Electronic – Why won’t this charge pump supply more than a couple of milliamps

The Thévenin's theorem unfortunately is not applicable since the circuit is not linear at all.

Thèvenin's theorem states that any linear circuit can be converted to a Thévenin equivalent circuit, comprising exactly one voltage source and one resistor. [wikipedia]

To calculate \$V_0\$ you force a zero current output, and solve the circuit, while to calculate \$R_0\$ you need to turn off the sources and measure the resistance seen in the output terminals. The first part can be easily done, but turning \$V\$ off will lead you to solve a completely different circuit, and it won't work.

But where is the problem? Well the fact is that the output resistance of your circuit is not constant and depends on the load resistance.

Try this: throw the schematic in a simulator, and plot the output resistance as a function of the load resistance: if the first is constant you will get a flat line, but this is not the case. But how can you calculate the output resistance?

Let's say you'd like a Thévenin equivalent of this circuit, so that the resulting schematic, including \$R_{load}\$, would be the series of \$V_0\$, \$R_0\$ and \$R_{load}\$. You can measure:

• \$V_0\$
• \$I_{out}\$, entering \$R_{load}\$
• \$V_{out}\$, across \$R_{load}\$

A simple calculation will lead to: $$ R_0 = \frac{V_0-V_{out}}{I_{out}} = \frac{V_0}{I_{out}} - \frac{V_{out}}{I_{out}} = \frac{V_0}{I_{out}} - R_{load} $$ As you can see the equivalent resistance is a function of the load, moreover the output current is a function of the load so the dipendance is not linear. Simulate the circuit and plot the first fraction of the formula, and see what you get. That will not be an usual parametric simulation since you need to wait for the circuit to stop oscillating. I'd choose five to ten values for \$R_{load}\$, make all the time domain simulations and plot the data with excel.

The peak ripple current for C1 is approximately I(out)/D where D= duty cycle. If the Duty cycle is say 50% at your 30 V output then the ripple for C1 is 3.3/0.5 = 6.6 A. As the duty cycle is reduced this gets worse. If the duty cycle was 10% = 0.1 then the current peak is 33 A.

If you then use your ESR value the power dissipated is about 0.4 W, much higher than you previously calculated.

If I look at 160 V capacitors on Mouser (I'm assuming you are using Al Electrolytics) then I see nothing generally available that could sustain the peak currents you need.

I'd suggest you use TI's Webench to work through a design and then look at the selected components. You will notice on many of the designs they use very low ESR capacitors and often have two or even three in parallel. For example they use Panasonic polymer caps often in the designs and they have very high ripple current ratings at very high frequencies.