# Problem with capacitor after bridge rectifier

I wanted to make a simple power supply so I made a bridge rectifier with 4 diodes rated 6 amps, then I connected the bridge to a transformer (220 volts to 9 volts (measured 9.4 with voltmeter) 50 Hz) and when I measured the voltage after the bridge the output was 9.1 volts.

Then I added a 4700uF capacitor (50 volts) in parallel and when I measured the output voltage it was 13.8 volts but when I disconnect it, voltage is 9 volts again. I first thought that maybe the voltmeter is broken but when tested with an small motor it was actually running faster with the cap.

Is it pulling more pressure on the transformer to draw 13 volts or something? Because I'm really confused right now. By the way will it help if I use an smaller cap? Because that would increase the ripple voltage?

• Your voltmeter is taking an average (or rms if it's a good one) of the rectified sine wave coming from the mains, when you add a capacitor, it gets charged up to the peak voltage and will stay there for a while without any real load. Without the capacitor you get a voltage that swings from 0V up to 13.something and looks a bit like this ∩∩∩∩∩∩∩∩, when you add the capacitor, instead of getting 100% ripple (that almost every voltmeter will average out to a single value), you get a much more flat, nearly ripple free output that sits at the maximum peak voltage that looks more like this ───────
– Sam
Jul 15, 2016 at 22:24
• You may want to learn how a simple linear power supply work and what is the effect of a filter capacitor after the bridge is. See for example this Adafruit article which is quite simple. Jul 15, 2016 at 22:32
– pipe
Jul 16, 2016 at 10:08

The answers already here are correct. I just wanted to add that since you are using a Full-Wave Rectifier (4 diodes), your ripple voltage is determined by:

$$V_{ripple}=\frac{V_{peak}}{2fCR}$$

or equivalently, $$V_{ripple}=\frac{I_{load}}{2fC}$$

Now, the thing here is that you want to keep the ripple voltage 'small.' A good number is within 10% of $V_{peak}$. From your post, I see $V_{peak}=13.8{\mathrm V}$

If you do the math, in order to keep the ripple voltage, say, 10% of $V_{peak}$, the maximum current you can draw from the rectifier is : $$I_{load}=2(1.38{\mathrm V})(50{\mathrm{Hz}})(4700\mu{\mathrm F})$$ $$I_{load}=0.65{\mathrm A}$$

That is the maximum current you want to draw if you want keep the ripple voltage at a maximum of $V_{ripple}=1.38{\mathrm V}$ or 10% of the peak voltage.

I saw in one of your comments that you were concerned about damaging components. You won't damage them as long as they are rated to handle the voltage/current they are being provided with. The capacitor should be fine (since it is 50V), but they may be a lot of current running through the circuit at startup (when the capacitor is fully discharged). So you want to make sure your diodes are rated to handle that much current, and also check the reverse voltage ratings for those diodes.

The reverse voltage across those diodes are theoretically the same as the peak voltage for a full wave rectifier with four diodes (in your case the diodes should handle more than the 13.8 volts you are getting at the output).

Your capacitor will charge to the peak bridge rectifier's output voltage, minus the drop through the diodes.

For a transformer with an output voltage of Voac, your capacitor should charge to somewhere around (Voac*sqrt(2))-1.4 where the 1.4 is the voltage lost across the two conducting diodes in your bridge rectifier.

• But the output voltage is more than (9×(sqrt 2)) which is 12.7 but i get 13.8,so where is that 1.1 extera volt is coming from (assuming voltage lost across diodes is 0)? And what i get from your answer is that it wont damage anything right? Jul 15, 2016 at 22:04
• poor regulation. at low power draw there will be "extra voltage" from the XFMR. draw a bit of power from your DClink and it will come into regulation. Also design tolerances could be pushing it up
– user16222
Jul 15, 2016 at 22:22
• @OM222O: dividing by $\sqrt 2\approx 1.41$ is the right factor for converting the peak voltage of AC into a DC voltage that will deliver the same power to a fixed resistive load -- but that may not be what your multimeter is doing when it sees a voltage that that always has the same sign. Instead it may just be doing a simple time average of the rectified voltage, and in that case the right factor to divide by is $\pi/2 \approx 1.53$. Jul 16, 2016 at 3:41

That's correct. You are storing the top voltage value in your capacitors so you will see an 1.41 times increase. Google top wave rectification.

Regarding their size, you need to make a trade off between output voltage ripple and input current/power factor/THD. How big is your load?

• As i use it as a power supply it varies a lot so i dont exactly know that but its between 0-3 amps , 0-9 volts.by the way if its only stored in capacitor i dont mind the increased voltage as i can change it but does it not damage / shorten the lives of any parts in the circuit? As long as its not doing any damage im happy to keep it there.thanks and thanks for your answer. Jul 15, 2016 at 21:28
• You don't seem experienced if I may say so. Aim for 20 % ripple at full load and you will have a starting point. U=C*dI/dt so Ut=CI. 0.2*13.8*(1/120)/3 = capacitance in Farad. Jul 15, 2016 at 21:48
• Sorry, you where in 50 Hz country. Substitute 120 for 100 (2*50) in the equation above. Jul 15, 2016 at 22:26

The capacitor, as others have told you, basically stores peak voltage with little ripple. Assuming that your load is mostly constant and that the transformer is a voltage source, the ripple consists of a long discharge time and a rather short charge time where the voltage from the transformer replenishes the ripple.

This short charge time will draw a lot of current and you want to avoid voltage loss, so your connections should be solid enough. Now the whole period time is 10ms and the recharge period will be a fraction of that: for discrete circuits, that's too short for heat death, so you only need to worry about the sustained power losses. For diodes (with their constant voltage drop) they are proportional to the current through them, for resistive material they are proportional to the square of the current. This includes the transformer coils.

So yes: using that capacitor will likely cause your transformer to get warmer than otherwise. But their specs for continued operation will support the kind of peak load that standard rectifying circuits with capacitors will cause. Killing a mains transformer with overload usually requires rather drastically exceeding the specs.