# How does this 4-transistor DC-DC 1.5V to 9V boost converter work?

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:

The circuit was posted by Popescu Marian at ElectroSchematics.com at the link below:

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:

1. How does this DC-DC boost converter work?
2. 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:

1. The output is not regulating 9V;
2. 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.

• For not understanding the circuit, you do a good job explaining it. How about C1, any idea how that component is utilized here? Considering your table, how does a slight load influence the output voltage? BTW: It does seem pretty power hungry for a no-load boost converter. Commented Sep 27, 2015 at 14:24
• I'd connect a fast 'scope up to Q2 base and emitter, and see what is actually going on there. It could be resonating in the mid to upper MHz range, and wasting large amounts of power. A simple ferrite bead on a few select leads may straighten it right out. Also consider that breadboards impose parasitic capacitance and inductance, which may aggravate the issue. Commented Sep 27, 2015 at 14:41
• Might I suggest getting a rechargeable 9v battery for practicality instead? Commented Sep 27, 2015 at 15:05
• @user2813274 - yeah, it would be much more practical, but it would spoil all the fun :D - In any case, thanks for pointing that out. Commented Sep 27, 2015 at 21:47

The oscillation is supposed to be by Q1/Q2 and C1, with off time controlled by $R1 \cdot C1 \approx~ 10us$. On time in the original design could be limited by inductor saturation more than anything else- the current increases linearly until the inductor saturates, the current spikes up and Q2 comes out of saturation, causing the pair to more-or-less snap off, or by the current supplied by C1 tailing off, causing Q1 collector current/Q2 base current to to drop and again Q2 comes out of saturation, or some combination of the two.