I've been studying power supplies and diodes, all books I've read discuss a great deal of things, but there's something I haven't found, and that is how to calculate how much current a bridge rectifier draws when it has its output filtered by a capacitor (and connected to a load).

Sedra/Smith's book mentions that since the filtering capacitor will work as a reservoir tank, the diodes will only conduct during a certain part of the input signal needed to fill back up the capacitor, the portion that the diode is conducting is referred to as "conduction angle" , dividing that conduction angle by 2 times pi, you get cycle percentage, or in other words what percentage of the cycle the diode is conducting.

This helps to determine what the text refers to Idmax and Idavg, which are the maximum and average currents through the diode during the period of time that the diode is conducting to charge up the capacitor.

So it basically means that the bigger the capacitor (the smaller the ripple voltage) the diode will conduct during a shorter period of time, and in that shorter period of time it will have to conduct more current (since it has less time to charge up the capacitor).

Those figures obtained by Idmax and Idavg, only tell you the current through the diode during the conduction intervals, which are useful to determine if certain diode will be able to withstand brief outbursts of current. What I need to know is how much total average current is going through the diodes, or in other words:

How can I calculate the current being drawn by the bridge rectifier (entire circuit) as it would be shown on a conventional DMM when measuring the current being drawn by the bridge, when its output is being filtered by a capacitor (and connected to a load)?

I've simulated several circuits and when measuring the current being drawn by the diode bridge is not equal to the current being drawn by the load, and that's to be expected since the diodes are not conducting all the time, but not being able to know how much current is being drawn at that point (between transformer secondaries and bridge rectifier) is keeping me from being able to determine the total current draw of the circuit, thus I'm not able to calculate the transformer's power rating needed to feed the circuit.

  • \$\begingroup\$ Sorry, I thought that the entire purpose of this is asking clear and detailed questions, or as detailed as possible, displaying that the person asking the question has done some research, thats on the FAQ, ive read it. Sorry if its too boring or long for you, either way someone already answered my question. \$\endgroup\$
    – S.s.
    Commented Dec 31, 2012 at 7:27
  • \$\begingroup\$ If you want to draw more attention on your question, there are a few things you can do: 1- check grammar, punctuation and formatting; 2- make the key points of your question stand out, using emphasis and bold; 3- make a short version of your question at the bottom. \$\endgroup\$
    – clabacchio
    Commented Dec 31, 2012 at 9:28

4 Answers 4


Just so we're on the same page, we're talking about the transformer + rectifier + capacitor circuit that has been used for linear power supplies for decades, right? Something like this:

                |  |    |     |
              d1-  -d2  |     |
                ^  ^    |     |
  +---\  /-R1---+  |    |     |
120VAC T1 24VAC |  |    |   (load)
  +---/  \---------+    |     |
                |  |    |     |
              d3-  -d4 ---c1  |
                ^  ^   ---    |
                |  |    |     |
                +--+----+-----+-- GND

I'm assuming what you're really trying to find out is: What VA rating should I specify when I buy a transformer for my system?

The I^2R heating of a coil inside that transformer is proportional to the RMS current through that coil. If you keep that RMS current low enough, the manufacturer guarantees that the transformer will not overheat and fail. (Most manufacturers specify that max RMS current indirectly, implying it from the VA rating of the transformer.)

quick, conservative calculation

Say you already know the peak number of electrons per second flowing through some diode (Id_max) and what fraction of the full 1/60 second cycle that diode has nonzero flow (D). Then I can get a quick estimate of the VA rating required for the transformer with

estimated_I_RMS = 2 * D * Id_max^2.

So, for example, if you somehow know that any one diode is conducting 1/17th of a full cycle -- in other words, d1 conducts 2/17 of one half cycle, then it has zero current while d2 conducts 2/17 of the next half cycle, and so nonzero current is flowing through the transformer 2/17 of the time.

Say, for example, you also know that Id_max is 2 A. At any instant, whenever (nonzero) electrons are flowing through any diode, exactly the same number of electrons per second are flowing through the transformer. So the maximum electrons per second through any diode (Id_max) is the same as the maximum electrons per second through the transformer (Itx_max).

Then I estimate the RMS current through the transformer as

2 * (1/17) * (2 A)^2 = about 0.47 A_RMS

so for a 24 VAC output transformer, I would need to specify

estimated_VA = Vrms * estimated_I_RMS = 24 VAC * 0.47 A_RMS = about 11.3 VA.

Of course, no one sells transformers that are exactly 11.3 VA, so I'd round up to a 12 VA or a 15 VA or a 20 VA transformer -- whatever my suppliers have in stock at some reasonable cost.

This is a conservative estimate -- the actual RMS current through the transformer is somewhat less than this estimate, but more than the RMS current through the load.

more details

To more accurately calculate the actual RMS current flowing through the transformer, I could divide up the complete cycle into 6 or so time slices, estimate the current flowing during each time slice -- that's pretty easy when it's zero -- and then do the root-mean-square (RMS) calculation: square each current, average each of those squared values, weighted by the time that current was flowing, and then that the square root of that average. It might be quicker and more accurate to run a simulation with thousands of time-slices than to work it out by hand.

There are many techniques for reducing the RMS current through the transformer while supplying exactly the same power to the load. Electric power companies love those techniques, because their customers are just as happy (the load gets exactly the same power), they get paid the same amount of money (for customers that pay per kWh), and they can spend less money for transformers and long power lines (because higher RMS currents require bigger, heavier, more expensive transformers and power lines). Those techniques go by the general name of "power factor correction".


Perhaps the simplest such technique is the "R1" resistor in the above diagram. Some systems use a more complicated "valley fill" circuit -- see Serial capacitors in electronic ballast of a fluorescent lamp . And many systems -- such as most computer power supplies -- use an even more complicated "active power correction" system.

  • \$\begingroup\$ Thank you David that was exactly what I was hoping to find, I somehow thought it would have to do something with averaging the Diode "on" time current, I wasnt sure how to do it. I never thought something as simple as this would be somehow so convoluted to find, after all a linear supply like the one you described is one of the most basic circuits, all the books Ive read so far which involve making a linear power supply mention how to calculate every parameter, but when it comes to calculating the transformer they just state "Choose a proper transformer depending on the circuit needs". \$\endgroup\$
    – S.s.
    Commented Dec 31, 2012 at 7:26
  • 1
    \$\begingroup\$ So, just out of curiosity, how did people do it back in the day when they didnt have simulations at hand? using a similar quick conservative calculation like the one you proposed or by doing the "slice" method? \$\endgroup\$
    – S.s.
    Commented Dec 31, 2012 at 7:32
  • \$\begingroup\$ I would like to know the origin of the estimation of: estimated_I_RMS = 2 * D * Id_max^2. If I approach the current on the diodes to a pulse train of a magnitude that equals to the peak current (Id_max), the rms of the current (Irms) is yielded by Irms = Id_max (D)^(1/2), being D the duty cycle ratio. \$\endgroup\$
    – user54681
    Commented Oct 7, 2014 at 1:29
  • \$\begingroup\$ @JulioNavas: You are right, I seem to have put a typo in that equation. \$\endgroup\$
    – davidcary
    Commented Oct 15, 2014 at 17:40

That would depend on your DMM I suppose. Unless you have a true RMS meter, it would likely only read the peak AC current, assume it's sinusoidal and do the usual root-2 conversion from Peak to RMS, ignoring the large off-time.

A true RMS meter will sample and do actual root-mean-square calculations over the period. A digital scope will do the same thing.

If you want to calculate the average of the AC current, you have to do the same thing - compute the on-time diode current, how long the on-time and off-time is per switching cycle and mathematically compute the average.

(It's probably faster to rely on the simulation to do this, FWIW...)

  • \$\begingroup\$ Hello, True RMS aside, my DMM example is just for clarification purposes, i didnt intend to ask about the DMM itself, it was just an example, more about the calculation needed to determine the average current VS the instanteneous conducting current of the diode. The DMM example was just to further exemplify my initial question, which I believe you partially addressed in your 3rd paragraph, about the computation of the on-time diode current. Thats exactly what I want to know, how to compute the average of the on-time vs off-time \$\endgroup\$
    – S.s.
    Commented Dec 15, 2012 at 21:26

In a full-wave rectifier set up with Rl as load and C for filter.

$$I_{diode,avg} = I_{load} \cdot \left[1+\pi\cdot\left(\frac{V_{in,peak}}{2\cdot V_{ripple,p-p}}\right)^{\frac{1}{2}}\right]$$


To calculate exactly, current peak or true RMS of diode bridge, is not an easy task. The problem is that equations are highly non linear. Since full wave diode bridge rectifier conducts only 20% of time (on average), during that time power from power network drown must be equal to power consumed on the load, that is average voltage on the output squared over resistance. So approximated formula can be to find integral during 20% of time of sine voltage with known amplitude and cosine current of unknown amplitude that has to be calculated and make it equal to averaged output DC voltage squared via load resistance. It is useful to numerically simulate instead of exact calculating, like in here:



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