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I'm having trouble analyzing this biasing circuit.

It's supposed to shift Vin up by 1.4 V, and everything works when I simulate it, but I can't get anything out of the manual analysis.

I know that with \$ V_{\text{in}}=0 \$ node A and B are supposed to be respectively at 1.4 V and 0 V since no current flows on R3, and with \$ V_{\text{in}}>0\$ node B is supposed to be at $$ V_B = V_{\text{in}}\frac{R_1 || R_2}{(R_1 || R_2)+R_3} $$ but I can't figure out how to get there.

Does anyone have any tips?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Don't assume that no current flows through R3. \$\endgroup\$ Aug 26 at 11:00
  • \$\begingroup\$ @ElliotAlderson thank you for the tip \$\endgroup\$
    – ale_zec
    Aug 26 at 13:52
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The first step is to find the potential at B, when \$V_{IN} = 0V\$. I'll start by redrawing the circuit with all known potentials (namely -3V and +3V) derived explicitly from an appropriate voltage source:

schematic

simulate this circuit – Schematic created using CircuitLab

In the left circuit I see that the source Vin, being zero, can be replaced by a direct connection to ground, 0V. I also notice that sources V3 and Vab are effectively in series, and can be replaced by an equivalent single source V4 of 1.6V. These simplifications give rise to the circuit on the right.

Now I want to find the voltage at B, using the superposition theorem. Here we determine the influence of each voltage source in turn, while the others are replaced by short circuits, and we combine the results to find B's potential. Starting with the contribution from source V4, I'll short circuit V2:

schematic

simulate this circuit

This is a simple potential divider, for which we can calculate the potential at B. If \$ V_{B4} \$ is the voltage contributed by source V4 to the voltage at B, then:

$$ V_{B4} = V_4 \times \frac{R_X}{R_1 + R_X} = 1.6V \times \frac{R_2 \parallel R_3}{R_1 + R_2 \parallel R_3 } = +9.91mV $$

Now I short V4, to find V2's contribution:

schematic

simulate this circuit

\$V_{B2}\$, the contribution at B due to V2 is:

$$ V_{B2} = V_2 \times \frac{R_Y}{R_2 + R_Y} = -3.0V \times \frac{R_1 \parallel R_3}{R_2 + R_1 \parallel R_3 } = -9.91mV $$

Now we have two expressions for the contributions to the potential at B, which we add to find the actual voltage there:

$$ V_B = V_{B2} + V_{B4} = +9.91mV + (-9.91mV) = 0V $$

That seems like a lot of work for "naught". I'm sure there's an easier way, but we did get to use the superposition theorem, which is nice.

From here on I want to treat Vin and R3 as a signal source with its own output resistance, as a distinct entity, separate from the rest of the circuit. And I want to model point B as a resistance connected to some voltage source, or, in other words, its Thevenin equivalent.

We already know the Thevenin equivalent voltage \$ V_T \$- it's 0V. To calculate the Thevenin resistance, we short-circuit all the voltage sources, (excluding Vin and R3, as I explained):

schematic

simulate this circuit

The Thevenin equivalent series resistance here is the combined resistance to ground, that is, the parallel combination of R1 and R2:

$$ R_T = R_1 \parallel R_2 $$

At last we can build the simplified circuit like this:

schematic

simulate this circuit

Again, since the Thevenin source voltage is 0V, it can be replaced by a short to 0V, as shown on the right. From there it is pretty clear:

$$ V_B = V_{IN} \times \frac{R_1 \parallel R_2}{R_3 + R_1 \parallel R_2} $$

Perhaps the most pertinent point in all this is what happens to \$ V_A \$. Looking back at your schematic, the very place we started, Kirchhoff tells us that:

$$ V_A = V_B + V_{AB} = V_{IN} \times \frac{R_1 \parallel R_2}{R_3 + R_1 \parallel R_2} + 1.4V $$

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  • \$\begingroup\$ Thank you, I'm not sure why I got stuck for so long on this now. Do you have any idea if there are ways to tell that 0 V at a glance, or is it just experience? The book I'm reading talks about it as if it was trivial \$\endgroup\$
    – ale_zec
    Aug 26 at 13:50
  • \$\begingroup\$ I initially wanted to point out that without R3 and Vin, B was already at 0V, kind of by intuition, but that answer would be grossly incomplete - however, since you ask, I did the potential divider arithmetic (which is a lot simpler than superposition) without Vin and R3, and found B was at 0V. From there, putting in Vin=0V and R3 makes absolutely no difference to B! But that would not be the general case (consider a different value of R1, for example, then B would not be at 0V) , so I went the whole nine yards. \$\endgroup\$ Aug 26 at 14:00
  • \$\begingroup\$ Okay that makes sense, since R3 is pretty much connected to gnd it doesn’t force any potential to B, which stays at 0. Thank you again \$\endgroup\$
    – ale_zec
    Aug 26 at 14:07

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