Few months back, I designed a switch mode power supply using Viper22A chip. Now, I am trying to improve that design. The new circuit is intended for universal input (85 VAC - 265 VAC). Output will be 5 V 1.5 A.

Here is a part of the circuit:


D4 is a TVS diode. I chose it over MOVs due to it's longer life and quicker response time. NTC to reduce inrush current during start. X3 is a common mode choke (2.2 mH).

C11 and C13 are X caps. However I couldn't find a good article on value selection. Any pointers will be very much appreciated.

Besides this, is there anything else which I should do to improve the reliability of this part of the circuit?

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    \$\begingroup\$ I am not so used to seeing a TVS being used at the mains side. As a TVS is very quick and the mains can have short nasty spikes, I think it might destroy itself very quickly. A MOV maybe slower but higher speed is not always better. The MOV might survive where the TVS might not. If there is a spike at the input the TVS takes it all, I see no fuse/fusable resistor for protection. Suppose there's a low-impedance pulse at the input, the TVS blows and fails as an open, now that nasty voltage can reach the rest of the circuit. As everyone else uses a MOV, I'd also stick to a MOV. \$\endgroup\$ Nov 22, 2016 at 13:15
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    \$\begingroup\$ My guess is that C11 and C13 are a compromise between spurious supression and (having a bad) power factor. I would use a value between 100 nF - 500 nF. \$\endgroup\$ Nov 22, 2016 at 13:17
  • \$\begingroup\$ @Fake - Thanks for the great advice on MOV vs TVS. I'll switch to MOV. Regarding fuse / fusible resistor - I am planning to use in line fuse holder before P7 so that an electrician or user can easily change the fuse in case of failure. Putting the fuse/fusible resistor on PCB means a trip to the service center. \$\endgroup\$ Nov 22, 2016 at 13:27
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    \$\begingroup\$ Unless your MOV is bidirectional, you should move it to after the bridge. Moreover, having it behind the CM filter will make the clamping action far more effective since the source impedance is higher. \$\endgroup\$
    – winny
    Nov 22, 2016 at 14:51
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    \$\begingroup\$ Sorry. I mixed up TVS and MOV. Beheind your CM. paralell to C13. \$\endgroup\$
    – winny
    Nov 22, 2016 at 19:47

1 Answer 1


IMPORTANT: If you use a TVS or a MOV, you MUST USE A FUSE IN SERIES WITH THE HOT LINE. This is non optional. MOVs tend to fail shorted, and so can TVS diodes. In the event this occurs, your options are use a fuse, or start a fire. Fuses are the better option.

1. What do these X and Y capacitors even do?

X capacitors, along with their cousins Y-capacitors, are both grouped together and known simply as 'safety capacitors'. In your application, which I assume is class II, you have no earth connection (and thus has a sufficient insulation barrier to qualify as class II).

The purpose of both X and Y safety capacitors is more or less the same. They are both there to reduce the amount of EMI that gets injected into the mains from whatever the power supply is doing. As you know, capacitors become less impeding as frequency increases, so these capacitors serve to effectively short higher frequency noise.

2. What is the big deal about noise?

You always see them present on the input stage of a switch mode power supply, yet they are generally absent on older 60/50Hz transformer-based power supplies. The switching harmonics from the bridge rectifier all get dissipated as tiny eddy currents and hysteresis losses in all that iron and never make it across back into the primary winding.

Switch mode power supplies generally employ square waves (or try to get as close to a square wave as possible) and the harmonic content of that alone is none trivial and a large portion of it can conduct just through the capacitive coupling of the SMPS transformer windings, none the less the ferrite core. Worse, the diodes are directly on the mains, and there is no transformer that could potentially mitigate the diode harmonics.

3. An X across mains is not the only option.

I'm going to assume you have a fully isolated (totally floating) 5V output, as is common for a low power class II power supply.

The problem is nothing is ever completely isolated. There is parasitic capacitance from everything to everything, and everywhere to everywhere. There is probably over a volt of AC potential between any point on your body and earth at this very moment (assuming you're floating and not grounded through something). This is why you get a hum on audio equipment if you touch an input jack with your finger.

Well, the situation is somewhat crappier in the case of your class II power supply. You know that little ferrite transformer? The one with two conductive windings right next to each other? Yeah - they're gonna couple capacitively, but the impedance will be quite high even to high frequency noise. This is going to turn everything on the isolated secondary side into an unintentional radiator. This may or may not be an issue, but one solution is to connect a Y capacitor between the primary ground (the neutral line in your case) and the output ground. If for some reason the polarity of the plug might get reversed, you can connect 2 Y capacitors to the output ground, almost as if it was your Earth.

The idea is to create a much lower impedance path for high frequency noise, shunting it back to the primary side instead of radiating it.

HOWEVER, there is a very important safety consideration here: you've created a path for current to leak across the isolation barrier, and the potential will be potentially dangerous. You must be careful and ensure the Y capacitors are not so large in value to allow dangerous levels of leakage current to flow, because that flow might be going through a person/dog/kitten/whatever.

4. Come on already, how do I pick that X capacitor value?!

It really just comes down to how noisy is your power supply, how much noise from the mains you want the supply to tolerate, and how good of a power factor you want. Power factor is almost always going to be lower priority though, since most countries require you to meet EMI standards above anything else.

A larger X capacitor will survive higher surge transients, will have a more potent effect on differential noise because it is lower impedance to a wider frequency range, and generally just improve things in the differential EMI area. The drawback is that, being across the line, it will constantly be drawing a small amount of apparent power. For example, a 0.47µF X capacitor across the line with 240VAC input at 60Hz (I know 50Hz would be more common, but lets make things worst case) will draw roughly 10W of apparent power at all times. If your SMPS is a 500W ATX power supply, then your power factor is 0.98. Great! If it is a 50W laptop power brick however, your power factor is now 0.8. Not so good. You'll probably want to choose something a little smaller. As the power levels decrease further, it becomes less realistic to achieve a decent power factor, but you also don't really have to care that much. You're building a 7.5W power supply. Let's say it draws 10W of real power under full load. Using a 0.1µF capacitor would result in 2.2W of apparent power draw assuming 240V 60Hz. But that's ok, it's just 2.2W.

The smallest X type capacitor you really see is 10nF, and they can get as big as several µF. I think 0.1µF is a reasonable choice for your application. You'll consume some apparent power, but this is on the order of 400mW in North America. 0.1µF is probably a bit larger than you really need, but with noise, its usually better to have too much noise reduction than not enough. Actually, its always better.

Unfortunately, there just is no real hard and fast rule here though. You can't really calculate the value, because the minimum value is based on the actual conducted EMI and what levels are acceptable for your application. On the other hand, the value is usually not that critical. I usually size it based on a reasonable power factor (but don't get greedy - sure, a power factor of .99 after the input sounds pretty sweet, but it doesn't matter if your brick fails FCC etc.) and, to quote Samuel L. Jackson (in Jurassic Park), I "hold onto my butt" and hope its enough to meet EMI compliance. Thus far, it's not really been an issue, as long as you've kept EMI in mind when designing the remainder of the supply.

Maybe I've just gotten lucky, but it's worked for me so far.

I would ad that a second X capacitor after the choke is the correct placement and will further reduce differential noise and help the common mode choke with common mode noise a little bit too. But it is definitely optional.

5. You didn't ask, but lets talk about MOVs vs. TVS diodes.

You CAN use a bidirectional TVS diode here, but remember, your breakdown voltage will need to be higher than potential sustained overvoltage conditions on mains, and peak voltage is what matters here, so 1.41*250VA = 350V, plus some healthy head room. So lets call it 380V. The thing is, most real surges you might see on the AC line are going to pack some joules behind them, and joules make TVS diodes pop like a bubblewrap scratching post in a room full of angry cats. Sure, you can get pretty beefy ones, but they get pretty expensive ($2+) at the voltages you need. And even then, their clamping voltage is basically the same as a MOV.

MOVs are preferred in the place you've used a TVS. MOVs engage in nanoseconds, and any surge that is powerful enough to make the difference between 10 nanoseconds and 30 picoseconds actually matter is probably too much for a TVS diode to handle anyway. Plus fir 50 cents, you can get a MOV rated for 250VAC line voltage that can absorb almost 200J - that's more energy than the impact of the baseball going 100 miles per hour. Let me reiterate, for 50 cents.

It's better to use a TVS after the diode bridge, but before the actual step-down conversion. Let beefier things soak up the heavy stuff, and then any tiny fast spikes that make it through can be safely handled by the TVS diode.

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    \$\begingroup\$ I wish every question of mine was answered by you. Thanks a lot and have a nice day. \$\endgroup\$ Mar 13, 2017 at 12:57
  • \$\begingroup\$ I think you said "apparent power" when you should have said "reactive power", and then your math appears to be incorrect. Power Factor is True Power : Apparent Power, where Apparent power is the hypotenuse of the triangle formed by reactive power and true power. At least according to this source: allaboutcircuits.com/textbook/alternating-current/chpt-11/… \$\endgroup\$ Jan 16, 2021 at 2:18

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