Power Supply Ripple Measurement

Hi,

  1. I am trying to measure the output voltage ripple of a 5 Volt , 1A power supply. The input to the power supply is 120V, 60Hz.

Can someone please suggest an appropiate Differential probe to do this measurement?

Plus I am expecting the ripple to be 2mV peak to peak. If I do not get 2mV peak to peak then what should I do to reduce it. I can not use Voltage regulator at the output.

  1. I am also looking at the following part

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It's a PWM ocntroller for power charger. I am trying to understand the Figure2 on page 2 and trying to determine the output ripple on paper.

I do not have the chip yet. Can someone advise how can I calculate the ripple by the data given in the data sheet?

I am sorry if I have asked a stupid question.

melissa

Reply to
walravenmelissa
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** You do not need one.
** So it's a switching supply.

Hook you scope ( with AC coupling) across the output with a load attached an see what you get.

** If you have such a problem, come back then and describe it.

** Not stupid, but outside the realm of "basic" electronics.

The actual p-p ripple voltage seen will depend on load current, circuit layout, stray inductances in PCB tracks plus the actual ESR the electros in the filter.

.... Phil

Reply to
Phil Allison

Hi,

What is the purpose of the transformer TX1 in Figure 1 of the schematic?

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melissa

Reply to
walravenmelissa

** O - M - G . . . . . :-0

I suppose this new Q is at least "on topic " for SEB.

... Phil

Reply to
Phil Allison

It is to isolate the load from the AC line. Also I would assume that it is to cut the voltage from the AC line down to a voltage more compatiable with the battery it is charging.

Notice the ground sysmbles on the left side is not the same as the chassies ground onthe right.

Reply to
Ralph Mowery

That question, coupled with the lethal voltages on the left of the schematic, is rather alarming.

What Ralph said: it isolates the output from the line (because you can't trust either pin of the line to be "safe"), and it probably has a many:1 turns ratio to help cut the voltage while keeping the power supply efficient.

Have fun. Learn. Don't electrocute yourself or burn your house down.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com
Reply to
Tim Wescott

The output ripple is the amount of voltage that the output capacitors (Co1 and Co2) sag when D6 is not conducting. To a first approximation, this is close to the output current divided by the switching frequency times the total output capacitance.

But there's a hell of a lot going on in that circuit, and first approximations may not work. If you can find an example circuit that's already laid out and has all the part numbers called out for it you'll get a huge boost in having a successful outcome.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com
Reply to
Tim Wescott

careful that the drawing of the chip in figure 2 is mirrored respect of reality (figure 5).

Bye Jack

--
Yoda of Borg am I! Assimilated shall you be! Futile resistance is, hmm?
Reply to
Jack

infact the ground symbols on the left don't mean ground at all. they'll be over 100V different from protective earth

--
  \_(?)_
Reply to
Jasen Betts

1

** The FAN301 switches at 45kHz, a frequency where the ESR of typical elect ros dominates the impedance ( ESR meters test electros at this kind of freq uency ) plus the overall impedance is rising with frequency due to stray in ductance in the leads and body of the cap.

To a first approximation, the ripple voltage is the *ESR* times the load cu rrent.

** For sure.

** See p32, further down the same pdf.

... Phil

Reply to
Phil Allison

Thats true. The ground symbles on the left side seem just to indicate a common point more so than a real ground like the chassies ground symble on the right.

Reply to
Ralph Mowery

Hi,

I have been testing this Switching power supply ( 12V, 1A) to find out that if it is carrying any differential and common mode noise and how much. The block diagram of the circuit is as follows

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The block diagram of the way, I am testing it is as follows

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The load was picking lots of noise from the supply. I was inserting common mode noise to the power supply.

But when I shorted the neutral of the power supply to the Ground of the DC output of the supply, the noise went really low almost zero.

I was wondering why did that happen.

melissa

Reply to
walravenmelissa

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