SSZTBK7 march 2016

In my first, gripping installment of this series, I explained qualitatively how charge pumps work – while skillfully leaving out any numbers. But numbers are what it’s all about in the end, and you can’t design anything properly until you know what components to use. In this installment, I’ll explain the effect that external components have on performance.

Figure 1 shows an unregulated charge pump that operates with a duty cycle
of 0.5 and whose output voltage is nominally twice the input voltage. In practice,
the output voltage of the charge pump is less than 2V_{I} and is given by
Equation 1:

Equation 1.
${\mathrm{V}}_{\mathrm{O}}=2{\mathrm{V}}_{\mathrm{I}}-{\mathrm{I}}_{\mathrm{O}}{\mathrm{R}}_{\mathrm{O}}$

where I_{O} is the output
current and R_{O} is the output resistance.

As long as the flying capacitor is big
enough, the output resistance, R_{O}, depends only on the r_{DS(on)}
of transistors Q1 through Q4. If you do the math (frankly, I don’t recommend it),
you end up with Equation 2
for the output resistance:

Equation 2.
${R}_{O}=\frac{1+\frac{-1}{{e}^{2fRC}}}{fC\left(1-\frac{-1}{{e}^{2fRC}}\right)}$

where f is the switching frequency; C
is the flying capacitance; and R is the resistance of the charge and discharge
paths, assuming they are all equal. (If Q1 and Q2 are the same and Q3 and Q4 are the
same, then R = r_{DS(on)(Q1)} + r_{DS(on)(Q4)}.)

which is math talk for saying that it looks almost like two straight lines, as shown in Figure 2.

The curve in Figure 2 is the key to understanding charge-pump operation. What it tells you is that if C is larger than some critical value, it has almost no effect on the output impedance. This critical value occurs when . Actually, the output impedance is 1.3 times the minimum value when , but in the same way that straight-line approximations are a useful simplification in Bode plots, it’s easier just to make sure that your flying capacitor is bigger than .

A few more mathematical gymnastics
show that for large values of flying capacitance, the peak current in the charge
pump during steady-state operation is equal to twice the output current. During
startup, when the flying capacitor charges up from 0 V to ≈V_{I}, the peak
current can be significantly higher. Well-designed charge pumps gradually increase
the gate-source voltage of the FET switches during startup, which acts as a
soft-start and limits the peak current.

Equation 3 calculates the output voltage ripple:

Equation 3.
${\mathrm{V}}_{\mathrm{O}\left(\mathrm{P}\mathrm{P}\right)}=\frac{{\mathrm{I}}_{\mathrm{O}}}{{\mathrm{f}\mathrm{C}}_{\mathrm{O}}}$

C_{O} is the output
capacitance.

Don’t forget that ceramic capacitors
exhibit a DC-bias effect, which means that the effective capacitance in a circuit is
often a lot less than the nominal value. Double-check the effective capacitance at
V_{I} for the flying capacitance and V_{O} for the output
capacitor.

In a regulated charge pump, an adjustable current source (or sink for an inverting charge pump) typically controls the output voltage, as shown in Figure 3.

Since this circuit only has three
switches, the output impedance tends to when C is very large. However, since the current source,
I_{DRV}, needs a certain minimum voltage across it, V_{DROP}, in
order to operate correctly, the maximum output voltage of the circuit in Figure 3 (assuming C is larger than the critical value) is given by Equation 4:

Equation 4.
${\mathrm{V}}_{\mathrm{O}\mathrm{M}}=2{\mathrm{V}}_{\mathrm{I}}-{\mathrm{V}}_{\mathrm{D}\mathrm{R}\mathrm{O}\mathrm{P}}-{\mathrm{I}}_{\mathrm{O}}{\mathrm{R}}_{\mathrm{O}}$

As long as the output voltage is less than this maximum value, the regulation loop of the charge pump will reduce the output impedance to such a small value that it has almost no effect on the output voltage in practice.

Note that the results are slightly different (although still the same shape) if the charge pump uses diodes for any of the switches or if the duty cycle is not 0.5. I’ll cover those results in the next installment.