There is a schematic of an pre-amplifier that use one-half of voltage source output biasing (T1 emitter is one-half of V1). How the resistive network was calculated to allow one-half output Q-point?

Original link: http://www.zen22142.zen.co.uk/Circuits/Audio/lvpreamp.htm

enter image description here

  • \$\begingroup\$ Have you biased transistors before? Is your question relating in particular to how the feedback network affects things? \$\endgroup\$
    – jramsay42
    Jan 30 '18 at 8:47
  • \$\begingroup\$ @jramsay42 Yes, I was biased CE with emitter degeneration resistor with variations but without using feedback in two-stage schematic solutions \$\endgroup\$
    – MaxMil
    Jan 30 '18 at 14:49

For \$Ic2 = 500 \mu A\$ and \$Ic1 = 1mA \: \: \beta_1 = \beta_2 = 100\$

And if we assumed \$ V_{E1} = 0.5\cdot V_{CC}\$ we alredy know that

\$R_{1} = \frac{Vcc- 0.5 V_{CC} - V_{BE1}}{I_{C2} + I_{B1}} = \frac{12V - 6V - 0.6V}{500 \mu A + 10 \mu A } = 10 \textrm{k}\Omega \$

$$R_{4} = \frac{0.5V}{505 \mu A} = 1\textrm{k}\Omega$$

$$R_2+R3 \approx \frac{0.5V_{CC}}{I_{C1}} = 6\textrm{k}\Omega$$

The \$T_2\$ base voltage is around \$V_{BE2}+V_{E2} = 0.5V+0.6V = 1.1V\$

So the voltage drop across \$R_3\$ should be larger or equal than this value (1.1V).

\$R_3 = \frac{1.5V}{1mA} = 1.5 \textrm{k}\Omega \$


\$ R2 = 6 \textrm{k}\Omega - 1.5 \textrm{k}\Omega = 4.7 \textrm{k}\Omega\$

And finaly

\$R_5+R_7 = \frac{1.5V - 1.1V}{5 \mu A} = 82\textrm{k}\Omega\$

And we done.

Now we could check the calculations by doing DC analysis.


simulate this circuit – Schematic created using CircuitLab

First KVL:

$$I_{E2}\cdot R_4 + V_{BE2}+\frac{I_{C2}}{\beta} \cdot (R_5+R_7) = (I_{E1} -\frac{I_{C2}}{\beta}) \cdot R_3 $$

And another KVL equation is:

$$V_{CC} - (I_{C2} + \frac{I_{C1}}{\beta}) \cdot R_1 - V_{BE1} =I_{E1} \cdot R_2 +(I_{E1} -\frac{I_{C2}}{\beta}) \cdot R_3$$

So, if we solve this two equation we can solve for a quiescent point.

  • \$\begingroup\$ For what BJT's Ic2 and Ic1 relate? \$\endgroup\$
    – MaxMil
    Jan 30 '18 at 15:44
  • \$\begingroup\$ What? I don't get it? \$\endgroup\$
    – G36
    Jan 30 '18 at 15:46
  • \$\begingroup\$ I see you set initial BJT currents for calculations, but trying to understand the first equation for R1 founds that numerator of R1 must contain Vbe1 not Vbe2. And where you get R4 0.5V in numerator? \$\endgroup\$
    – MaxMil
    Jan 30 '18 at 15:49
  • \$\begingroup\$ I corrected the error. As for 0.5V I simply assumed Ve1 = 0.5V as we usually do in BJT's circuits. \$\endgroup\$
    – G36
    Jan 30 '18 at 15:54
  • \$\begingroup\$ In my calculation I found that Ic2 = 100 uA is already too high. You start with Ic2 = 500 uA. In the article it says: T2 collector current is about 70uA which confirms my 100 uA being too high. Why would you start with Ic2 = 500 uA? \$\endgroup\$ Jan 30 '18 at 15:59

There is some interaction going on between the collector current of T2, the emitter resistor R4 and that voltage divider R2, R3.

If we assume that the current through R5 and R7 is so small that there's almost no voltage drop across them then the voltage across R3 is equal to Vbe_T2 + V_R4.

That Vbe_T2 will be quite constant at around 0.6 V.

But what will the voltage across R4 be? It depends on the current through it of course. But since we want the voltage across R2 + R3 to be Vcc/2 is means the voltage across R3 is Vcc/4. From that Vcc/4 subtract the Vbe of T2, that leaves 3 V / 4 - 0.6 V = 0.15 V across R4. So the current through R4 will be 0.15V / 1.5k ohm = 100 uA.

Then we need to check if the rest of the circuit can support the values we just calculated. At the collector of T2 we will have 3 V - 100 uA * 12 kohm = 1.8 V Since T2's emitter is at 0.15 V that leaves a Vce of 1.65 V which is enough.

Base of T1 is at that same 1.8 V, subtract Vbe to get to emitter voltage: 1.8 V - 0.6 V = 1.2 V That is a bit lower than the 1.5 V we wanted, so you would think that T1 would be off (Vbe is too low) and the circuit cannot work.

However due to the feedback the circuit will automatically find a point where it can work.

What happens is that when T1 is off, the voltage across R2, R3 is lower than we want and that also means that the voltage across R4 will be less. So T2 will have a much smaller Ic than we predicted. This makes the voltage at the collector of T2 rise. Which is what we need to turn T1 on. That in turn will increase the voltage across R2, R3 and R4 until a certain balance situation is found where the voltages and currents will be close to what we predicted.

You could re-do the calculation using a larger Vbe, for example Vbe = 0.65 V instead of 0.6 V, that would make Ic of T2 a bit smaller and decrease the voltage across R1 (which was the issue we had above) then perhaps you get closer to the real solution.

When designing such a circuit I do the hand calculation as explained above and then use a circuit simulator to find the exact values and check if my calculation was sane (close enough I mean).

  • \$\begingroup\$ Why you wrote: "we want the voltage across R3 + R4 to be Vcc/2"? I found that the Vcc/2 must be on T1 emitter as Q-point value. Or it's shown for example to demostrate how it's working theoretically? \$\endgroup\$
    – MaxMil
    Jan 30 '18 at 14:56
  • \$\begingroup\$ we want the voltage across R3 + R4 to be Vcc/2 Because you wrote: How the resistive network was calculated to allow one-half output Q-point? For me that means that you want the output to be DC biased at Vcc/2. Then that means that the voltage across R3 + R4 will be Vcc/2. It is my starting point for the loop calculation. \$\endgroup\$ Jan 30 '18 at 15:32
  • \$\begingroup\$ Yes, it's all so. But I see that if output is on T1 emitter then V_R2 + V_R3 = Vcc/2. But I can't understand how it relates to V_R3 + V_R4? \$\endgroup\$
    – MaxMil
    Jan 30 '18 at 15:38
  • \$\begingroup\$ Oh, geez, typo ! The line should have been we want the voltage across R2 + R3 to be Vcc/2 ! So forget R3 + R4, it should have said R2 + R3 \$\endgroup\$ Jan 30 '18 at 15:53

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