1 resistor.

In order to set VOUT node to the midpoint of ±12V, which is 0VDC, your first aim should be setting \$V_{C_{Q3}}\$ to about -0.6VDC.

Also, the common emitter amplifier formed by Q3-R1-R2-R5-R6 has a gain of R1/R5 and equals the total gain of the whole amplifier circuit.

  • Quiescent current for biasing diodes can be anything between 2-10mA. Let's select 5mA.
  • Gain is \$K_v=R1/R5\$ and input voltage is 70mVrms = 0.1Vp. Since the peak value of output signal can be VCC=12V before clipping, \$K_{v(max)}= 12 / 0.1=120\$ for unclipped maximum swing --not necessary to set that high though.


Let's set the gain to 2.


simulate this circuit – Schematic created using CircuitLab

So \$R1/R5=2\$. In order to have -0.6VDC on collector of Q3, voltage drop on R1 should be \$V_{R1}=(+12V - (-0.6V) - 1.2V)=11.4VDC\$. Let's select \$I_{bias}=5mA\$, so \$R1=11.4V/5mA=2.2k\$ so R5 will be 1k1. Since the voltage drop on R5 will be \$V_{R5}=1.1k \cdot 5mA = 5.5V\$, the voltage on the base node of Q3 should be \$V_B=-12 + (5.5 + 0.6) = -5.9V\$. For this purpose, I chose R2=33k and R6=12k.

According to simulation results, quiescent output DC level is 0.3VDC which is tolerable. If you set R1 = 5k1 and R5 = 2k5 then the quiescent DC output on VOUT node will be less than 0.1VDC, but the current passing through the biasing diodes will be reduced to about 2.5mA which is fairly enough.

If we select \$K_v=10\$ and \$I_{bias}=2mA\$ then we can find that (with the calculations above) R1=5k6, R5=560R, R2=33k and R6=2k7.

You can tune the circuit according to your gain needs.

  • \$\begingroup\$ I see that maximum diode current is 10mA, but in original scheme its 35mA and the schema positioned as working. May be current through the diode more than 10mA? Why Kv(max) = 170? May you show the calculation of all resistors which values you write? \$\endgroup\$ – MaxMil Oct 26 '16 at 8:52
  • \$\begingroup\$ I don’t fully understand transistor biasing with two power sources! To fully understand how I can calculate resistors values share your calculations for bias resistors for BC184! \$\endgroup\$ – MaxMil Oct 26 '16 at 9:21
  • \$\begingroup\$ Ooops, Kv(max)=120, not 170. Sorry --- anyway, I've edited my post. \$\endgroup\$ – Rohat Kılıç Oct 26 '16 at 9:24
  • \$\begingroup\$ How you choose R5 value? The voltage drop on R5 may not be founded as for R1! \$\endgroup\$ – MaxMil Oct 26 '16 at 9:35
  • \$\begingroup\$ @MaxMil please read the post carefully. R5 can be selected from required/selected gain, Kv = R1/R5. If you select Kv=2 then R5=R1/2. If you select Kv=10 then R5=R1/10. That's it. \$\endgroup\$ – Rohat Kılıç Oct 26 '16 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.