SWMPS

[zachtheterrible]

I just tried using my 5v SWMPS for my cell phone as a powersupply for a circuit that I'm building. The SWMPS has a ripple voltage at around 625HZ. Unfortunately it is wreaking havoc on my circuit.

Just wondering if there is any way of suppressing this? I tried decoupling capacitors with absolutely no change. Would an inductor in series w/ the supply work?

 

[Optikon]

I just tried using my 5v SWMPS for my cell phone as a powersupply for a circuit that I'm building. The SWMPS has a ripple voltage at around 625HZ. Unfortunately it is wreaking havoc on my circuit.

Just wondering if there is any way of suppressing this? I tried decoupling capacitors with absolutely no change. Would an inductor in series w/ the supply work?

My guess is your decoupling capacitor is just simply not enough capacitance to significantly reduce your ripple. Likely you need (again a guess) 10's or 100's of uF to be effective. It depends on your load current.


Use equation i=C*(dv/dt) to figure out how how much C you want to reduce ripple to an acceptable level. The MOST important thing is however, to use low esr capcitors. The ripple current acting with the esr will create a ripple voltage (which you are trying to reduce)

Series inductors will help by making a low pass filter but they must be able to carry the rated current and not saturate.

 

[stevez]

Zach - how much ripple do you have? Did you mean 625 mV - or is the ripple voltage at a frequency of 625 Hz?

 

[zachtheterrible]

It is 625 hz. not mv.

I'll have to measure the peak to peak, but it is very small.

Optikon, in your equation what does dv and dt mean?

As for an inductor, the current that this circuit consumes is very minimal, on the order of 20-50 mA I would guess. I should also measure that.

 

[audioguru]

Hi Zach,
The current for your 5V project is low, so the ripple on the output of the power supply probably also is low.
Maybe the problem isn't with the power supply but with your project.

Maybe your circuit uses transistors that are supposed to be turned off at times, but are not biased correctly so are almost turned on and power supply noise turns them on.

Maybe your circuit uses logic ICs that usually have good noise immunity, but your circuit has a logic low voltage too high, so that power supply noise adds to it resulting in failures to detect the logic low.

Please post your circuit so we can see why it is so sensitive to power supply ripple. :lol:

 

[zachtheterrible]

Its not that audio, it runs fine off of 4.5v from AA batteries. The circuit that I'm doing is an RX module, HT12E decoder, and a 555 to drive a piezo buzzer when the decoder output is activated.

I did some calculations, and I'm not sure if they're right but here they are:

@ 625HZ:
100uf cap- 2.5 ohm capacitive reactance
1000uf cap- .25 ohm capacitive reactance

100uh ind- .39 ohm inductive ractance
1000uh indu- 3.9 ohm inductive reactance

I'm thinking that I'll have to use capacitors instead of inductors because it would take a ridiculously large inductor to introduce any sufficient reactnace to 625HZ. Unless my calculations are off? I'm still struggling with converting uF and uH to F and H. I don't have a scientific calculator unfortunately.

A low ESR 10000uF capacitor would probably work because it would only have .025 ohm capacitive reactance.

 

[Optikon]

Its not that audio, it runs fine off of 4.5v from AA batteries. The circuit that I'm doing is an RX module, HT12E decoder, and a 555 to drive a piezo buzzer when the decoder output is activated.

I did some calculations, and I'm not sure if they're right but here they are:

@ 625HZ:
100uf cap- 2.5 ohm capacitive reactance
1000uf cap- .25 ohm capacitive reactance

100uh ind- .39 ohm inductive ractance
1000uh indu- 3.9 ohm inductive reactance

I'm thinking that I'll have to use capacitors instead of inductors because it would take a ridiculously large inductor to introduce any sufficient reactnace to 625HZ. Unless my calculations are off? I'm still struggling with converting uF and uH to F and H. I don't have a scientific calculator unfortunately.

A low ESR 10000uF capacitor would probably work because it would only have .025 ohm capacitive reactance.

As audio said, your design is perhaps sensitive to ripple (which most likely can be made immune) if you ran it off of batteries and it works, it is because the batteries do not have a ripple voltage associated with them.

Regarding ripple voltage, the main thing to understand is that when your transformer is not charging the output capacitor and supplying load current, the capacitor all by itself is supplying the load current. Over a small interval of time , this discharges the cap causing the voltage to droop (the cause of the ripple voltage).

For a given ripple voltage:
1) a larger capacitor can supply higher currents for a half cycle of 625 Hz (or whatever it is)

OR

2) a larger capacitor can supply lower currents but for longer before the voltage drops enough to meet the ripple spec.

To reduce ripple magnitude:
1) larger cap - provides more storage for load current.
2) higher running frequency - replenish the capacitor more often per unit time.
3) lower the load current - draw less from the capacitor to begin with.

All these help. You mentioned big inductors. keep in mind a 10,000 uF cap may not be all that much smaller!

Good design = ripple current is not an issue at the expense of a giant output capacitor.

Better design = ripple current is designed to be such a value that the load has no problems with it. Output cap is only as big as it needs to be to meet ripple spec.

Best design = load is immune to any power supply ripple and in some cases, an unregulated supply can be used. A small smoothing capacitor is usually a good idea though..

Which way do you want to go with all this?

 

[audioguru]

Hi Zach,
I bought some 5V/2A switching tiny wall-warts for only a couple of bucks each. They have a very small ripple voltage that is easily filtered with a small cap because they switch at 30kHz.
Change your 625Hz power supply to a linear regulated one or make one. :lol:

 

[zevon8]

These supplies are often similar to all battery chargers in that they rely on the battery itself to smooth the output ripple. Try placing somthing like a 1000uF cap with a smaller ( say 1uF tantalum ) in parallel on the supply.

dV/dT refers to the greek letter "delta" , in this case meaning
"change". So the formula
I=C*(dv/dt)

becomes

I=C*(( the change in voltage of ripple) / ( the time period of the ripple voltage))

the time period of your ripple voltage is 1/625 , or 0.0016

I= C* ( mV of ripple ) / ( 0.0016 )

you will have to insert what your load current is for I, and what is your peak to peak ripple voltage. From this the result will be the value of capacitor you need in uF. Generally you add 20% atleast for headroom.

 

[zachtheterrible]

Hi Zach,
I bought some 5V/2A switching tiny wall-warts for only a couple of bucks each. They have a very small ripple voltage that is easily filtered with a small cap because they switch at 30kHz.
Change your 625Hz power supply to a linear regulated one or make one. :lol:

I just bought another phone charger switcher supply off ebay for a couple bucks, and they seem much smaller in the picture than mine, so I'll bet that the frequency is much higher like yours. Maybe I'll get it in the mail today and itll solve my problem :lol:

Regarding ripple voltage, the main thing to understand is that when your transformer is not charging the output capacitor and supplying load current, the capacitor all by itself is supplying the load current. Over a small interval of time , this discharges the cap causing the voltage to droop (the cause of the ripple voltage).

I guess switching supplies are quite a bit different than normal wallwart supplies. In a normal supply, its basically just a full-wave bridge rectifier that supplies DC, and I thought that the capacitor just smooths the output out. From what you're saying, it seems that in a switching supply, the capacitor supplies the load all by itself because the transformer supplies it with pulsing DC? I'm kind of confused here :roll:

I=C*(( the change in voltage of ripple) / ( the time period of the ripple voltage))

So how exactly do I find C? Isn't the equation set up to find I?

Thanks for all the help everyone :lol:

 

[Nigel Goodwin]

I would be VERY surprised if your switch-mode supply ran at only 625Hz, they normally run in the ten's of Khz - running at such a low frequency would negate most of the reasons for using switch-mode!.

Are you ABSOLUTELY SURE! that the ripple isn't generated by your own circuit?, perhaps even by poor layout and instability?.

If you can get your PSU to pieces, try scoping the output of the switch-mode transformer, and see what frequency you have there! - be careful to avoid the live mains primary, even at your paltry mains voltage :lol:

 

[zevon8]

I guess switching supplies are quite a bit different than normal wallwart supplies. In a normal supply, its basically just a full-wave bridge rectifier that supplies DC, and I thought that the capacitor just smooths the output out. From what you're saying, it seems that in a switching supply, the capacitor supplies the load all by itself because the transformer supplies it with pulsing DC? I'm kind of confused here :roll:

In general terms, the output of a switching supply is filtered the same way as a mains frequency wallwart supply. The difference is that due to the much higher frequency, a smaller capacitor is used, since it only has to "fill in " a shorter period of time.

I=C*(( the change in voltage of ripple) / ( the time period of the ripple voltage))

So how exactly do I find C? Isn't the equation set up to find I?

You could solve the equation for C, just plug in the values you know, the only one you don't know is C.

Example:

load current ( I ) = 50mA
ripple voltage ( dV ) = 100mV
ripple voltage time period ( dT ) = 1/625Hz ( divide by 2 for 1/2 cycle ) =0.8mS

0.05 = C ( 0.1 / 0.0008 )
0.05 = C ( 125 )
0.05 / 125 = C
C = 0.0004
C = 400uF

so you need 400uF, given a standard value, use 470uF

Change the values for load current and ripple voltage to what you have and do the same calculation, and you figure out your capacitor.

I hope I didn't bork the math up

 

[zachtheterrible]

Thanks for explaining that equation zevon, itll be usefull.

I measured the peak to peak on the ripple and its only .03V!! The ripple is indeed 625HZ, i measured it without any load.

i'm not going to worry about dealing with this SWMPS. Ill wait till I get the one i ordered in the mail and then worry about that one. Hopefully itll be different than mine.

 

[audioguru]

Without a load then the ripple will be very small because the output capacitor in the supply isn't being discharged by anything. You need to know how much ripple there is when it is loaded. :lol:

Maybe the frequency is normally very low without a load. The pulses won't be as frequent because it makes an output pulse then waits for the output voltage to drop when it pulses again. The frequency is probably much higher at its rated load.

Maybe your supply doesn't even work with "only" your 50mA load. The cell phone it was made to power probably draws much more.
Maybe your supply is actually a battery charger. When it senses a low current it might reduce its output or give low current pulses to a Li-Po battery that was discharged too far. :lol:

 

[zachtheterrible]

Good points audio. I'll bet one of those is the reason.

Ah well, only a couple days till i get my other swmode.

 

[Sebi]

Maybe the SMPS without load go to some "sleep" mode and this give the strange low freq.ripple... Try it with min. 10% load.

 

[Oznog]

You may not have ripple, but rather feedback oscillation. 625Hz is probably not the SMPS freq so I'm thinking oscillation.