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The board on an old battery charger went bad, but the transformer was good. So I've got about a 12.3VDC output when I connect the transformer to a bridge rectifier. However, if I decide to wire in a capacitor, the voltage jumps to over 20v. When I hook up a 12v load (light bulb) it goes back down to around 13v.

  1. Why is this voltage increase happening with the cap?

  2. If this 20v isn't my "actual" voltage, what is? Is it the voltage I see with a load attached? If load decreases the voltage in the circut, how do I know I'm not sending too big a voltage to my load?

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    \$\begingroup\$ The true voltage is whatever it is at the time. It doesn't have to stay the same when you change the load any more than you have to be able to lift a very heavy block as high as a lighter block. Output impedance and saturation, in particular that are responsible for this in a transformer. \$\endgroup\$
    – DKNguyen
    Commented Dec 13, 2019 at 17:18
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    \$\begingroup\$ Is your 12V light bulb actually an automotive bulb? If so, then it is actually designed to operate at higher voltages than 12V. When your engine is running and the battery is charging the voltage will likely be in the 13.5V to 14.5V range. \$\endgroup\$
    – Elliot Alderson
    Commented Dec 13, 2019 at 17:34

1 Answer 1

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enter image description here

Figure 1. Bridge rectifier and smoothing capacitor. Image uncredited on EE.SE.

Without the smoothing capacitor you get the dotted-line full-wave rectified waveform. You are reading the average value of this, 12.3 V.

The smoothing capacitor, when added in, is charged to the peak value of the voltage and maintains it during the dips in the rectified voltage. The average voltage will be close to the peak voltage, 20 V.

When I hook up a 12v load (light bulb) it goes back down to around 13v.

The capacitor's value isn't high enough to maintain the voltage at 20 V with such a large load as your lamp. It does hold it up 0.7 V higher than without the capacitor.

Why is this voltage increase happening with the cap?

Explained above.

If load decreases the voltage in the circut, how do I know I'm not sending too big a voltage to my lamp?

You use the meter to check this. With 13 V you are running a little "hot".

Note that incandescent lamps are tolerant of peaking voltages due to the thermal characteristics of the filament. Electronic devices may be destroyed if the peak voltage exceeds the device's rated voltage.

Note: 'V' for volt, 'A' for ampere, etc. SI units named after a person have their symbols capitalised but are lowercase when spelled out.

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  • \$\begingroup\$ Thanks Transistor. I really appreciate it. If it wasn't for guys like you, I wouldn't be able to keep learning this stuff and getting better at electronics. \$\endgroup\$
    – Jared Cravens
    Commented Dec 13, 2019 at 19:39

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