How can I measure the output of a voltage booster circuit?

Question

I made this circuit, which was described on-line as a "transformerless Joule Thief." I was able to light four LEDs of varying forward-voltage in series, but I couldn't figure out how to use a multimeter to measure the the total output voltage. What components could I add to this circuit to give an accurate output voltage measurement?

Link to my Falstad circuit

What is the oscillation frequency? Probably it is above what is measurable with a multimeter due to its frequency response. — devnull, May 13 '22 at 14:33
A multimeter? Circuit needs to be loaded by the LEDs for you to get a meaningful result. — winny, May 13 '22 at 14:35
I'd advise using a resistor, perhaps 1k? but depending on the LED and inductor also I suppose, going between the PNP collector and the NPN base. Better efficiency for one thing. And it relaxes tension between the PNP collector and NPN base, avoiding unnecessarily excessive NPN base current, without upsetting other goals. — jonk, May 13 '22 at 18:04
@jonk Efficiency went from 81% to 94% with the simulation below and the power at the LEDs doubled (2.1 x). No change in the oscillation frequency. It obviously depends on many factors, but that is definitely an important resistor... — devnull, May 13 '22 at 18:26
@devnull It makes me cringe to see a PNP's emitter to power+, NPN emitter to power-, and one of their collectors tied directly to the other's base. — jonk, May 13 '22 at 18:36
@MicroservicesOnDDD If that's a serious question and not a joke, you may need to ask it. I'd have a hard time writing out all the reasons in comments. — jonk, Jul 07 '22 at 19:35
@jonk -- I'm finding constructs for power are very different from constructs for signal. I'll try to come up with an example I can post so you can show me what you mean. I'm sure I still have a lot to learn. — MicroservicesOnDDD, Jul 08 '22 at 00:36
@jonk -- I like this Joule Thief because it seems to be fairly efficient and it often achieves no negative current at all on the base-emitter junction of both bipolars. — MicroservicesOnDDD, Jul 08 '22 at 00:54

JRE · Answer 1 · 2022-05-13T15:15:15.587

An oscilloscope would be your best bet.

If you don't have an oscilloscope, connect a capacitor in parallel with the LEDs. From the junction of the LED and the inductor to ground. That will get you a (somewhat) more reasonable result out of the voltmeter.

100nF to 1µF would be a place to start.

For some idea of what you're dealing with, here's the output voltage of a (somewhat similar and similarly crude) voltage booster I built a couple of years ago:

The peaks are when the LED is lit up. The lowest horizontal part of the trace is when the inductor is charging. The middle level is when the inductor doesn't have enough energy left to raise the voltage to the minimum needed for the LED bulb I was using - that's the 1.5V of the D cell I was using for power. (Yes, LED bulb. That was a nominal 12V LED light bulb intended for use in a track light.)

score 3 · Accepted Answer · answered May 13 '22 at 15:07

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how to use a multimeter to measure the the total output voltage

If by "total output voltage" you meant the peak value of the pulses, you could do something like this:

The diode must be a fast one (not the regular types used in 50/60Hz rectifiers). The DC voltage across the capacitor will be somewhat 400mV below the peak voltage. More accuracy is possible if you are willing to use an active circuit (opamp) powered by another source.

answered May 13 '22 at 15:07

devnull

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Adding the diode and capacitor as-shown and connecting the multimeter across the capacitor in a closed circuit measured about 11V with 4 LEDs in series. With 3 LEDs it was about 8.5V. With 2 LEDs it was about 6V. With 1 LED it was about 3.25V. After charging the LED with no LED or 1, 2, 3 or 4 LEDs in series, when I opened the circuit the capacitor voltage always measured about 11V. – Annette Gates May 14 '22 at 14:02
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Used a 1N4001 diode. – Annette Gates May 14 '22 at 14:09
This is a diode for a line rectifier, with a very slow recovery time. I wouldn't expect the peak detector to work well at 11kHz with this diode. Regarding the first comment, I could not understand the difference you meant after "after charging the LED". – devnull May 14 '22 at 14:12
Sorry, I meant "charging the capacitor", so after closing the circuit momentarily, then opening it the multimeter showed 11V – Annette Gates May 14 '22 at 14:16
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I got close to the same measurements with a 1N4148 diode. – Annette Gates May 14 '22 at 14:16
Without load, it is expected that the output rises a lot. Depending on the inductance and how fast the NPN turns off, it could reach tens of Volts, and even damage the transistor. The capacitor smoothed this peak. The datasheet I have here doesn't even mention the 1n4001 recovery time (I've found different figures online, all above a few us), which is < 10ns for the 1n4148. – devnull May 14 '22 at 14:22

hacktastical · Answer 3 · 2022-05-13T18:12:26.563

The LED will clamp the boost output voltage to its forward voltage. That is, if your LED is a blue or white type, the max voltage will be clamped to about 3 - 3.5V, peak.

The challenge with using a meter is that they usually measure average, ‘fake RMS’ (scaled average) or ‘true RMS’ voltage. These measurements are influenced by the measured pulse shape. Low duty-cycle pulses measure low average or RMS voltages.

In this circuit the LED isn’t conducting all the time. When the transistor is on the voltage will be about 200-500mV (Vce(sat)), which will drag down the average a lot. This will give the impression that the LED isn’t getting hardly any voltage. That’s wrong - it is, and probably a lot of peak current too. It’s just at a low duty cycle.

If your meter has a peak setting that might be useful. But to really know what the circuit is doing you need an oscilloscope and a current probe so you can observe dynamic behavior and do accurate analysis.

How can I measure the output of a voltage booster circuit?

3 Answers3