Explanation of effects in a simple DC power supply - Page 1
Forum section: electronics / projects

author: Mangozac
author_rank: Frequent Contributor
timestamp: August 02, 2012, 01:58:56 am
content: Hi all,I’ve joined this forum as there seems to be some very smart, practical electronics enthusiasts (and professionals) frequenting. I’ve had an interest in electronics since I was very young and have recently graduated with my bachelors degree. I have recently been doing some study of power conversion and have got a question that nobody has been able to explain to me as yet.Please refer to the following schematic depicting a simple DC power supply composed of a diode bridge and filtering capacitor. I’m using a resistor as the load.Resistors R2 and R3 are simply for current measurement. The source voltage is a sine wave with a peak of 141.4V (so an RMS value of 100V). Pretty simple huh? So our really basic analysis of this says that we should get 141.4V DC across C1 and therefore 141.4mA through R1 so that R1 is dissipating 20 Watts. In reality that will be a little less due to the forward voltage drops of the diodes and voltage ripple, but for simplicity’s sake we’ll just ignore that.The following LTSPICE plot shows the voltage, current and power waveforms for the output:The following measurements were taken:V(out): 137.92V (average and RMS)I(R1): 137.92mA (average and RMS)P(R1): 19.023W (average of instantaneous values)The power can also be calculated as the product of the V and I RMS values: 137.92 x 0.13792 = 19.022VA.Since the load is resistive this is exactly as one would expect. Note that the slightly lower values than those indicated earlier caused by the DC voltage ripple.Now let’s look at the input to the circuit:Here we can see the 100V RMS source voltage, the supply current and the supply power. Note how the current is slightly leading the voltage waveform and is composed of narrow, high current spikes. This is because the filter capacitor is only charged at the peaks of the supply sine wave.Another set of measurements can be made:V(src): 99.97V RMSI(R3): 473mA RMSP(src): 19.427W (average of instantaneous values)Notice though that the input current is much higher than the output current. If we multiply V(src) x I(R3) we get a value of 47.29VA. In a pure AC circuit one would say that this difference between this apparent power (VA) and the true power (W) is the reactive power relating to the capacitor. The capacitor would store this energy and then return it to the source, resulting in no actual power dissipation. This is why we are normally only interested in the real power. In this case we would say that the circuit has a power factor of 19.427/47.29 = 0.41. That’s really bad in the scheme of things. Rather than being rated for the 138mA that the load is drawing the power supply cables would have to be oversized to handle 473mA. Not really a problem here but at higher levels of power it can require quite large cable sizes which cost more.The inclusion of the diode bridge changes the analysis though. The diodes prevent the capacitor from returning its stored energy to the source (and it is instead supplied to the load between peaks of the rectified voltage). And this is where my confusion stems from: if the supply voltage and current are in phase and the capacitor is unable to return power to the supply then how can a power factor other than 1 exist? I understand that this reactive power is not being dissipated, but surely the current caused by it will have real effects that need to be accounted for i.e. regarding the supply cable size?I thought that perhaps both my calculations/understanding and LTSPICE could be wrong, so in an effort to rule this out I actually physically constructed a test circuit with a diode bridge, capacitor and lamp for a load. Using an oscilloscope and professional power meter I was able to confirm exactly the same results as the simulations indicated.If you’ve made it this far thanks for taking the time to read this. I’d be very appreciative if anybody is able to explain this analysis - or point out where I'm going wrong!

author: IanB
author_rank: Super Contributor
timestamp: August 02, 2012, 05:44:58 am
content: I think you really do have a bad power factor, and that the definition of power factor is a little more complex than the way you are thinking of it.It is because of the poor power factor of circuits like this that larger power supplies these days have power factor correction built in to them to make them appear more like resistive loads to the supply.

author: cyberfish
author_rank: Regular Contributor
timestamp: August 02, 2012, 06:10:47 am
content: I'm not REALLY sure about this, since power electronics is not my specialty, but I'll give it a try -Calculating power factor from VI relative phase only makes sense when you have voltages and currents that are sine waves with the same frequency. In your case, voltage is a perfect sine wave, but current is most definitely not.But we can make the current a sum of sine waves with different frequencies and phases using fourier transform.If you do that, you'll get a fundamental frequency that's the same as the voltage wave. For this component, power factor is 1.But you also have current in all the harmonic frequencies, and those are not in phase with the voltage at all. That's what kills your power factor.In this case, I don't know of a simple way to calculate total power factor, except doing the integration over one cycle and use the definition of power factor (real/apparent power).

author: Mangozac
author_rank: Frequent Contributor
timestamp: August 02, 2012, 06:47:19 am
content: You know what? I was only ever taught that power factor is related to the phase angle between the voltage and current. None of my textbooks even mention that it is also dependent on thetotal harmonic distortion!This is making much more sense now! See, I told you it was something simple I was missing! This is the key to understanding it:Quote from: cyberfish on August 02, 2012, 06:10:47 amSo can you tell me if my other comment still stands: if the circuit has a power factor of less than one then the supply wiring needs to be adequately rated to handle the current that results from theapparentpower?

author: cyberfish
author_rank: Regular Contributor
timestamp: August 02, 2012, 07:00:54 am
content: QuoteProbably because in most real AC systems, that's not really relevant, and when we think of power factors, we usually think about those kind of systems. The system you have here is obviously not really an AC system.QuoteYes, because the wire doesn't care about voltage. It only cares about current flowing through it heating it up. That's (Irms)^2 x R.

author: IanB
author_rank: Super Contributor
timestamp: August 02, 2012, 07:36:39 am
content: Here's a way of looking at it with a DC circuit. I think I've got the following analysis right.First, imagine a 10 V DC supply delivering a constant 1 A into a load. The power is 10 W and the power factor is 1.Now, suppose that the load is drawing 2 A pulses on a 50% duty cycle. The average power delivered is still 10 W, but now the RMS current has increased from 1 A to 1.4 A. Therefore the load is running at 14 VA and the power factor is 0.71.

author: cyberfish
author_rank: Regular Contributor
timestamp: August 02, 2012, 08:13:20 pm
content: QuoteThis part is incorrect. You can only do the sqrt(2) multiply if the signal is perfectly sinusoidal. Otherwise you have to do the integration (definition of RMS).

author: Mangozac
author_rank: Frequent Contributor
timestamp: August 02, 2012, 10:42:48 pm
content: Quote from: cyberfish on August 02, 2012, 07:00:54 amTrue, but it still seems like a very important bit of theory to neglect.Quote from: cyberfish on August 02, 2012, 07:00:54 amGood to know my understanding is correct there!Quote from: cyberfish on August 02, 2012, 08:13:20 pmYou are correct that multiplying by sqrt(2) is only valid for pure sinusoids, but for my own edification I just did the integration (double checked with a quick EXCEL discrete analysis) and for a 50% duty cycle IanB is actually correct. That may have just been a fluke on his partbut it is important to note that the integration calculation needs to be done.Thanks for your help guys it's much appreciated!

author: IanB
author_rank: Super Contributor
timestamp: August 03, 2012, 01:26:46 am
content: Quote from: cyberfish on August 02, 2012, 08:13:20 pmWell now, I left the calculation as an exercise for the reader, but I did assume you would do the working and check my arithmetic before saying I was wrongWe have I = 2, therefore I^2 = 4, and at a 50% duty cycle I^2(ave) = 0.5 x 2^2 = 2. Therefore I(RMS) = sqrt(0.5 x 2^2) = sqrt(2) =1.4 A.

author: IanB
author_rank: Super Contributor
timestamp: August 03, 2012, 01:31:28 am
content: Quote from: cyberfish on August 02, 2012, 08:13:20 pmApparently you can also do it for a 50% duty cycle DC square wave. Try it and see...

author: cyberfish
author_rank: Regular Contributor
timestamp: August 03, 2012, 06:47:31 pm
content: I stand corrected... but did you intentionally choose 2 to screw people over

author: IanB
author_rank: Super Contributor
timestamp: August 03, 2012, 06:55:05 pm
content: Quote from: cyberfish on August 03, 2012, 06:47:31 pmNo, it was just an accident

author: cyberfish
author_rank: Regular Contributor
timestamp: August 03, 2012, 07:01:39 pm
content: Clarifying for future Googlers:To calculate the RMS of anything but a sinusoidal signal, the full integration must be done. Multiplying by sqrt(2) is only valid for sinusoidal signals.In this case, the answer just happens to be the same coincidentally (but in most other cases it won't be).

author: megajocke
author_rank: Regular Contributor
timestamp: August 04, 2012, 01:31:45 am
content: Quote from: Mangozac on August 02, 2012, 06:47:19 amYes, that's correct. Or the transformer feeding the rectifier, like is often done. However, the transformer series impedance makes the input RMS current lower than what you'd get with a pure voltage source so a simulation done with a pure voltage source would be pessimistic when it comes to power factor. It would also overestimate the output voltage under load.cyberfish: Or RMS/average and peak/RMS ratios for a 50% duty on-off square wave. These ratios are sqrt(1/D) where D is the duty cycle from zero to one. So the RMS value is actually the geometric mean of the average and peak values!

author: vk6zgo
author_rank: Super Contributor
timestamp: August 04, 2012, 03:41:30 am
content: Quote from: megajocke on August 04, 2012, 01:31:45 amYes,the transformer will modify the current waveform considerably.The bridge rectifiers at the input of a switchmode power supply may look very much like the simulation above.In fact,there was some discussion when they first became commonly used,whether there would be an effect upon power factor.Much of the discussion I have seen in Technical books applies to the older techniques,although there must be a fair bit of more recent material covering the OP's situation.Classically,transformer power supplies were of two types:Capacitor input LC filters.Choke input LC filters.(A few RC filters were used for low current purposes.)The first capacitor of the capacitive input filters may have exhibited the same behaviour,as in the OP's simulation,always allowing for modification by the transformer's contribution.They were commonly used with thermionic rectifiers,either vacuum,or gaseous type,which contributed a fair amount of series resistance,further modifying the capacitor charge current.In Choke input filters,the charging current for the first capacitor has to flow through the inductor,which will oppose a very sharp rise time of the current.High power supplies are often 3 phase,which further complicates things.

author: Mangozac
author_rank: Frequent Contributor
timestamp: August 04, 2012, 11:14:20 am
content: Very good points regarding the effects of the source impedance. Indeed I was aware that the simulation lacks an appropriate source impedance, however the measurements I made on the physical circuit were really not much better in terms of power factor.To be clear, my investigation is primarily related to creating DC direct from the mains voltage for power conversion, so does not have the luxury of a transformer to smooth things out. The values I chose for the circuit above were simply for simplification.

author: IanB
author_rank: Super Contributor
timestamp: August 04, 2012, 04:24:48 pm
content: Check out this video if you haven't already seen it:http://www.eevblog.com/2012/05/04/eevblog-273-power-factor-correction-with-the-mc34262/

author: megajocke
author_rank: Regular Contributor
timestamp: August 04, 2012, 05:02:06 pm
content: Quote from: Mangozac on August 04, 2012, 11:14:20 amWell, you do have 0.2 ohms in there from your current measurement resistors. That might not be too far from what a power line looks like, depending on where you live. In any case, below a few hundred watts the power line impedance can probably be neglected.If you didn't know it already, you can get a "current probe" to use on wires in LTspice by holding down the ALT key. ALT-clicking on components plots power. If a resistor affects your circuit too much and you need stable references to currents in formulas, immune to circuit changes, one traditional way is to put in a voltage source of 0 V.

author: Mangozac
author_rank: Frequent Contributor
timestamp: August 05, 2012, 07:32:43 am
content: Quote from: IanB on August 04, 2012, 04:24:48 pmNo, I hadn't seen it. Thanks!Quote from: megajocke on August 04, 2012, 05:02:06 pmGood points - I agree.Quote from: megajocke on August 04, 2012, 05:02:06 pmI knew about ALT-click for component power, but not for current through a wire. Thanks for the tip!