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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

schematic

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.

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  • \$\begingroup\$ Using s-notation and nodal, I get the multiplied capacitance AND a series resistance that is \$R_1\mid\mid R_2\$. \$V_a=V_o=V_i\frac{Z_C}{R_2+Z_C}\$, \$I_{V_i}=\frac{V_i-V_a}{R_1}+\frac{V_i-V_a}{R_2}\$ and \$Z_{in}=\frac{V_i}{I_{V_i}}=\frac{1+R_1\,C_1\,s}{\left(1+\frac{R_2}{R_1}\right)\,C_1\,s}\$. Set \$K=1+\frac{R_2}{R_1}\$ and then you get \$Z_{in}=\frac{1}{K\,C_1\,s}+\left(R_1\mid\mid R_2\right)\$. So that's the proof. But it includes a series resistance, too. \$\endgroup\$
    – jonk
    Apr 6, 2021 at 4:38
  • \$\begingroup\$ By the way, I think the Wiki page on the opamp capacitance multiplier gets the series resistance approximation wrong. (Also note their resistor numbering is different than yours.) \$\endgroup\$
    – jonk
    Apr 6, 2021 at 4:51
  • \$\begingroup\$ \$ V_a=V_o=V_i\frac{Z_C}{R_2+Z_C} \$ Could you make it more verbose, please? I don't understand how you eliminated R1. \$\endgroup\$
    – bimjhi
    Apr 6, 2021 at 7:16
  • \$\begingroup\$ \$V_i\$ is a voltage source. The divider to ground goes through a resistor and a cap. You can work out what \$V_a\$ is from that, as shown, ignoring \$R_1\$ because it doesn't impact that divider. \$\endgroup\$
    – jonk
    Apr 6, 2021 at 7:23
  • \$\begingroup\$ Ok, I understood the part with the voltage divider. Then I googled for the explanation of \$ Z_{in}=\frac{V_i}{I_{V_i}}=\frac{1+R_1\,C_1\,s}{\left(1+\frac{R_2}{R_1}\right)\,C_1\,s} \$ but there wasn't a detailed solution. Could you clarify this part too, please? Thanks. \$\endgroup\$
    – bimjhi
    Apr 6, 2021 at 8:27

2 Answers 2

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

schematic

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.

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    \$\begingroup\$ Got the same result with a pen and paper. Thank you so much! I hope sometime there will be a way to get by without pens and paper though. \$\endgroup\$
    – bimjhi
    Apr 7, 2021 at 22:30
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    \$\begingroup\$ @bimjhi I understand that sentiment. In any case, intuition is developed by doing lots and lots of these kinds of problems. Gradually, the basic topologies sink in and their priorities (what's more important and what's less so) also sinks in. Then you have your intuition. The concept of density didn't just arrive out of thin air to someone. (Well, it did but it didn't.) The idea of mass had to arrive and then the idea of the ratio. Before that, many just thought that sharp things "cut water" and sunk and blunt things didn't and therefore floated. Physical instinct isn't natural. \$\endgroup\$
    – jonk
    Apr 8, 2021 at 1:53
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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.

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