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I would like to measure the power of a very unusual ac waveform. Please say if this could work...

  1. Construct a black pvc tube with a very sensitive light to voltage converter TSL257 in one end, and a tiny lamp in the other end, seal both ends.

  2. Drive the lamp with the waveform to be measured and record the lamp intensity voltage from the TSL257.

  3. Drive the lamp again with a battery to the same intensity and calculate the power ( Volts @ Load x Amps )

Would this method be similar to the old "Hotwire" ammeters?

Thanks.

Transistor
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William Mingin
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    What waveform? Voltage? Current? Because power consists of both a current and voltage waveform, but tapping into the waveform to measure it distorts the waveform and thus the thing you are trying to measure. So you don't want to tap into it too much, and a lighbulb might be too much. Not only that, it is also unclear how you plan to simultaneously drive the lamp" with the voltage AND current waveform you are trying to measure. You can do one, you can do the other, but you can't do both. For this reason I would say it won't work. – DKNguyen Jan 06 '21 at 16:29
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    You can't throw something together with opamps? Opamps are cheap. You also do not say what voltages and currents are involved. – DKNguyen Jan 06 '21 at 16:30
  • Why do you want to do it this way, rather than using conventional methods? – Bruce Abbott Jan 06 '21 at 17:05
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    congratulation, you've re-invented the optocoupler, now you only need to calibrate it, as it's not really clear how your brightness relates to your power (it probably doesn't like you think it does) and explain why this immensely complex solution is better than measuring the voltage and current directly. – Marcus Müller Jan 06 '21 at 17:32
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    Yes, a lamp -to-photodetector can give r.m.s. equivalence for "weird" waveforms...with a few caveats. Fundamental frequency must be high enough that the lamp's temperature doesn't change much from cycle-to-cycle. And frequency must be low enough that parasitic reactance of the lamp is tiny compared to filament resistance. I'm assuming you use a filament-type lamp...a LED won't do. – glen_geek Jan 06 '21 at 17:58
  • The Voltage is small, 500 mVpp Current a few mA pp, Frequency about 800 kHz. My digital multimeter can't handle the crazy waveform ( non-sinusoidal ) or the frequency. The lamp I'm using is a grain of wheat incandescent. The sensor seems to be able to detect illumination from the bulb even before I can see the glow. – William Mingin Jan 07 '21 at 18:57
  • Thank you all for your answers. I constructed the device and it works quite well. I measured the power from a 60 hz sine wave on my DMM and compared it to the results obtained with the incandescent lamp/detector and the results matched almost exactly. I can measure unusual waveforms down to a few milliwatts...good enough. Thanks again. – William Mingin Jan 22 '21 at 16:39

1 Answers1

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A significant difference to a hot-wire ammeter is that the resistance of the bulb varies with the temperature, whereas the material of an ammeter is chosen to have a very low temperature coefficient of resistance.

To first order, this won't matter, as you are driving the lamp to the same temperature in the calibration phase. However, not knowing the resistance, and having it vary from run to run, could cause you difficulties in driving the lamp consistently.

Detectors of this type have been used for RF measurements, where the working temperature of the lamp is chosen to make its resistance 50 Ω. DC is applied to heat the lamp, and the resistance is servoed to 50 Ω rather than controlling the light output or temperature. When RF is applied, the increased power causes a drop in the DC required to maintain the same temperature.

For very unusual waveforms, say with crest factors exceeding 5:1, this might be better than using commercial RMS to DC converters.

Neil_UK
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