# AC to DC; how does the magnitude of current/voltage change?

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I know an AC current can be rectified to a DC current with a diode bridge, but does the magnitude of the voltage and/or current change? If so, how.

For example, if one had an AC current at 2 A and 12 V, and one converted it to DC, what would the current and voltage of the new DC current be? Btw, let's assume that the process is 100% efficient.

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There is always a small voltage drop across the diodes.

Also, depending on the frequency and size of the capacitor, there will be small "bumps" in the DC output.

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AC voltage is measured in something known as RMS volts, which stands for Root Mean Square. AC current (I'll assume you're using wall current) is a sine wave, and the average value of the voltage is the voltage given. So 12V AC means that the average positive voltage is 12V.

When you rectify AC to DC, your voltage changes from the average to the peak voltage. So instead of the average of the sine wave, you need the peak. To calculate this, you need to multiply the voltage by the square root of 2 (~1.41). So 12V AC will become 17V DC. To get the amps available, assume energy is conserved and use the watts in to calculate the amps out. Basically 12VACx2A = 17VDCx?A. So you get roughly 1.41A.

You will need to use capacitors to smooth out the pulsed DC (the rectifier basically takes the absolute value of the voltage function), in order to get a decently constant voltage. This will also affect your peak amps, but not your average amps.

Remember that this is ideal. In reality you need to compensate for:

Diode voltage drop as Neon said.

Misc supply losses (ex. resistive losses in a transformer).

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