1. Home
  2. Electronics Design
  3. Active power factor correction

Active power factor correction

Hi

I am simulating an active power factor correction switching regulator. I am having a bit of trouble getting a good algorithm going for it though.

Here is my circuit. The input is rectified AC, 60Hz, ~160V vin = 160*|sin(2*pi*60*t)|

|------[inductor]--[L_ESR]---|---[diode->|]---|-------| | | | | | | | | [vin] [contro]-[fet] [cap] [load] | | | | | | [C_ESR] | | | | | |--------------------|-------|----------------|-------| | [gnd]

This is the code I am using to simulate the circuit: (StepSize is 1E-9)

if (SwitchOn) { //left half inductorCurrent += (vin - inductorESRVoltage) * StepSize / inductance;

inductorESRVoltage = inductorCurrent * inductorESR;

//right half capacitorVoltage += loadCurrent * StepSize / capacitance; vout = capacitorVoltage + loadCurrent * capacitorESR; loadCurrent = vout / loadResistance; capacitorCurrent = loadCurrent; } else { inductorCurrent += (vin - inductorESRVoltage) * StepSize / inductance; if (inductorCurrent < 0) { inductorCurrent = 0;//diode } inductorESRVoltage = inductorCurrent * inductorESR + vout; capacitorVoltage += (inductorCurrent - loadCurrent) * StepSize /

capacitance; loadCurrent = vout / loadResistance; vout = capacitorVoltage + loadCurrent * capacitorESR; capacitorCurrent = inductorCurrent - loadCurrent; }

The PWM controlling the proportion of time on/off for the Switch in the above code is here:

Microloop()//called at 100kHz { duty = (desiredVout - vout)*(k * vin - inductorCurrent); //the (desiredVout - vout) sets the error for the output voltage //(k * vin - inductorCurrent) sets an error current for the input, to //make sure the input current is proportional to input voltage. I have //been using a k of 0.25 if (duty >= 95) { duty = 95; } else if (duty < 0) { duty = 0; } }

The problem with my simulated controller as it stands is that the inductor current has a lot of ripple in it, but the magnitude of the inductor ripple is proportional to the input voltage. The output voltage regulates well though. Do I need to make changes to my circuit or algorithm to make this work better?

Scott

Reply to
Scott Ronald
Loading thread data ...

Note the lack of an input side EMI filter and that the input side bridge circuit is modeled.

This is basically right as a straight line modeling of the switching. You can do a bit better if you do one of the summing actions twice per loop and divide the amount by two. This helps if you have one short time constant and one longer one.

inductance;

A lossless diode is hard to find in real life.

I haven't looked carefully enough but the above looks incorrect to me. The servo loop of your PWM circuit needs to be much slower than the PWM action and a lot faster than the mains frequency. You seem to have it at the PWM frequency.

The input side ripple at the switching frequency is normal. You can't get rid of it in the topology you have. The frequency content below that should be just the mains frequency.

Reply to
MooseFET

[snip]

What did you think the inductor current should look like?

Actually, your controller is working exactly as it should.

The only refinement would be to add a smallish capacitor between the rectifier and the inductor, to keep the high-frequency ripple current from being conducted back out onto the input line. Select it for high impedance at 60/120 Hz, and low impedance at the switching frequency.

Actually, there's one other refinement. You'll want to low-pass filter the (desiredVout - vout) term in your duty cycle calculation. A cutoff frequency in the range of 10 to 20 Hz is typical. You will in fact want to see some 120-Hz ripple on your output capacitor -- otherwise, you'll be distorting the shape of the average line current waveform too much relative to the sinewave voltage and thereby deviating from "perfect" power factor correction. If the load can't tolerate this ripple, add a second regulator to get rid of it. Actually, in most applications the load on the PFC circuit is indeed one or more buck regulators, so this tends not to be an issue.

-- Dave Tweed

Reply to
David Tweed

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.