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quicksilver

Sun May 16 2010, 05:21PM

Registered Member #1408 Joined: Fri Mar 21 2008, 03:49PM
Location: Oracle, AZ
Posts: 679

Why don't bleeder resistors form a short circuit when in parallel on a filter cap? It just seems Illogical for that not to take place...
I made a small DC supply using a small variac rectified and the filter cap is a real danger as well as a nuisance for certain applications. I was going to put a bleeder on it when this question arose: I just don't get it. Is it that the formula for a bleeder's value is such that a short can't take place?

CT2

Sun May 16 2010, 05:57PM

Registered Member #180 Joined: Thu Feb 16 2006, 02:12AM
Location: Ontario, Canada
Posts: 187

A short circuit would mean no resistance, a bleed resistor has resistance (ovbiously) and therefore isn't a short. The resistor does constantly bleed the capacitor, but the resistors value is high enough that it doesn't actually dissipate much power. Its value is picked such that the RC time constant is a safe value for your application. For example if you want the capcitor to be discharged 2 seconds after power is turned off, you would have to pick a resistor value to go with your capacitor value such that R*C = 2 seconds (it will only be discharged to 37 % by this point however). And then you would need to use the voltage across the resistor to determine the current, and I^2*R to pick the power rating of the bleed resistor.

quicksilver

Sun May 16 2010, 06:19PM

Registered Member #1408 Joined: Fri Mar 21 2008, 03:49PM
Location: Oracle, AZ
Posts: 679

Thank you for taking the time. I really appreciate it.
It occurred to me later that since there was resistance a true short could not take place but often I still think of the analogy of "running water" when conceptualizing electricity; so that circuits would all be simplistically linear.

radiotech

Sun May 16 2010, 11:56PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
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Posts: 1546

There is another reason for a bleeder resistor. The DC value for full wave rectifiers is 0.636 * peak AC . However with no bleeder resistor,
if a capacitor is used, the voltage across it would rise to the peak of the AC under no load conditions which could overvolt any load suddenly switched on. In addition more expensive capacitors (higher voltage) woud be needed.

quicksilver

Mon May 17 2010, 01:56PM

Registered Member #1408 Joined: Fri Mar 21 2008, 03:49PM
Location: Oracle, AZ
Posts: 679

I didn't know the numeric element but I did see that phenomenon in testing with a meter. I thought it was odd at first; so used to seeing on the meter, what is projected in the circuit design. .....But it was there. I did know the need for higher value - high quality filter caps. Mains AC current has so many issues that even though I had a slight background in commercial electrical work, I still find more subtle issues, the more I explore.
EE students have a serious advantage over those who haven't had that background. I am just now studying that material that 1st year engineering students would know.

Steve Conner

Mon May 17 2010, 02:30PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

radiotech wrote ...

There is another reason for a bleeder resistor. The DC value for full wave rectifiers is 0.636 * peak AC . However with no bleeder resistor,
if a capacitor is used, the voltage across it would rise to the peak of the AC under no load conditions which could overvolt any load suddenly switched on. In addition more expensive capacitors (higher voltage) woud be needed.

That's only true for a choke input filter, and those are hardly used nowadays.

radiotech

Mon May 17 2010, 04:33PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

The average value is used for rectified AC converted to DC amps
for things like chemical cells and motor efficiency calcs.

An electric heater will get hotter on AC than it will on fullwave rectified DC. 1.11 times hotter.

Proud Mary

Mon May 17 2010, 06:47PM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Steve McConner wrote ...

That's only true for a choke input filter, and those are hardly used nowadays.

I still use them, not because they're the best, nor that they are simple enough for me to understand, nor again that this kind of circuitry lends itself well to a 40W soldering iron, but simply because I can, and that I love the look of those glossy, black, oil-filled Parmeko chokes. My idea of a bleeder resistor is something which can augment the central heating in the winter time.

Steve Conner

Tue May 18 2010, 10:36AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Well, if you've got some of those Parmeko chokes, it would be a shame not to put them to use

It's hard to get chokes designed for choke input filter service nowadays. Audiophiles love the Parmeko stuff, I've seen a pair of Neptune output transformers go for over Â£500 on Ebay.

Of course, buck and forward converter SMPS are closely related to the choke input filter, and they need a minimum load or a bleeder resistor. Different century, same problem

Radiotech: You're (deliberately?) confusing RMS and average. The electric heater heats according to the RMS value, an electrochemical cell or battery responds to the average value.

The 1.11 factor you give is the RMS-to-average ratio for a DC supply fullwave rectified from single-phase AC and unsmoothed. If it measures 240V DC on your Simpson multimeter, which being a moving-coil instrument reads the average, then it'll actually be 266V RMS.

radiotech

Tue May 18 2010, 05:32PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

Radiotech: You're (deliberately?) confusing RMS and average.

moi?
The clearest statement I have seen is in this paragraph shown:
And this Simpson model 20 could be used to measure " power effectiveness"' , it being a thermocouple type.

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