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# How can I make a rectified AC Voltage measurement using a DMM?

### Question :

How can I make a rectified AC Voltage measurement using a DMM?

### Réponses :

When measuring the DC (average value) content of rectified AC, the signal must be applied to the DC input of the DMM.

For the case of half-wave rectified AC, the DC content, or average, is Vpeak / PI.  As an example, a 20.6V RMS transformer is connected to a silicon diode which is connected to a 10k resistor. The resistor is used to cause enough current to flow in the diode to create a voltage drop across the diode and have rectification occur.

The peak voltage is sqrt (2) * 20.6 = 29.1Vpk. The drop across the diode is approximately 0.6V. The peak voltage at the DMM input will be 29.1 – 0.6 = 28.5Vpk. The average, or DC, voltage is 28.5V / PI = 9.08VDC. We set the DMM range to 10V, connect the signal to the DMM and the DMM shows an overflow. Some DMMs are not sophisticated enough to show an overflow. These DMM may show an errant reading. Why the errant reading or overflow?

The peak voltage at the DMM is 28.5V. The DMM is expecting 10VDC maximum. The input amplifier of the DMM becomes saturated causing an overflow on a sophisticated DMM or an errant measurement on a less sophisticated DMM. To alleviate this problem, move the range up from 10V to 100V. The reading on our DMM is now 9.06VDC; very close to the expected value.

For full wave rectified AC, average or DC = 2 * Vpeak / PI.

For our example, we would expect DC to be 2 times the peak voltage, less the 0.6V diode drop all divided by PI. DC = (2 * 29.1 – 0.6) / PI = 18.3VDC. For this case, the peak voltage of 28.5 and the DC equivalent of 18.3 are both on the same range. For this example, we would not have the problem of the peak voltage saturating the input of the DMM when the range is set for the expected DC measurement.

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