AC inverter problem

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AC inverter problem

Post by arlojanis »

I have a 1000 watt Vanner inverter. The voltage reads 85 volts on an analog meter and 110 volts on a true AC meter. Is this normal?
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Re: AC inverter problem

Post by Bill_G »

What do both say when measuring shore power?
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Re: AC inverter problem

Post by Wowbagger »

Your inverter is not a "true sine" device, and likely your meter isn't a true RMS meter.

Many inverters do not create a real honest-to-Fourier sine wave, but what is known as a "modified sine wave". The waveform goes from a negative voltage for a time, to zero for a time, to a positive voltage for a time, back to zero for a time, and back to negative (lather, rinse, repeat). This gives you a peak voltage higher than the RMS voltage, which a simple square wave (positive voltage, negative voltage, back to positive and repeat) will not give you. However, the ratio between peak voltage and RMS voltage (the "crest factor") is not the same as a sine wave - the sine wave will have a higher peak than a modified sine wave for the same RMS power.

Now, your meter: measuring true RMS power is hard. You either a) have to heat a resistor with it and measure the power dissipated (yes, believe it or not there are still meters that do that), or you sample the voltage really fast and "do the math" - compute the square root of the sum of the voltage readings squared divided by the number of samples (that is what RMS stands for: Root Mean Squared). If you have a fancy DMM it may do this.

Since getting true RMS is hard, many meters cheat: they measure peak voltage, then they *assume* the signal is a simple sinusoid and figure RMS is .707 of the peak voltage (put another way: they assume a crest factor of 1.414) That's great if the signal is truly a sine wave, but in the case of a "modified sine" signal, that isn't the case (the crest factor isn't 1.414) and so they LIE to you.

Given what you've said: I'll bet your analog meter is one of the "assume crest factor of 1.414" and is lying to you.
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