So I found this 9V 1A AC power supply laying around and decided to make a DC source. I rectified and added a capacitor to make things a bit even. After adding a capacitor my voltage boosted from 9V to 14V. Can somebody explain why this happened for me? (maybe this has something to do with frequency?) According to theoretical graph I should get aprox the same voltage even after adding the capacitor. And voltage varies depending on how much capacity the capacitor has (aprox. from 12-16V). Maybe there's some sort of equation to determine the actual output voltage depending on caps? It would be great to get 12V out of this thing.

P.S.: Personally I find this boost weird because this circuit doesn't have any switches and coils to boost the voltage, so I need an explanation, thank you in advance!

a schematic


3 Answers 3


The rectified AC waveform catches the peaks. The input 9VAC is RMS (Root-Mean-Square average) equivalent -- the actual amplitude of the sinewave is about 40% higher than the RMS average (square root of 2 is 1.414). So on your picture the 9V equivalent is about 70% of the way between 0V and the peaks.

The numbers don't work out exactly to the ideal square-root-of-two crest factor, because there is some voltage drop across the two diodes that are on, and also because there is some variation in the line voltage.

The reason RMS is used to describe AC voltages, is because the amount of power (heat) delivered to the resistive load is the same as it would be for a 9V DC 1A source.

Edit: explaining the observed difference in load voltage measurement for different load capacitor condition, explaining why DMM gives wrong measurement for full-wave rectified waveform...

Voltage is not actually being boosted in this circuit. When the capacitor is removed, the full-wave rectified signal doesn't sustain the peak voltages. As Ignacio Vazquez-Abrams mentions, the DMM may not be measuring the waveform correctly, especially in the case where there was no capacitor -- assuming you measured with the DMM's DC Voltage setting, without the capacitor the full-wave rectified waveform would be confusing the measurement. The 9V DC measurement reported by the DMM matches the rated 9V AC RMS equivalent, so maybe the DMM was somehow measuring the RMS value. Then when you added a capacitor, the waveform peaks were sustained long enough for the DMM to start measuring accurately. Sadly, it is possible for measurement equipment to "lie" to us under some conditions. Happens to the best of us sometimes.

The DMM is just a electronic machine, it's not a magic box that always gives the right voltage measurement. Most DMM's use a measurement technique called dual-slope integration, where a capacitor is first rapidly charged up to the voltage being sampled, and then the sampling capacitor is discharged through a constant-current source. The DMM counts how long it takes to discharge the capacitor back to zero. The value of that counter is what the DMM displays. Calibration depends on the current source, the comparator offset voltage, and the quality of the sampling capacitor. This technique is cheap to implement and it works great, as long as the input signal doesn't change very quickly. But when connected to that full-wave rectified signal, the sampling capacitor doesn't stay at the peak voltage. It's indeterminate where the sample interval begins and ends, so it's hard to know exactly how many counts the DMM might report. So if the C is omitted, then it's not really a DC circuit, so the DMM DC measurement isn't valid.

You also asked about using different value of C. The bridge rectifier is not regulated, its output voltage can vary with different load impedance. Changing the load capacitance C affects the capacitor's reactance Xc, which also affects load impedance.

$$ Xc = \frac{1}{2 * pi * frequency * C} $$

A lower load impedance, just like load resistance, will draw more current at a given voltage. But unlike resistance, the current and voltage waveforms may be out of phase. So it's possible to have voltage across a capacitor even with zero current, and it's possible to have current through an inductor even with zero voltage (under some conditions).

Putting a reactance in parallel with a resistance is a bit more complicated than putting resistors in parallel, because the voltage and current waveforms are in-phase for the resistor but are 90 degrees out-of-phase for the capacitor. In AC circuit analysis, we use complex numbers and phasor notation (yes, that really is a thing) to model these AC circuit elements. If you think of impedance as a vector, with the length of the vector acting similar to resistance in Ohm's law, and reactance acting at right angles to resistance, then putting the resistor and capacitor in parallel gives the total load impedance Z. Although it's possible to go deeper into the maths, there's another important point worth mentioning:

This circuit isn't regulated. If you want to get 12V DC output, you can't just select a capacitor value and expect this to always give 12V DC output regardless of how much load current is drawn. This circuit is a good building block to start with, but the full-wave rectified output voltage will vary with changes in the utility line-in voltage as well as the load current. If you really want it to be regulated, add a regulator circuit such as 78M05 (or 78M12 if you really need 12V). In that case you'll need the full-wave bridge to provide a bit more then 12V so that the regulator has some headroom to work with (but not too much, because the linear regulator works by wasting the unwanted energy.)

AC circuits theory can be kind of mind-bending at first, because there are all these surprising mathematical things like imaginary numbers and Euler's law, that turn out to actually work in real life. The comment about how the capacitor "evens out the peaks" is... kind of true... but it's a major oversimplification. As you've discovered, a qualitative statement like that doesn't help you determine how much capacitance you need to achieve your design goal of making a 12V DC power supply.

I'm not going to be able to fully explain AC circuits theory here, but here are at least some interesting breadcrumbs:

See https://en.wikipedia.org/wiki/Electrical_reactance

See https://en.wikipedia.org/wiki/Electrical_impedance

See https://en.wikipedia.org/wiki/Phasor

  • \$\begingroup\$ and why does voltage peak is different with different capacitance? Is it because it evens out these peaks better? give me some reference and I'll google that, because right now I don't know where to start. Till now I just worked with DC circuits. \$\endgroup\$
    – Melectro
    Commented Jun 15, 2015 at 5:37
  • \$\begingroup\$ Thank you! You're the best! Now I get why and what happened. This page is pure gold. \$\endgroup\$
    – Melectro
    Commented Jun 15, 2015 at 10:54

Your DMM doesn't measure the peak voltage, it measures the apparent voltage (which is close to but not exactly the average). And as the graph shows the average voltage did increase. Note that the actual discharge will depend on the load, but a DMM is almost no load at all.


So I found this 9V 1A AC power supply laying around and decided to make a DC source. I rectified and added a capacitor to make things a bit even. After adding a capacitor my voltage boosted from 9V to 14V. Can somebody explain why this happened for me?

There are a few things going on here.

First off AC voltage is usually measured in RMS. For a sinusoid the RMS voltage is about 0.7 times the peak voltage.

A DMM in DC mode will usually measure some approximation of the mean voltage (on a high end bench meter potentially a very good approximation). After rectifying a sinewave with a perfect rectifier that is not capacitively loaded (you don't specify what type of diodes you used so we don't know how close to perfect your rectifier was) the mean value of the resulting waveform will be roughly 0.6 times it's peak.

So why did you get 9V on your DC meter before bringing in the capacitor? The answer is most because the output voltage of transformers is typically stated at full load. Output voltage of transformers under light load can be quite a bit higher than the output voltage under full load.

Now what happens when we add the capacitor? The circuit now operates in two modes, for part of the cycle current flows from the input and the output voltage follows the input, but for part of the cycle the capacitor supplies the load. While the capacitor is supplying the load the voltage supplied to the load falls. Eventually the next peak of the AC waveform comes along, the rectified input voltage reaches the output voltage and the output voltage starts following the input voltage again.

The average voltage seen at the output in a rectifier-capacitor-resistor circuit depends largely on the rate of discharge of the capacitor. If a tiny capacitor is used then it won't store any singificant charge and the voltage will look much like it did when the capacitor was discharged. If a massive capacitor is used and the rectifier is perfect then the average output voltage will be very nearly equal to the peak voltage of the supply (again this suggests that the actual RMS voltage of your AC supply was more than 9V)

According to theoretical graph I should get aprox the same voltage even after adding the capacitor.

The same peak voltage yes but clearly not the same average voltage (which is generally what DMMs measure)

Take a look at the scope traces on http://web.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/64.57.html they show a half wave rather than a full-wave setup but they clearly show how changing the capacitor value changes the shape of the output voltage wave and hence it's average voltage.

Maybe there's some sort of equation to determine the actual output voltage depending on caps?

You can't really analyse this circuit with traditional "AC analysis" techniques because of the nonlinearity presented by the diodes. You have to look at it in the time domain, calculate equations for the two "modes" and then work out where it switches between those modes.

To a first approximation for large capacitances (RC time constant much larger than cycle time) you can just assume that nearly all the time is spent in the discharging state and the capacitor is charged to the peak voltage one per half-cycle.

It would be great to get 12V out of this thing.

You won't create a stable 12V supply this way, the average output voltage is dependent on the load and in any case most loads don't just care about the average voltage, they also care about the voltage remaining stable.


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