Proving the formula for capacitance multiplier

Question

I'm looking into the mathematical justification of how a capacitance multiplier works. Here is the circuit.

I have to prove that resulted grounded (multiplied) capacitance is (R2 * C)/R1 here by means of mathematics. It was checked in LTspice in that question:

Why does two-BJT transistor-based capacitance multiplier show bad performance

It turned out that a normal capacitor discharged over 0.22 sec whereas the equivalent of this circuit with R2 being 100k and R1 being 10 Ohms did so over 2200 sec

^{simulate this circuit – Schematic created using CircuitLab}

So I do nodal and loop analysis for this circuit.

Nodal analysis based on Kirchhoff's current law for Node1:

\$ C\frac{d(V_{a}-0V)}{dt} + \frac{V_{a}-V_{b}}{R_{2}} + 0A = 0 \$

\$ C\frac{dV_{a}}{dt} + \frac{V_{a}}{R_{2}} - \frac{V_{b}}{R_{2}} = 0 \$

Nodal analysis based on Kirchhoff's current law for Node2:

\$ V_{o} = V_{a} \$ (1)

This is because of op-amp "golden rules":

No current flows into the +/− inputs of the op amp. In a circuit with negative feedback, the output of the op amp will try to adjust its output so that the voltage difference between the + and − inputs is zero (V+ = V−).

Loop analysis for pictured loop based on Kirchhoff's voltage law:

\$ V_{b} - R_{1}I_{1} - R_{2}I_{2} = 0 \$ (2)

Voltage drop on R2:

\$ V_{b} - V_{a} = R_{2}I_{2} \$ (3)

\$ V_{b} - V_{a} - R_{2}I_{2} = 0 \$ (4)

\$ V_{b} - V_{o} - R_{2}I_{2} = 0 \$ (5)

From (2) and (4)

\$ V_{a} = R_{1}I_{1} \$ (6)

Voltage drop on R1:

\$ V_{b} - V_{o} = R_{1}I_{1} \$ (7)

From (1)

\$ V_{b} - V_{a} = R_{1}I_{1} \$ (8)

From (3) and (8)

\$ R_{2}I_{2} = R_{1}I_{1} \$ (9)

From (6) and (9)

\$ V_{a} = R_{2}I_{2} \$ (10)

From (4) and (9)

\$ V_{b} - V_{a} - R_{1}I_{1} = 0 \$ (11)

\$ V_{b} = V_{a} + R_{1}I_{1} \$ (12)

Getting back to capacitance:

\$ C\frac{dV_{a}}{dt} + \frac{V_{a}}{R_{2}} - \frac{V_{a} + R_{1}I_{1}}{R_{2}} = 0 \$

\$ C\frac{dV_{a}}{dt} = \frac{R_{1}I_{1}}{R_{2}} \$

\$ \frac{dV_{a}}{dt} = \frac{R_{1}I_{1}}{CR_{2}} \$

\$ I_{1} = \frac{q_{1}}{t} \$

\$ C = \frac{q_{2}}{V_{a}} \$

Now I'm stuck here.

What has to be done further? Regards.

Answer 1

Your diagram was too complicated, so I've redrawn it:

^{simulate this circuit – Schematic created using CircuitLab}

Note that this now highlights the fact that \$V_i\$ is a voltage source. And rather than "busing around" that source I've separated it out. I also just tossed down a couple of node labels.

Here, we can say that there is a nice voltage divider in \$R_2\$ and \$C_1\$:

$$V_a=V_i\cdot \frac{Z_{C_1}}{R_2+Z_{C_1}}=V_i\cdot \frac{\frac1{s\,C_1}}{R_2+\frac1{s\,C_1}}=V_i\cdot \frac1{1+R_2\,C_1\,s}$$

But on the ideal opamp assumption for this follower arrangement, we can also say that \$V_b=V_a\$.

Also, we know \$I_{R_1}=\frac{V_i-V_b}{R_1}\$ and \$I_{R_2}=\frac{V_i-V_a}{R_2}\$. (Of course, we can replace \$V_b\$ with \$V_a\$ for \$I_{R_1}\$.)

To work out the impedance that \$V_i\$ sees, we just need to calculate:

$$\begin{align*} Z&=\frac{V_i}{I_{R_1}+I_{R_2}}\\\\ &=\frac{V_i}{\frac{V_i-V_a}{R_1}+\frac{V_i-V_a}{R_2}}\\\\ &=\frac{V_i}{\frac{V_i-V_i\cdot \frac1{1+R_2\,C_1\,s}}{R_1}+\frac{V_i-V_i\cdot \frac1{1+R_2\,C_1\,s}}{R_2}}\\\\ &=\frac{1}{\frac{1-\frac1{1+R_2\,C_1\,s}}{R_1}+\frac{1-\frac1{1+R_2\,C_1\,s}}{R_2}}\\\\ &=\frac{R_1\left(1+R_2\,C_1\,s\right)}{\left(R_1+R_2\right)\,C_1\,s}\\\\ &=\frac{1+R_2\,C_1\,s}{\left(1+\frac{R_2}{R_1}\right)\,C_1\,s}\\\\ &=\frac{1}{\left(1+\frac{R_2}{R_1}\right)\,C_1\,s}+\frac{R_2}{1+\frac{R_2}{R_1}}\\\\ &=\frac{1}{\left(1+\frac{R_2}{R_1}\right)\,C_1\,s}+\left(R_1\mid\mid R_2\right) \end{align*}$$

At this point, it's pretty clear that the unitless \$1+\frac{R_2}{R_1}\$ multiplies \$C_1\$ and that the first term is the impedance of this multiplied capacitance. But because of the sum, it's in series with the second term which is the parallel resistance of \$R_1\$ and \$R_2\$. That makes sense as both resistors have \$V_i\$ on one side and \$V_a\$ (virtual or otherwise) on the other side.

So the two resistors really are effectively in parallel with each other, but in series with the multiplied capacitance.

None of this qualifies as an "intuitive approach." (Unless one considers pursuing a chain of algebraic manipulation as intuitive.) But you said you were looking for "the mathematical justification" and that you "have to prove that [...] by means of mathematics." So that's the direction I went here.

Answer 2

To get an intuitive feeling look at it like this:

When a V is applied at Vb (NODE3), some current will flow through R2 into the capacitor. The Opamp's '+' input will be at some voltage. The opamp will drive its output so that NODE2 is at the same voltage.

Now, the voltages across R2 and R1 are equal (left side == input V, right side == opamp '+' or '-'). Therefore the current that flows in R1 is R2's current times R2/R1.

So, of the current that flows into the input, just a fraction R1/(R1+R2) flows into the capacitor; the rest flows through R1 into the opamp's output. Thus the voltage build-up across the capacitor is a fraction R1/(R1+R2) that it otherwise would be. This is equivalent to replacing it with a component (1+R2/R1) times larger.