# Calculating capacitor charging time in a boost converter

1. Jun 4, 2009

### bitrex

This is close enough to a homework question so I've put it in this section, though maybe it should be posted in the EE section. I've been looking at the equations for boost converters today, and I'm interested in the problem of how long it would take a bank of capacitors to charge through an ideal boost converter, assuming there is not any load on the output. I couldn't find an equation for it - so far I've considered the problem like this: the equation for a boost converter operating in continuous mode is:

$$\frac{Vs}{L} * DT + \frac{Vs-Vo}{L} * (1-D)T = 0$$

The portion that describes the current through the inductor when the boost converter switch is open, allowing the current stored in the inductor to flow into the capacitor is:

$$\Delta I_{L} =\frac{Vs-Vo}{L} * (1-D)T$$

Since the capacitor is in series with the inductor, current will "flow" through the capacitor charging it based on:

$$I_{c} = C\frac{dV}{dT}$$

So I do this: $$\int \frac{dV}{dT} = \int \frac{Vs - Vo}{CL}(1-D)t$$ =

$$V(t) = \frac{1}{2}\frac{Vs - Vo}{CL}(1-D)t^{2}$$

Vs is the source voltage and Vo is $$\frac{Vs}{1-D}$$

Does this look legit?

Last edited: Jun 4, 2009
2. Jun 4, 2009

### Staff: Mentor

Good post, bitrex. But I need to ask about what you mean by "charging up a bank of capacitors". Do you mean you are starting up the boost converter into a bank of capacitors as the load, or do you mean that the boost converter is already running into a light load, and you close a switch to a bank of capacitors as the full load? Either way, I think the transient equations will be different from what you've worked out above.

In the first case, the input storage cap for the boost converter will be charging up at the same time as you are trying to charge up the output load caps. In the second case, considerations about current limit and input storage capacitor size versus the output load capacitor bank come into play....

3. Jun 4, 2009

### bitrex

For the sake of argument for now I'm assuming that I'm running the boost converter from an ideal voltage source with zero output impedance, so that the input source doesn't come into play. The situation on the output is the first situation you describe, the capacitors are the total load for the boost converter. Again, assuming for the sake of argument that the inductor in the supply could handle the current and the input circuit doesn't come into play. In the site I'm looking at (http://services.eng.uts.edu.au/~venkat/pe_html/ch07s3/ch07s3p1.htm [Broken]) there are equations that describe how to calculate the output voltage in continuous mode (from the conservation of energy, the first equation in my post above) and the output ripple with a load connected, etc. but doesn't go into detail about how long it would take the capacitor bank to charge since it's assumed that this will be a design for a continuous power supply that's supplying current to a load, and the bank will be continuously "topped up." I'm more interested in the theory of how to go about calculating the charging time of a large capacitance that is then discharged through a load at once for uh, experiments.

Last edited by a moderator: May 4, 2017
4. Jun 4, 2009

### Staff: Mentor

For charging up a capacitor bank, your boost converter will be in current limit for much of the charge-up. Do you know how to set the cycle-by-cycle current limit on a boost converter?

5. Jun 4, 2009

### bitrex

I haven't read much about boost converter design, but the way to limit the current into the capacitor bank seems straightforward enough - something like a small resistor in series with the output of the converter and a transistor optocoupler across it, and calculate the value for the voltage drop across the resistor at the current limit to be the turn on voltage of the optocoupler. When the voltage and therefore current goes above the limit the optocoupler could be used to inhibit the PWM and let the current in the inductor fall. I think if a feedback system like this were used and if the capacitor bank were in current limit for much of the charge up, then the circuit would behave kind of like a constant current source and the capacitor charging would be essentially linear?

6. Jun 4, 2009

### Staff: Mentor

Pretty close. You are correct about the small value resistor to sense the switch current, but you usually size it for a Vbe drop at the limit current, and put a transistor BE across it so that the transistor turns on at that current limit. Then use the collector current from the transistor to shut off that cycle of the boost. This is called cycle-by-cycle current limit, because it acts quickly and consistently.

So yes, for much of the charging of the output caps, it will just be the current limit giving a relatively constant charge time.