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GOR3 10th January 2013 03:04 AM

Help with Ripple Current
 
I am building a power supply to provide 30VDC at 2A. The filter capacitor is 20,000ĶF following a full-wave bridge. I am trying to determine if the filter capacitor can withstand the ripple current. I have calculated the ripple voltage using Vripple = I/2fC; where, I is the current drawn in amps, f is the line frequency in Hertz, and C is the capacitance in Farads. Thus:
Vripple = 2/2*60*.02 = .83. This voltage is peak-to-peak.
To approximate Vrms, I used Vrms = Vpp/2*sqrt(2). Thus: Vrms = .83/2*sqrt(2) = .29.
Hereís my question: Is calculating the ripple current as simple as Ohmís Law? That is, is the ripple current equal to Vrms/Z ? The capacitor, actually two Panasonic TS-UP 10,000ĶF capacitors in parallel, each have an ESR of .04Ω at 120Hz. I assume the effective ESR is halved by using the capacitors in parallel. If so, the ESR is about .02Ω. I calculated Z using Z = sqrt ((1/2πfC)^2 + ESR^2). Thus: sqrt((1/2π*120*.02)^2 + (.02)^2) = .069. Using Ohmís Law, the ripple current is .29/.069 or 4.2A.
Iím not asking for someone to check my math; rather, is my approach correct?
Many thanks,
George

megajocke 10th January 2013 04:45 AM

It doesn't seem to be too far off. I'd use this approach:

The RMS current in the transformer winding is the same as the RMS current on the DC side of the rectifier before the filter capacitor. That current is the output current plus the capacitor current. Also, the capacitor current has no DC component and the output current has no AC component. Because of this, (transformer RMS current)^2 = (dc output current)^2 + (capacitor RMS current)^2.

For typical values of transformer impedance at this power level and a large capacitor (ripple lower than 40% or so) you end up with the rule-of-thumb value of transformer RMS current = 1.8 * dc output current. For this case, by inserting this into the previous equation, the capacitor RMS current becomes sqrt(1.8^2-1^2) * dc output current = 1.5 * dc output current.

So you end up with about 3 A (RMS) capacitor ripple current and 3.6 A (RMS) transformer secondary current. Also, be careful about output voltage which is going to be lower than sqrt(2) * nominal transformer RMS voltage even considering diode voltage drop. It ends up being about 90%-95% of that because of the distorted current waveform causing voltage drop in the transformer. I like to simulate these kind of circuits but you need a good estimate of transformer impedance (winding resistance and leakage inductance) to get meaningful results at all.

But anyways, 3 A or 4.2 A ripple current should not be a problem for a typical 20 mF capacitor/capacitor bank.

DF96 10th January 2013 10:30 AM

For an approximate answer the approach may work. There are two weaknesses:
1. the conversion from Vripple to Vrms assumes a sinewave, when in fact it is a rough approximation to a sawtooth.
2. the calculation of RMS ripple current from Vrms assumes that ESR dominates over capacitive reactance.

You could check your model using PSUD2.

Mooly 10th January 2013 10:54 AM

1 Attachment(s)
Hardly a shining example of (my :)) simulation skills but the peak currents are as expected very high. This is the voltage across the 0.1 ohm which I guessed was comparable to resistive and ESR losses is a real circuit. Top trace is the voltage across the cap i.e. the supply.

wintermute 10th January 2013 11:13 AM

1 Attachment(s)
Quick sim in LTSpice shows an rms ripple current of 4.75A however this was simmed with NO series resistance on the voltage source, so in the real world I imagine it should be lower due to resistance in the secondary of the transformer.

Putting a series resistance of 0.1 ohms in the voltage source dropped the current to 4A. I'm not sure what a reasonable figure would be to add in (as the series resistance) to make it more realistic.

edit: haha mooly, we both had similar ideas I was a bit slower!

Tony.

megajocke 10th January 2013 03:08 PM

Try 27.5V RMS in series with 0.63 ohms for something resembling a 100VA 25V transformer with 10% regulation. (10% * 25V / (100VA / 25V) = 0.63 ohms) Such a configuration should give close to 30V at full load and nominal line if I remember correctly.

A transformer with tighter regulation can be rated at a lower nominal voltage for the same output voltage. Though, if 30V is desired voltage after a linear regulator, something close to a 30V transformer will probably be needed considering low line, ripple and dropout.

edit: Regarding an earlier simulation in the thread, 20 mF of capacitors will hopefully have a much lower ESR than 0.1 ohms... Probably more like on the order of 0.01 ohms.

Mooly 10th January 2013 04:58 PM

Quote:

Originally Posted by wintermute (Post 3319310)
edit: haha mooly, we both had similar ideas I was a bit slower!

Tony.

You managed to get some data figures into yours... I'd have to reach for the text books to remind me how to do that :D

gootee 10th January 2013 06:02 PM

Quote:

I am building a power supply to provide 30VDC at 2A. The filter capacitor is 20,000ĶF following a full-wave bridge. I am trying to determine if the filter capacitor can withstand the ripple current. I have calculated the ripple voltage using Vripple = I/2fC; where, I is the current drawn in amps, f is the line frequency in Hertz, and C is the capacitance in Farads. Thus:
Vripple = 2/2*60*.02 = .83. This voltage is peak-to-peak.
To approximate Vrms, I used Vrms = Vpp/2*sqrt(2). Thus: Vrms = .83/2*sqrt(2) = .29.
Hereís my question: Is calculating the ripple current as simple as Ohmís Law? That is, is the ripple current equal to Vrms/Z ? The capacitor, actually two Panasonic TS-UP 10,000ĶF capacitors in parallel, each have an ESR of .04Ω at 120Hz. I assume the effective ESR is halved by using the capacitors in parallel. If so, the ESR is about .02Ω. I calculated Z using Z = sqrt ((1/2πfC)^2 + ESR^2). Thus: sqrt((1/2π*120*.02)^2 + (.02)^2) = .069. Using Ohmís Law, the ripple current is .29/.069 or 4.2A.
Iím not asking for someone to check my math; rather, is my approach correct?
Many thanks,
George
Quote:

Originally Posted by megajocke (Post 3319026)
It doesn't seem to be too far off. I'd use this approach:

The RMS current in the transformer winding is the same as the RMS current on the DC side of the rectifier before the filter capacitor. That current is the output current plus the capacitor current. Also, the capacitor current has no DC component and the output current has no AC component. Because of this, (transformer RMS current)^2 = (dc output current)^2 + (capacitor RMS current)^2.

For typical values of transformer impedance at this power level and a large capacitor (ripple lower than 40% or so) you end up with the rule-of-thumb value of transformer RMS current = 1.8 * dc output current. For this case, by inserting this into the previous equation, the capacitor RMS current becomes sqrt(1.8^2-1^2) * dc output current = 1.5 * dc output current.

So you end up with about 3 A (RMS) capacitor ripple current and 3.6 A (RMS) transformer secondary current. Also, be careful about output voltage which is going to be lower than sqrt(2) * nominal transformer RMS voltage even considering diode voltage drop. It ends up being about 90%-95% of that because of the distorted current waveform causing voltage drop in the transformer. I like to simulate these kind of circuits but you need a good estimate of transformer impedance (winding resistance and leakage inductance) to get meaningful results at all.

But anyways, 3 A or 4.2 A ripple current should not be a problem for a typical 20 mF capacitor/capacitor bank.

The downloadable spreadsheet at the following link should do what you need. It doesn't use any approximations because it solves the differential equations numerically.

http://www.diyaudio.com/forums/power...ml#post3287619

It also has spice model parameters for a transformer. Additionally, the transformer model was per-unitized so it can be scaled, such that you can enter a different Volt-Amps rating, or a different output or input voltage, or a different line frequency, and the spreadsheet will calculate a new set of spice model parameters for you.

There is also a section with the measurements on which the model is based, with instructions for measuring a different transformer.

Earlier in that thread, many more spice-simulation specifics are given, along with downloadable .asc files for LT-Spice.

Osvaldo de Banfield 10th January 2013 06:19 PM

The calculation is right, but you forgot to add the audio currents through this capacitors, in effect, the audio current at the speaker (or any other load) terminals ALSO flow through the capacitor.

GOR3 10th January 2013 06:30 PM

Many Thanks
 
Thank you for your thoughtful replies.
Tom, the spreadsheet is astonishing. I look forward to using it.
All the best,
George


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