Diode High Voltage Rectifier

Hello I just built a rectifier but I don't understand the voltage that I've got.?

Hello I just built a rectifier but I don't understand the voltage that I've got. I used the following components: a 12 Volts transformer, that when connected to the power grid it's giving me the right value; 1N4004 diodes associated as Graetz bridge; and a electrolytic capacitor of 25V and 3300uF. When I'm reading the voltage on the capacitor I'm getting 16.8 Volts. How come that I such high value?

Thank you very much.

Ok - so let me get this straight. Did you measure the RMS voltage of the output of your transformer, and get a reading of 12V? And then, when you measured the output of your rectifier, you got 16.8V? If that's the case, you have nothing to worry about, and your circuit is working perfectly

Essentially, what happened is that the RMS voltage (the typical number your multimeter will give you if you set it on AC voltage mode, and what you probably measured) is NOT equal to the maximum voltage coming out of your transformer.

Now, if we measure the instantaneous voltage coming out of your transformer as a function of time, we will see something that looks like:

V(t) = V_amplitude * sin ( w t)
where:
t = time
V(t) is voltage as a function of time
V_amplitude = voltage amplitude of your AC signal (a constant)
w = angular frequency = (frequency) * 2 * pi = (60 Hz) * 2 * pi ~ 377 radians/sec

We can clearly see that the maximum and minimum voltages reached are + V_amplitude and -V_amplitude volts (since -1 <= sin(x) <= +1)

The RMS (root-mean-square) voltage is a kind of "average" voltage for an AC signal. (There is a particular algorithm to calculate it - see http://en.wikipedia.org/wiki/Root_mean_square for a more detailed explanation, and some useful properties of the RMS voltage. For example, the RMS voltage and RMS current can be used to calculate the AVERAGE power of a signal). For an AC signal, the RMS voltage will be less than the peak voltage of your signal. The RMS voltage and the voltage amplitude of a SINUSOIDAL AC signal (for example, mains power) are related by:

V_amplitude = sqrt(2) * V_rms
where V_amplitude = voltage amplitude of your signal
V_rms = RMS voltage of your signal

Your AC signal from the "12V transformer" has an RMS (root-mean-square) voltage amplitude of 12V. However, the ACTUAL amplitude of your signal is:

V_amplitude = sqrt(2) * V_rms = sqrt(2) * 12 = 17.0V

Thus, your AC signal coming from your transformer is oscillating between a maximum of +17.0V and a minimum of -17.0V.

Now.. onto the rectifier. You're using a full-bridge (Graetz bridge) rectifier, which makes your signal, currently oscillating between negative and positive voltages, a completely positive voltage signal. (Here's a good visualization: http://www.juliantrubin.com/encyclopedia/electronics/rectifier_files/350px-Rectified_waves.png). Other than diode voltage drops, your signal is now oscillating between +0V and +17.0V. The electrolytic capacitor then "smooths" out the signal, turning it essentially DC. Every cycle of your AC signal, the capacitor will be charged up to the maximum voltage. In the case of zero load current, your output voltage will actually be the maximum input voltage, +17.0V, which is more or less what you measured. If you have a heavier load current, the capacitor will discharge between cycles, giving you a signal with "ripple" (i.e. it oscillates slightly), and a lower average voltage. Ripple can be minimized by using larger capacitor values; however, there are some drawbacks (for example, the PEAK current in your diodes and transformer will be higher).

One other thing to note: the transformer windings have a significant resistance, and as you increase the load current, the voltage coming out of your transformer will also decrease. Also, transformers are usually rated according to their output RMS voltage at MAXIMUM current rating, so the RMS voltage when you have no load may be significantly higher than the rated output voltage! (maybe even 1.3 to 1.4 times higher)

So basically, to sum it up: your 12V transformer is outputting a signal with MAXIMUM voltages that are higher than 12V, the diode bridge is turning your signal (originally going between negative and postive voltages) completely positive, and your capacitor is being charged up to the maximum voltage of your signal. This is completely normal and nothing to worry about (it actually tells you it IS working properly...)

Hope it helps!

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