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Old 9th August 2012, 06:16 PM   #1
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Default Unregulated power supply output impedance

I suppose it is a simple question, but I haven't been able to find the answer.

When designing an unregulated power supply, how do we determine if the output impedance is low enough? Less than half the input impedance of the total load?

And do we include the transformer's secondary resistance in the determination of output impedance?

Thanks.
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Old 9th August 2012, 06:28 PM   #2
cbdb is offline cbdb  Canada
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Is it a Ac to Dc supply using a bridge rectifier and caps? If so, the output impedance is the same as the cap impedance, until the charge pulse happens, then the transformer impedance becomes part of the equation but it gets complicated. The cap impedance is a good approximation. And the output impedance should be closer to 1/10 (idealy as low as possible) of the load.
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Old 9th August 2012, 06:46 PM   #3
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Yes I meant an AC to DC supply, basically Transformer -> Rectifier -> Passive filters (RC/LC) kind of power supply.

I suppose for mid to high frequency, the cap's ESR is a good approximation. But for low frequencies where the current draw is beyond what the caps can store, the output impedance became the DC resistance of all components in series?
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Old 9th August 2012, 07:11 PM   #4
Elvee is offline Elvee  Belgium
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Quote:
Originally Posted by Navyblue View Post
But for low frequencies where the current draw is beyond what the caps can store, the output impedance became the DC resistance of all components in series?
No, it then becomes a sampled system, where the sample frequency is 100 or 120Hz, depending where you live.

Basically, it will obey the rules of sampled systems, ie. the transformer's impedance and upstream impedances will be multiplied by the ratio of the sampling period to the aperture time, but with some quirks: the average aperture time isn't constant and depends on the load, and the sampling itself is not purely binary, it is in fact a multiplication by a complex function, also varying with the load.
And because it is a discrete time system, funny things will happen when the stimulus frequency is a multiple or submultiple of the sampling frequency.

Unless you are familiar with Z-transform and similar tools, your best bet is to use very large filter caps...
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Old 9th August 2012, 07:17 PM   #5
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Quote:
Originally Posted by Navyblue View Post
I suppose it is a simple question, but I haven't been able to find the answer.

When designing an unregulated power supply, how do we determine if the output impedance is low enough? Less than half the input impedance of the total load?

And do we include the transformer's secondary resistance in the determination of output impedance?

Thanks.
For AC it is 1/2pi.f.C, with f in Hz and C in Farad.
For DC derived from 60Hz with full wave rectifier is 15.10 power -3/C. C is the Recervoir value in Earad.
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Old 9th August 2012, 07:23 PM   #6
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Originally Posted by Elvee View Post
Unless you are familiar with Z-transform and similar tools, your best bet is to use very large filter caps...
Meaning, the 1:10 impedance ratio can not be practically achieved? And, how large is "very large"? I suppose there is some kind of rules of thumb with respect to average current draw?
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Old 9th August 2012, 07:33 PM   #7
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why do you care about the output impedance of the Unregulated PSU?
Trying to calculate efficiency, losses, or some sort of distortion?

Thanks.
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Old 9th August 2012, 07:44 PM   #8
DF96 is offline DF96  England
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Are you asking about AC impedance or DC impedance? AC impedance is roughly given by the last cap in the supply. DC impedance (i.e. how much does it droop under load) is given by a messy combination of transformer secondary effective resistance (multiplied by charging pulse duty cycle) and storage capacitor value.

Effective secondary resistance R's = Rs + Rp x (Vs/Vp)^2

For 50Hz supply and fullwave rectification the reservoir caps get recharged every 10ms. They then droop under load by roughly I x 0.01/C, but the average DC droop will be half this so effective DC resistance due to the reservoir cap is 0.005/C (e.g. 5 ohms for 1000uF).

You need to know/measure/guess the charging duty cycle, otherwise use a factor of 5. Then add on any DC resistance from RC smoothing or choke resistance.

So Rpsu = 5 R's + 0.005/C + Rdc

This is only a rough estimate, but it will tell you whether you are close or way off the mark.
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Old 9th August 2012, 07:47 PM   #9
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Originally Posted by liching1952 View Post
For AC it is 1/2pi.f.C, with f in Hz and C in Farad.
For DC derived from 60Hz with full wave rectifier is 15.10 power -3/C. C is the Recervoir value in Earad.
Thanks.

How about DC derived from 50Hz supply?
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Old 9th August 2012, 07:56 PM   #10
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Quote:
Originally Posted by DF96 View Post
Are you asking about AC impedance or DC impedance? AC impedance is roughly given by the last cap in the supply. DC impedance (i.e. how much does it droop under load) is given by a messy combination of transformer secondary effective resistance (multiplied by charging pulse duty cycle) and storage capacitor value.

Effective secondary resistance R's = Rs + Rp x (Vs/Vp)^2

For 50Hz supply and fullwave rectification the reservoir caps get recharged every 10ms. They then droop under load by roughly I x 0.01/C, but the average DC droop will be half this so effective DC resistance due to the reservoir cap is 0.005/C (e.g. 5 ohms for 1000uF).

You need to know/measure/guess the charging duty cycle, otherwise use a factor of 5. Then add on any DC resistance from RC smoothing or choke resistance.

So Rpsu = 5 R's + 0.005/C + Rdc

This is only a rough estimate, but it will tell you whether you are close or way off the mark.
Thanks a lot. Let me chew on that for a while, I have a feeling that I haven't done asking.
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