Amps or Recievers Postulate or Question

[MacManNM]

This is correct. What I was wondering, is why in the world is anyone talking about half wave rectification supplies for any type of power amp. It has been known for decades that this significantly increases the eddy losses in the core, the ripple power losses in the caps, and the peak diode junction temperatures. You should have stated up front that your examples were not consistent with currently accepted amplifier design practices..

Comments on the following...note that I did not copy all of the correct postings, but some that are interesting..

Incorrect. It is twice the power..10dB is twice as loud.

3 dB is indeed measured as twice as much power. But it is not perceived as twice as loud, no matter what the relative distance or level is. It requires 10dB increase of power at any power level, to be perceived as twice as loud.

See the last statement. The relation is intensity independent.

No. it is 10dB, it is not a matter of perspective.

No. See above.

Incorrect for modern power amps. However, included here for continuity, as the statement half wave has since been introduced.

Incorrect. Line frequency plays a big role in how long the cap has to wait until the transformer voltage starts to exceed the voltage the capacitors were discharged to.

All are tradeoffs. Good amps with small caps are possible as well as large caps. Most designers do not understand well enough e/m field theory to design extremely good amps with small cap banks. A limit to a power amp is of course the input current, as ya can't get more than what the breaker will allow..for long.

Correct,

Incorrect. The caps at the primary supply are fed by rectified sine, with the current occuring near the peak of the voltage wave.

Incorrect. The cap voltage will decay between charge pulses based on the average current that is drawn from the cap. It will rise up, riding the ac, when the ac exceeds the voltage of the cap at the instant. If there is insufficient power capability for recharging, the cap voltage will start a downward trend, looking like a sawtooth of sorts, until the conduction angle reaches equilibrium where the power input matches the power draw.

An interesting point here. The 10 Khz ripple will be on the supply capacitors, and when the diode conducts to recharge the cap, this ripple voltage will be superimposed on the transformer secondary, and through to the primary, where is will go into the AC power line. Under no circumstances, is this ripple current considered when discussing line cords..given that this is 166 times line frequency, what prevents this from coupling to the input ground loop..remember, the faraday coupling constant is proportional to frequency, if you have hum at a 1 millivolt level, you will have 10K freq at 166 mV..audible..no?

As was pointed out, it is 1.414, or 1/.7071.

Cheers, John

Well, when I listen to ~ 1kHz test tone and I increase the amplitude by a factor of 2, it sounds twice as loud to me.

DC rail voltage cannot be 1.414 x peak amplitude. You cannot have a DC rail higher than the peak value, speaking from 0 to peak as in a rectified sine wave(which in reality is peak to peak). Now if you are speaking of peak, prior to the rectified signal voltage, then you multiply by 1.414 and you are correct, but speaking in the other context, multiply by 0.707 will get you in the ballpark.

 

[jneutron]

Well, when I listen to ~ 1kHz test tone and I increase the amplitude by a factor of 2, it sounds twice as loud to me.

Increasing the amplitude by a factor of 2 is a power increase of 4. Is that what you meant?
Published work spanning at least the last 30 years specifically details the fact that it requires 10dB power increase to be discerned as twice as loud. In that 30 years, nobody has published information to the contrary.
I would tend to trust published peer review studies of perceived music levels over an informal one person test at 6dB level. You may be onto something in that human hearing may react differently to pure sines, but the topic here has been music, broad bandwidth audio content.

DC rail voltage cannot be 1.414 x peak amplitude. You cannot have a DC rail higher than the peak value, speaking from 0 to peak as in a rectified sine wave(which in reality is peak to peak). Now if you are speaking of peak, prior to the rectified signal voltage, then you multiply by 1.414 and you are correct, but speaking in the other context, multiply by 0.707 will get you in the ballpark.

You should re-read my post a bit slower.

I quoted you exactly (cut and paste, actually), as saying:

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..

When I see the words "DC line voltage", as you stated, i think output DC voltage, as the transformer does not put out DC.

The transformer provides a rms voltage to the rectifier. The peak of this is 1.414 times the rms value. The capacitors will charge to this 1.414 times rms value with no load. For this value to go to .7 of peak requires a very large ripple current, with an AC voltage of 30 % of the peak superimposed on the cap..this is way too large.

I believe you meant "AC line voltage", and were referring to the transformer output voltage? If you meant "DC rail voltage", you are speaking of ripple voltages that are not useable in power amps.

Cheers, John

 

[mtrycrafts]

....Mtrycrafts, I'm disappointed others besides you and me don't ask more questions....don't they have a desire to learn?.....

It appears that is the case. What can I say. One can also learn a lot from mistakes as well, a very good learning tool. Learning is difficult, maybe that is the hindrance

 

[MacManNM]

Increasing the amplitude by a factor of 2 is a power increase of 4. Is that what you meant?
Published work spanning at least the last 30 years specifically details the fact that it requires 10dB power increase to be discerned as twice as loud. In that 30 years, nobody has published information to the contrary.
I would tend to trust published peer review studies of perceived music levels over an informal one person test at 6dB level. You may be onto something in that human hearing may react differently to pure sines, but the topic here has been music, broad bandwidth audio content.


You should re-read my post a bit slower.

I quoted you exactly (cut and paste, actually), as saying:


When I see the words "DC line voltage", as you stated, i think output DC voltage, as the transformer does not put out DC.

The transformer provides a rms voltage to the rectifier. The peak of this is 1.414 times the rms value. The capacitors will charge to this 1.414 times rms value with no load. For this value to go to .7 of peak requires a very large ripple current, with an AC voltage of 30 % of the peak superimposed on the cap..this is way too large.

I believe you meant "AC line voltage", and were referring to the transformer output voltage? If you meant "DC rail voltage", you are speaking of ripple voltages that are not useable in power amps.

Cheers, John

I'm saying that if you take this signal:

And multiply by 1.414 (as you suggest) you will never get 14.14 RMS out of it. The best you can hope for is ~* 7.07 RMS period. Ergo, DC rail can't be any higher than 7.07 (arb) I hope you understand what I'm talking about. Perhaps I need to be more clear in my posts.

As far as perception, people say you can't see 800nm light, yet somehow I can see it, and I do everyday. Everyone is different. And yes, I meant power.

 

[PENG]

I'm saying that if you take this signal:

And multiply by 1.414 (as you suggest) you will never get 14.14 RMS out of it. The best you can hope for is ~* 7.07 RMS period. Ergo, DC rail can't be any higher than 7.07 (arb) I hope you understand what I'm talking about. Perhaps I need to be more clear in my posts.

As far as perception, people say you can't see 800nm light, yet somehow I can see it, and I do everyday. Everyone is different. And yes, I meant power.

Assuming this is a rectified (full wave)from a pure sine wave, the peak is 10, so rms = 10/1.414=7.07 but it should be able to charge caps to just under rmsX1.414, i.e. just under 10.

I am curious, why not?

 

[MacManNM]

Assuming this is a rectified (full wave)from a pure sine wave, the peak is 10, so rms = 10/1.414=7.07 but it should be able to charge caps to just under rmsX1.414, i.e. just under 10.

I am curious, why not?

In theory without any semiconductor (or any other) loss it would be able to charge the caps to 10. Now to keep it there (with draw) is a different story.

 

[PENG]

In theory without any semiconductor (or any other) loss it would be able to charge the caps to 10. Now to keep it there (with draw) is a different story.

I forgot to mention that I was referring to no load condition, when there would be leakage only. Under load, the caps won't be able to keep it at 10 as it would discharge.

Back to the line frequency thing, whether it is 50, or 60 Hz, Vrms=0.707*Vp, that's independent of the frequency. In terms of its effect on how fast it can charge up a cap completely, surely the cap will get charge more often at a higher frequency but the a.c. (you call it pulse) would have its peak drop off to 0 quicker as well. I have to think about it some more, may be do some math if I can remember all those calculus.

 

[jneutron]

I'm saying that if you take this signal:

And multiply by 1.414 (as you suggest) you will never get 14.14 RMS out of it. The best you can hope for is ~* 7.07 RMS period. Ergo, DC rail can't be any higher than 7.07 (arb) I hope you understand what I'm talking about. Perhaps I need to be more clear in my posts.

As far as perception, people say you can't see 800nm light, yet somehow I can see it, and I do everyday. Everyone is different. And yes, I meant power.

You stated "DC line voltage", which is a term of no standard meaning in this contect..the picture you provide is far easier to talk about..thanks for it.

Now, the verbage:What you have drawn is a full wave rectified output of a 7.07 volt RMS AC waveform. The peak value of a 7 volt AC waveform is 10 volts. If you apply your waveform to a capacitor, it will charge to a value of 10 volts. This voltage is 1.4 times the AC RMS voltage. This is illustrated in the link I provided earlier.

When the capacitors are loaded, the peak voltage at the capacitor will drop to 10 volts minus the diode drop, which will be around 1 volt for normal silicon at high injection levels.

I recall an article years ago where Pease mentioned the ability to "see" ultraviolet light. As I recall, he stated that it was flourescence of the vitreos(sp) fluid.

Peng: it is not calculus. If you discharge the capacitors with a constant current, the voltage will drop linearly with time. This allows us to simply draw a straight line from the top of the waveform (minus say a volt diode drop), with a slope determined by the equation (dv/dt = I/C)..1 ampere pulled out of 1 farad will cause a drop of 1 volt per second..two amps will result in a drop of two volts per second.

Cheers, John

 

[MacManNM]

You stated "DC line voltage", which is a term of no standard meaning in this contect..the picture you provide is far easier to talk about..thanks for it.

Now, the verbage:What you have drawn is a full wave rectified output of a 7.07 volt RMS AC waveform. The peak value of a 7 volt AC waveform is 10 volts. If you apply your waveform to a capacitor, it will charge to a value of 10 volts. This voltage is 1.4 times the AC RMS voltage. This is illustrated in the link I provided earlier.

When the capacitors are loaded, the peak voltage at the capacitor will drop to 10 volts minus the diode drop, which will be around 1 volt for normal silicon at high injection levels.

I recall an article years ago where Pease mentioned the ability to "see" ultraviolet light. As I recall, he stated that it was flourescence of the vitreos(sp) fluid.

Peng: it is not calculus. If you discharge the capacitors with a constant current, the voltage will drop linearly with time. This allows us to simply draw a straight line from the top of the waveform (minus say a volt diode drop), with a slope determined by the equation (dv/dt = I/C)..1 ampere pulled out of 1 farad will cause a drop of 1 volt per second..two amps will result in a drop of two volts per second.

Cheers, John

I gues we were saying the same thing, i just couldnt articulate it well enough.

800nm is near IR, not UV. 266nm would be UV, and it does flouress eveything, due to to the greater amount of energy at that wavelength.

 

[jneutron]

I gues we were saying the same thing, i just couldnt articulate it well enough. .

The pic helped..thanks..

800nm is near IR, not UV. 266nm would be UV, and it does flouress eveything, due to to the greater amount of energy at that wavelength.

Uh, yah..I know..

Actually, not everything flouresces as a result of UV energy, but apparently, the fluid within the eye can..

Cheers, John

 

[PENG]

Peng: it is not calculus. If you discharge the capacitors with a constant current, the voltage will drop linearly with time. This allows us to simply draw a straight line from the top of the waveform (minus say a volt diode drop), with a slope determined by the equation (dv/dt = I/C)..1 ampere pulled out of 1 farad will cause a drop of 1 volt per second..two amps will result in a drop of two volts per second.

Cheers, John

John, you said its not calculus yet you cited "dv/dt......"

Regardless, I think you misunderstood my last point. Of course you are right about what you said above and calculus is not needed to prove anything, but I was talking about the effect of line frequency on how long it takes to charge up those blocking caps. Mac said it had an effect. I was simply saying that I had to think about it before I could fully agree. And yes, some calculus (at least not Fourier) may be helpful, if not needed, to prove it mathmatically. I bought into Mac's claim at first, then I started to wonder if frequency really matters all that much.

Whether it is 50 or 60 Hz, as long as it is a sine wave, Vrms=square root 2*Vpeak. A higher frequency wave will reach its peak more frequently but it also dips below it more frequently. In the end I am not sure if a 60 Hz sine wave will charge up a cap faster than a 50 Hz one. That's all I am saying for now until I have time to think it through. Until then I am back sitting on the fence.

Thanks anyway.

 

[jneutron]

John, you said its not calculus yet you cited "dv/dt......".

Well, yah ok...ya got me..

Regardless, I think you misunderstood my last point. Of course you are right about what you said above and calculus is not needed to prove anything, but I was talking about the effect of line frequency on how long it takes to charge up those blocking caps. Mac said it had an effect. I was simply saying that I had to think about it before I could fully agree. And yes, some calculus (at least not Fourier) may be helpful, if not needed, to prove it mathmatically. I bought into Mac's claim at first, then I started to wonder if frequency really matters all that much.

Whether it is 50 or 60 Hz, as long as it is a sine wave, Vrms=square root 2*Vpeak. A higher frequency wave will reach its peak more frequently but it also dips below it more frequently. In the end I am not sure if a 60 Hz sine wave will charge up a cap faster than a 50 Hz one. That's all I am saying for now until I have time to think it through. Until then I am back sitting on the fence.

Thanks anyway.

Here's a pic I drew..

Top 2 waves need no explanation.

Third and fourth..

The third pic represents the supply being very heavily loaded. I have detailed the portion of the waveform where the diodes are conducting and recharging the capacitors. The fourth is the same, but for a lighter load on the caps. If you look, I have drawn a straight line, this represents a constant current draw from the capacitors. In reality, this line will have wiggle to it, a 10 khz draw will look like a staircase with sloped risers where draw happens, flat where the opposite polarity channel is drawing..

Note that the supply voltage drops along the sloped line until the bridge output voltage comes back up to it. This is where the bridge starts to conduct again, recharging the capacitor. If the system feeding the bridge is unable to provide enough current to charge, this load will bring the supply voltage downward until the conduction wtime width is enough to establish equilibrium.

For a 50 hz system, the time from red line to blue line will be slightly longer, the drop will be a bit more.

For most supply systems, the cap bank will return to the voltage at the red line time..

Also consider the way this system pulls current from the wall outlet. It is only doing this from the blue line to the red line..this is haversine draw.

Cheers, John

 

[mtrycrafts]

Here's a pic I drew..

Top 2 waves need no explanation.

Third and fourth..

The third pic represents the supply being very heavily loaded. I have detailed the portion of the waveform where the diodes are conducting and recharging the capacitors. The fourth is the same, but for a lighter load on the caps. If you look, I have drawn a straight line, this represents a constant current draw from the capacitors. In reality, this line will have wiggle to it, a 10 khz draw will look like a staircase with sloped risers where draw happens, flat where the opposite polarity channel is drawing..

Note that the supply voltage drops along the sloped line until the bridge output voltage comes back up to it. This is where the bridge starts to conduct again, recharging the capacitor. If the system feeding the bridge is unable to provide enough current to charge, this load will bring the supply voltage downward until the conduction wtime width is enough to establish equilibrium.

For a 50 hz system, the time from red line to blue line will be slightly longer, the drop will be a bit more.

For most supply systems, the cap bank will return to the voltage at the red line time..

Also consider the way this system pulls current from the wall outlet. It is only doing this from the blue line to the red line..this is haversine draw.

Cheers, John

Thanks for the pictures. So then the caps are always discharging and charging? How would the discharge line shift if the cap didn't fully charge on a 1/2 cycle? Would the discharge point be below the peak value at the voltage point it was only able to charge to? Discharge line coming from the peak point would indicate a charged cap?

 

[jneutron]

Thanks for the pictures. So then the caps are always discharging and charging?

Yes, always.

How would the discharge line shift if the cap didn't fully charge on a 1/2 cycle? Would the discharge point be below the peak value at the voltage point it was only able to charge to?

Yes, it would be below.
This is why I said ""If the system feeding the bridge is unable to provide enough current to charge, this load will bring the supply voltage downward until the conduction wtime width is enough to establish equilibrium."" If the supply voltage really gets dragged down, the blue line will shift to the right, and the diodes will conduct for a longer time. If the supply voltage was dragged down to zero volts, the diodes would be conducting all the time. This is very difficult for the diodes also..

Discharge line coming from the peak point would indicate a charged cap?

Yes. I took liberty with the drawing of course, as the last quarter to half volt at the peak will be mushy because the diode is a non linear device. But the overall details are consistent to high power operations.

Cheers, John

 

[mtrycrafts]

Yes. I took liberty with the drawing of course, as the last quarter to half volt at the peak will be mushy because the diode is a non linear device. But the overall details are consistent to high power operations.

Cheers, John

Thanks again. Then this would tell me that the cap is charged up by the time the peak voltage is reached or as it is reached, less than 8 ms, but as you indicated, the caps always discharge and charge, I suppose when it is delivering any power to the load.