One tip...
Let your rails be +/-90V... That makes 180Vdc...
No bleed resistors...
Now, turn on your amplifier... And turn off it...
Wait 5 seconds...
If you touch with your fingers on those rails
ZZZTT!!.... You're ash!...
With bleeder's, those 180Vcd rails rapidly went to 0V (zero)...
My 2 cents...
Let your rails be +/-90V... That makes 180Vdc...
No bleed resistors...
Now, turn on your amplifier... And turn off it...
Wait 5 seconds...
If you touch with your fingers on those rails ZZZTT!!.... You're ash!...
With bleeder's, those 180Vcd rails rapidly went to 0V (zero)...
My 2 cents...
audioDIYer said:One tip...
Let your rails be +/-90V... That makes 180Vdc...
No bleed resistors...
Now, turn on your amplifier... And turn off it...
Wait 5 seconds...
ZZZTT!!.... You're ash!...
With bleeder's, those 180Vcd rails rapidly went to 0V (zero)...
My 2 cents...
When I work on a P/S, I usually unplug the amp while its playing; but, I always drain the caps with a 50 ohm wirewound resistor anyway.
One big surprise, though. Really old caps, even after draining them, can, hours later, mysteriously regain a charge. So, before you stick your hand in that spare parts box, bear in mind that you could be reaching for a really inappropriate coffee subsitute.
I don't understand why this should be any more complicated than a 5k-10k bleeder soldered across each rail. It takes a while for the PSU to discharge, but at least you can be certain that when you come back to your amp a week later it's not saving a surprise for you.
If it is too slow, then you can use a 50-100 ohm to speed up the process when you are doing work on the amp.
Am I missing something?
If it is too slow, then you can use a 50-100 ohm to speed up the process when you are doing work on the amp.
Am I missing something?
Hi Ryan,
Using bleeder resistors is simply good engineering practice. They will tend to dissipate some heat, but not ridiculous amounts.
Hi Daniel,
The bleeders do not draw enough power to make a big difference in sound at all. If you heard a big difference when removing them, something is very, very wrong. "Houston ... we have a problem"
Think about this a moment. A proper bleeder resistor will draw less current than the standing bias current in an amplifier. Hmmmm. What's going on here??
1. They discharge the supplies to make it safe for a technician and also to ensure a known state on power up.
2. Some regulators are less noisy with a minimum load current.
Your "bleeder resistors" can take the form of a resistor, a current source or guarantied bias current in an output stage until the supplies drop close to zero. The end results are the same.
-Chris
Nope.
Using bleeder resistors is simply good engineering practice. They will tend to dissipate some heat, but not ridiculous amounts.
Hi Daniel,
The bleeders do not draw enough power to make a big difference in sound at all. If you heard a big difference when removing them, something is very, very wrong. "Houston ... we have a problem"
Think about this a moment. A proper bleeder resistor will draw less current than the standing bias current in an amplifier. Hmmmm. What's going on here??
Well, two things off the top of my head here.
1. They discharge the supplies to make it safe for a technician and also to ensure a known state on power up.
2. Some regulators are less noisy with a minimum load current.
Your "bleeder resistors" can take the form of a resistor, a current source or guarantied bias current in an output stage until the supplies drop close to zero. The end results are the same.
-Chris
anatech said:
There was my mistake. Thank you! I had used 1/2 watt carbon during testing. These, of course, aren't great at shunting or passing very high frequencies. So it didn't. Well, there goes the remainder, a little screech, right into the caps, and it certainly did go into the amplifier too. Oops!!!
Perhaps a higher rated and higher quality type (metal film??) resistor could shunt more evenly?
Its funny. When I wrote up the design, I had Metal Oxide 10k written down. However, I managed to purchase 10R instead. I didn't use those for bleeder resistors.
So, there's pretty much how the carbon got onto the power supply. So, I'll just go buy the right thing and it'll probably work just fine.
It was just kind of a dumb mistake. The 1/2 watt carbon are in the "pass" components box, filed under "fails to pass digital hetrodynes from CD" use. And input circuit in-series resistor is about all they're good for. It certainly was a horrible choice for a shunting component. Oh well, I learned.
Thanks again!
Oh gosh! I'd bet that I shouldn't use 1/2 watt carbon for the amplifier's speaker output zobel's resistor either, because it won't shunt RF very well that way. Is that about right?
So, there's pretty much how the carbon got onto the power supply. So, I'll just go buy the right thing and it'll probably work just fine.
It was just kind of a dumb mistake. The 1/2 watt carbon are in the "pass" components box, filed under "fails to pass digital hetrodynes from CD" use. And input circuit in-series resistor is about all they're good for. It certainly was a horrible choice for a shunting component. Oh well, I learned.
Thanks again! Oh gosh! I'd bet that I shouldn't use 1/2 watt carbon for the amplifier's speaker output zobel's resistor either, because it won't shunt RF very well that way. Is that about right?
danielwritesbac:
We are all still missing some important piece of information. Bleeder resistors do not of and by themselves make noise, at least not the kind of noise that you report.
Bleeder resistors can increase ripple in power supplies with undersized filter capacitors, bleeder resistors can decrease headroom on regulated power supplies. Bleeder resistors can reduce the conduction angle and therefore the peak current in the primary rectifiers, but bleeder resistors do not normally make high levels of noise especially when paralleled with a large capacitor.
Please tell us what your supply voltage is and primary capacitor value (did I miss that post?). Something relevant is missing from our understanding and therefore our ability to help you.
Bleeder resistors that disconnect from the circuit when it's operating normally are fine, but this doesn't answer the question about where the noise is coming from.
We are all still missing some important piece of information. Bleeder resistors do not of and by themselves make noise, at least not the kind of noise that you report.
Bleeder resistors can increase ripple in power supplies with undersized filter capacitors, bleeder resistors can decrease headroom on regulated power supplies. Bleeder resistors can reduce the conduction angle and therefore the peak current in the primary rectifiers, but bleeder resistors do not normally make high levels of noise especially when paralleled with a large capacitor.
Please tell us what your supply voltage is and primary capacitor value (did I miss that post?). Something relevant is missing from our understanding and therefore our ability to help you.
Bleeder resistors that disconnect from the circuit when it's operating normally are fine, but this doesn't answer the question about where the noise is coming from.
Hi Daniel,
Andrew is right about this. Also, carbon composition types have very low inductance and are used in RF work. Carbon film resistors are not as noisy, but may have some small amount of inductance. Metal film types can be very inductive, or not. This depends on the manufacturer and the value.
Most bleeder resistors are wire wound. The inductance is not important in this application within reason. For your application, wire wound would be an excellent choice. Metal oxide types are also a favorite resistor classification of mine. Not too inductive and reasonably quiet. They have good surge properties, but wire wound are the best for that.
Finally, hermanv is dead right. Resistor type of presence would have nothing to do with any noise you may have there. You are chasing teh wrong path for this problem.
-Chris
Andrew is right about this. Also, carbon composition types have very low inductance and are used in RF work. Carbon film resistors are not as noisy, but may have some small amount of inductance. Metal film types can be very inductive, or not. This depends on the manufacturer and the value.
Most bleeder resistors are wire wound. The inductance is not important in this application within reason. For your application, wire wound would be an excellent choice. Metal oxide types are also a favorite resistor classification of mine. Not too inductive and reasonably quiet. They have good surge properties, but wire wound are the best for that.
Finally, hermanv is dead right. Resistor type of presence would have nothing to do with any noise you may have there. You are chasing teh wrong path for this problem.
-Chris
anatech said:
Hi Chris! Thanks! Um, I seem to lack for a baseline.
I need a starting place.
Howabout using some of the supplies that I have on hand?
That's eight Mallory (cornell) SEK 2200uf 50v caps and a 48vct (24+24vAC) transformer and four MR 400 series fast switch diodes.
That makes a power supply board pretty much immediately. There's 10,000uF per rail. However SEK caps of that size don't cover high frequencies.
I think of putting some higher-zoot cap first, right after the diodes. If there's a mistake in the bypass cap practice, its much less of a mistake there (farthest away from the amplifier). Maybe a 22uF like Nichicon KZ or Elna Tonerex? Maybe a 4.7uF like Mallory SK series?
Either will shunt what the big caps cannot.
As the smaller cap is first, then the big caps don't need to shunt the HF--a good bargain in not requiring them to do something they can't do.
Testing went well throughout that part.
Now I'm lost on what to do next.
Next up, and at the opposite end--the output end, is the usual spot to put small ceramic caps and an RC.
Perhaps just an RC?
Anyway, 10,000uF makes a boost, which also has a sound. This sound needs masked. That's an RC network needed. I don't know what values are recommended.
Now, that's where I'm really lost. Help?
I don't think you mentioned how many watts your amplifier delivers. At any thing more than 50 watts, I would guess that 10,000 uF (8,800 actually?) is on the small side.
In your case, the ripple added by the bleeders apparently puts you over the edge. 2,200 Ohms across 8,800 uF causes about 21mV of ripple for a full bridge or 42mV for half wave, this is added to the 60/120Hz droop caused by the load of the power amp.
(For constant current: Amps * Seconds = Volts * Farads. So 21mA times 8.33mS = 21mV times .0088 Farads)
For a 50 watt amplifier, droop due to load is around 2 V p-p so you can see that the few microvolts of resistor Kelvin noise is lost in the main ripple.
In your case, the ripple added by the bleeders apparently puts you over the edge. 2,200 Ohms across 8,800 uF causes about 21mV of ripple for a full bridge or 42mV for half wave, this is added to the 60/120Hz droop caused by the load of the power amp.
(For constant current: Amps * Seconds = Volts * Farads. So 21mA times 8.33mS = 21mV times .0088 Farads)
For a 50 watt amplifier, droop due to load is around 2 V p-p so you can see that the few microvolts of resistor Kelvin noise is lost in the main ripple.
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