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Old 4th December 2008, 08:11 PM   #1
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Default Location of voltage-dropping power resistor

Because I have a slightly "larger" transformer (voltage wise) than that need for an application, I'm going to use a 1-2 Watt resistor (aka "power resistor") to drop about 7VDC. It will be located in/near the PSU rectifier, but before a typical 3-pin regulator.

The value of the power resistor will be ~160 ohm.

Is it "best" to place this power resistor after the smoothing caps (aka main rectifier caps), or between diode bridge and the smoothing caps?

A possible reason why "between" may be better is discussed in this topically unrelated thread on the PFM forum. (Note the position of R1 in the diagram).

Thx for any feedback you can provide!
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Old 4th December 2008, 08:25 PM   #2
infinia is offline infinia  United States
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Hi
No doubt about it... the best performance is between rectifier and smoothing caps. Forms lowpass fuction with largest C. Also greatly reduces peak charging and minimizes cap esr requirements.
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Old 4th December 2008, 08:25 PM   #3
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It has to go after the caps as that's the only way you can guarantee the voltage fed to the reg won't creep up to the full unwanted amount over times of light loading.
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Old 4th December 2008, 09:41 PM   #4
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Quote:
Originally posted by infinia
No doubt about it... the best performance is between rectifier and smoothing caps.
Quote:
Originally posted by richie00boy
It has to go after the caps ...
Can't have it both ways ... or can I?
FYI: I'm probab. only going to use ONE smooting cap (~2000 uF).
I could place a small R, say 1 ohm, between diodes and C. And, then, the 160 ohm power R after C (before reg.).
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Old 4th December 2008, 10:07 PM   #5
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I agree with richie00boy. You need a small bypass cap on the output side of the resistor, across the regulator input.
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Old 4th December 2008, 10:31 PM   #6
lineup is offline lineup  Sweden
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One Vbe multiplier transistor
would give a more constant voltage drop.
Actually it will work like a zener.

10 x Vbe is ~ 6.5 volt

I guess the current drawn is like 45 mA. (7V/160ohm)
One TO-126 transistor could do the job.
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Old 4th December 2008, 11:41 PM   #7
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If the system is lightly loaded then the 3 terminal regulator should be able to handle the higher voltage and low current. I'll go with "infinia" on this one and put the resistor between the rectifier and the big caps.
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Old 5th December 2008, 01:15 AM   #8
john65b is offline john65b  United States
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Why not a pair of CL60 thermistors between the rectifiers and caps (one on pos rail and one on neg rail) ? It will drop 5v - 6v each rail and it serves as a soft start too??
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Old 5th December 2008, 01:23 AM   #9
paulb is offline paulb  Canada
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Use a CRC. One cap after the diode, then the R, then another cap. A pi filter.
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Old 5th December 2008, 02:03 AM   #10
john65b is offline john65b  United States
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I thought a pi filter was a CLC??

Whoops, I thought this was a question on reducing voltage rails to an amp...just saw the regulator reference on the first posting...
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