While searching for DC-DC boost converters to replace 6F22 9V batteries by 2 NiMH rechargeable AAs for my cheap multimeters, I came across this circuit:
Unfortunately, the circuit was posted without any explanation whatsoever.
As most boost converters, I imagine that this one works by using an oscillator (Q1 and Q2) in conjunction with the inductor (L1) to store and release energy with the right timing and regulates the output with help of D2 (the red LED), Q3 and Q4.
But I'm just guessing. I was hoping that someone could give me a more authoritative answer regarding the circuit workings.
My questions are:
- How does this DC-DC boost converter work?
- Do those transistor configurations have a name? What are they, so I can look them up and learn more about them?
I built the circuit on a breadboard, but I don't think it's working correctly yet. Using some inductor calculators out there, I figured that the inductor value is about 18uH. I replaced it with a 10uH 20% 2.5A PANASONIC (ELC09D100F) inductor (the closest value that I had handy - next up is 100uH). I've also replaced C1 with a 270pF ceramic capacitor.
When powering the circuit with a variable voltage power supply and no load, what I get is some sort of voltage amplification, according to the table below (approximate values from memory):
+----------------+---------+ | Voltage | Current | +----------------+---------+ | input | output | input | +-------+--------+---------+ | 1V | 6V | 50mA | | 2V | 8V | 200mA | | 3V | 10V | 450mA | | 4V | 12V | 600mA | +-------+--------+---------+
Well, that doesn't sound right, for two reasons:
- The output is not regulating 9V;
- The input current seems a bit excessive. It is almost all going through the inductor, which gets really hot to touch.
Can you tell what may be wrong with my breadboard circuit. If that's not enough detail for debugging, that's ok, I'll check it myself once I learn how the circuit works.