Choose an amplifier for me

[Dan Banquer]

Chose an amp for me

I wonder how an amp does this magic?

Air can be had with lots of phase shift as would the depth be affected, or the recording and speakers or room acoustics. But an amp? Something funny is going on in there.

It also can be a function of noise. Remember: when you are running fairly efficent speakers your average power is lower so noise becomes a more critical factor. Right?
d.b.

 

[PENG]

Well, you answered the question. It is on the DC side, so it is not limited by the line frequency as Mac is implying. The implication is clear, especially when he included the European line frequency.
We were not discussing how long it took due to the amount of discharge, nor did he even mention the RC time constants, just the limiting factor of the line frequency of 60 Hz, 16ms.

Now that he clarified this, he's right, the line frequency shoud make a difference because on the DC side, the waveform without cap effect, is pulsating (from 0 to peak to 0) even with full wave rectification. As you mentioned before though, the line frquency is 60 Hz (in NA) fixed anyway.

 

[mtrycrafts]

Now that he clarified this, he's right, the line frequency shoud make a difference because on the DC side, the waveform without cap effect, is pulsating (from 0 to peak to 0) even with full wave rectification. As you mentioned before though, the line frquency is 60 Hz (in NA) fixed anyway.

So, if the RC time constant is a few microseconds, why couldn't the cap charge during those microsecond intervals where the line voltage is above the cap rated voltage? Of course it will charge. And, it will charge as long as the wave form is above the cap voltage on each waveform, no? After all, the voltage will be above for a considerable amount of time, no?

If it cannot, how does it charge when the frequency demand is above the line frequency? How does it charge at 10kHz with full power?

 

[mtrycrafts]

When the rectifier diodes change the AC to DC they change it to a 120 Hz positive or negative going wave. The positive DC upply will look like the two halves of the sinusoidal wave only they are going from about zero volts to whatever the peak positive is, and the negative DC supply does just that but only going in the negative direction.
In any case I wish I could draw the wave forms, becasue a picture would explain a thousand words.
Now that we sort of have that explained, you must remember that recharging a cap is not only a function of current it is also a function of voltage. so not only does the cap have to refill it's resovoir of current it also must climb back to the nominal voltage. In essence John is right: it does take at least one full cycle to recharge, that's about 17 mS. He is also saying that if we had a higher line frequency the recharge time would be even less. Think of it this way: if we take the reciprocal of 60 hZ we get about 17mS. if the line frequency was at 120 Hz than we could be recharging those caps at about 8.5 mS. For those of you who are math challenged the reciprocal is 1 divided by a number, so when I say the reciprocal of 60 Hz it is 1 divided by 60 Hz which comes out to about 17 mS. So the reciprocal of a known frequency is also the duration in time it gets to do one full cycle.

Does everyone understand?
d,b,

But if that RC time constant to recharge is in micro seconds, you have a considerable amount of time where the line voltage is swinging above the cap value, no?

And, any signal above 120 Hz for a duration would deplete the caps? Where does the RMS power coming from, the line voltage alone? Maybe talking with Chris Russel will clear this up for me as one question leads to others.

 

[Dan Banquer]

I think what we are trying to tell you here is that this just is not about RC time constants. In fact the RC time constant factor appears to be somewhat minor as opposed to frequency of the rectified voltage, especially when you think about the resistance of the secondary of a transformer.
d.b.

 

[PENG]

But if that RC time constant to recharge is in micro seconds, you have a considerable amount of time where the line voltage is swinging above the cap value, no?

And, any signal above 120 Hz for a duration would deplete the caps? Where does the RMS power coming from, the line voltage alone? Maybe talking with Chris Russel will clear this up for me as one question leads to others.

Most of the time, the output voltage at the amp output don't come close to the d.c. rail voltage so the caps don't really get drained to begin with, until the going get tough.

 

[MacManNM]

When the rectifier diodes change the AC to DC they change it to a 120 Hz positive or negative going wave. The positive DC upply will look like the two halves of the sinusoidal wave only they are going from about zero volts to whatever the peak positive is, and the negative DC supply does just that but only going in the negative direction.
In any case I wish I could draw the wave forms, becasue a picture would explain a thousand words.
Now that we sort of have that explained, you must remember that recharging a cap is not only a function of current it is also a function of voltage. so not only does the cap have to refill it's resovoir of current it also must climb back to the nominal voltage. In essence John is right: it does take at least one full cycle to recharge, that's about 17 mS. He is also saying that if we had a higher line frequency the recharge time would be even less. Think of it this way: if we take the reciprocal of 60 hZ we get about 17mS. if the line frequency was at 120 Hz than we could be recharging those caps at about 8.5 mS. For those of you who are math challenged the reciprocal is 1 divided by a number, so when I say the reciprocal of 60 Hz it is 1 divided by 60 Hz which comes out to about 17 mS. So the reciprocal of a known frequency is also the duration in time it gets to do one full cycle.

Does everyone understand?
d,b,

That's exactly what I was trying to say, thanks. Your ability to articulate is surpassed only by your knowledge and your ability to design amps.

 

[MacManNM]

One must also remember that the DC line voltage is usualy somewhere around .707 x the rectified peak voltage, usually a little less though. This means it will take a good portion of the waveform to charge them. Also depending on the size of the power supply (size of the caps), it may well take several cycles to charge the caps completely.

 

[MacManNM]

Ok Dan, since you are so good at splainin' stuff, why don't you teach the kids how a switching PS works!

 

[mtrycrafts]

Most of the time, the output voltage at the amp output don't come close to the d.c. rail voltage so the caps don't really get drained to begin with, until the going get tough.

Where is that point of 'getting tough?' Yes, but, the caps still get recharged when they are drained any amount, no? How fast can it or does it get recharged.

 

[PENG]

Where is that point of 'getting tough?' Yes, but, the caps still get recharged when they are drained any amount, no? How fast can it or does it get recharged.

From time to time the caps will discharge and be recharged but not continuously because the music do not always demand the output voltage of the amp(s) to get close to the supplied voltage (D.C. side). It is really hard to say "how fast" as it depends on so many varying parameters.

The caps should start getting recharged as soon as the voltage across them drops below the a.c. voltage peak but as others explained, still only on point on wave (waveform of the a.c. voltage) basis only. Just for information, Vrms=VpeakX0.707 or Vpeak=VrmsX1.414. Under no load condition, the caps should be able to hold the voltage close to Vpeak, that is square root 2 (or 1.414)XVrms less any small voltage drop in the rectifier circuit.

 

[mtrycrafts]

From time to time the caps will discharge and be recharged but not continuously because the music do not always demand the output voltage of the amp(s) to get close to the supplied voltage (D.C. side). It is really hard to say "how fast" as it depends on so many varying parameters.

The caps should start getting recharged as soon as the voltage across them drops below the a.c. voltage peak but as others explained, still only on point on wave (waveform of the a.c. voltage) basis only. Just for information, Vrms=VpeakX0.707 or Vpeak=VrmsX1.414. Under no load condition, the caps should be able to hold the voltage close to Vpeak, that is square root 2 (or 1.414)XVrms less any small voltage drop in the rectifier circuit.

Yes, I see that the only time the caps can recharge, if needed, is during those two halfs cycles of the line frequency and when the voltage is more.

Talking to Chris Russell at Bryston, secondary resistance of the transformer is a small fraction of an ohm. So depending on the amount it needs to charge, the cap size, is how fast it will charge. And it seems that some will be recharged much quicker than the cycle time and deeper discharges will take longer, depending on inrush current limitations as well

 

[jneutron]

My goodness, power supply thinkings??

Wow, some of the ideas...

Take a look at this:

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/rectbr.html

Look at the RC example..see where that output voltage tracks the peaks of the ac waveform? That happens every 8.33 milliseconds this side of the pond.

The linear downslope there represents the discharging of the supply caps during the point in the ac waveform when the transformer voltage is below that of the cap bank. By increasing the capacitance, that slope goes more horizontal, meaning the supply voltage droops less..the caveat is that when the ac voltage gets about a volt above where the caps discharged to, there will be a large inrush of current again.

If you look at the percentage of time that the cap bank is tracking the ac input, you can see how wide the haversine pulse will be. Bigger caps, the lower the conduction timeframe and larger the current spikes.

While the harmonics are even order on the cap bank side of the circuit, they are ODD order harmonics on the line side....60, 180, 300hz up.

Design tradeoff issues for a large power amp:

1. Use a small cap bank. This eases up on the bridge surge requirements, as the turn on inrush is not so bad..Problem is, the output pulls the cap bank down faster, meaning that the supply ripple is larger. This forces the designer to raise the supply voltage to compensate. This in turn means the output devices have to drop more voltage causing increased power dissipation. Also, the smaller cap bank may dissipate too much as a result of ripple current.

2. Use an extremely large cap bank. This allow lower rails, lower output dissipation, lower ripple. This makes it harder for the diodes at turnon, as the inrush current peak can exceed the transient current capabilities of the bridge. For example, the GI GBPC35xx series bridges run 35 amps rms, 400 amp single cycle (16ms) surge current superimposed on the 35 amps, or 660 amps for surge greater than 1 ms and less than 8.3 ms.

I am suprised by the amount of incorrect information that is being bandied about here..

Cheers, John

Edit note:

The data on the GI GBPC35xx series bridges, from edition 10, has the IFSM spec stating two half cycles of 8.3 mSec, for a total of 16.6. That data is incorrect..For the life of me, I've no idea why they spec'd it that way. It is a nonsensical thing. Nobody else does it or specs it that way. It SHOULD read a single cycle of 8.3 mSec superimposed on the operating current, which is the way the rest of the world does it.

The idiot responsible for updating that entire catalog of 640 pages should be shot, or at least flogged..

Course, I only had two weeks to review and correct the entire catalog, so..oops..

 

[Dan Banquer]

Choose an amp for me.

Hi John;
I see your clouding the issue with facts again. Inrush current can be eased with any number of techniques.
BTW I recently got one of those Kill A Watt meters for cheap money off the net. They plug into the AC line and measure, Volts, VA, Watts, Frequency, Power Factor, and Kilowatt hours. Not bad fro 28.00. They claim 0.2%accuracy.
d.b.

 

[jneutron]

Hi John;
I see your clouding the issue with facts again. Inrush current can be eased with any number of techniques.
BTW I recently got one of those Kill A Watt meters for cheap money off the net. They plug into the AC line and measure, Volts, VA, Watts, Frequency, Power Factor, and Kilowatt hours. Not bad fro 28.00. They claim 0.2%accuracy.
d.b.

Went to download the manual, but they are currently in an upgrade cycle, none available on the web. Did you get one with the unit?

Is the meter capable of measuring to .2% accuracy for haversines, or is it designed to duplicate the reading of only pure sine draw regardless of lag angle? Will switchmodes or bridge supplies or dimmers read properly?

Cheers, John

 

[Dan Banquer]

Choose an amplifier for me.

There wasn't much of a manual as I recall, and I don't know what to tell you about switchers and haversines. It appears that this was designed and marketed for household appliances but it certainly can be used for linear power supply draw. Some initial testing has been very informative.
Sorry I couldn't be much help here.
d.b.

 

[WmAx]

There wasn't much of a manual as I recall, and I don't know what to tell you about switchers and haversines. It appears that this was designed and marketed for household appliances but it certainly can be used for linear power supply draw. Some initial testing has been very informative.
Sorry I couldn't be much help here.
d.b.

I came acoss this comparison of the Watt-A-Meter on an inductive vs. mostly resistive load:

http://www.safehomeproducts.com/SHP2/data/manuals/Kill-A-Watt_Meter_Report.pdf

Certainly appears to be an excellent value. You can get it for under $25 shipped on ebay, new in box. Investigating the source of this product, it seems to actually be a re-badged product from Prodigit, an established test equipment manufacturer, marketing under the P3 name as a consumer product in the U.S. Refer to the Prodigit consumer product 2000CS.

-Chris

 

[PENG]

I am suprised by the amount of incorrect information that is being bandied about here..

Cheers, John

You should be surprised if there were no incorrect information when the subject involve a little beyond introduction to physics and electrical engineering.