I recently put together my first amplifier. As much as I like the sound and am please with what I have done, I can take only a small portion of the credit. I owe it all to Forum members.
I was working from Sanyos datasheet but would now like to know more about supply capacitors and bleed resistors for a chip amp power supply. I have been doing my homework. Some of this is unclear to me. I would like some help to understand how supply capacitor and bleed resitor values will affect my amplifier in the area of sound, safety and reliability.
Thanks
I was working from Sanyos datasheet but would now like to know more about supply capacitors and bleed resistors for a chip amp power supply. I have been doing my homework. Some of this is unclear to me. I would like some help to understand how supply capacitor and bleed resitor values will affect my amplifier in the area of sound, safety and reliability.
Thanks
You could choose the minimum capacitance (per rail) to use, for unhindered bass response down to frequency f, with:
C ≥ imax / ( πf∙(Δv - (ESR∙imax)))
which was derived in the post at http://www.diyaudio.com/forums/power-supplies/216409-power-supply-resevoir-size-169.html#post3320547 .
and gives the capacitance value, C, that would be required in order to supply the current for one half-cycle of a sine signal of frequency f (in Hz), with a 0-to-peak amplitude of imax Amperes, while causing the voltage across the capacitor to dip by no more than your choice of Δv Volts. Note that no re-charging pulses are assumed, so far.
imax = √(2 ∙ P_rated_rms / R_load) (that's a square root; the √ symbol is a bit hard to read)
With the C equation above, you could first let ESR = 0, then calculate C, then find the ESR for that C at frequency f, then recalculate C with the ESR you found, and you might need to iterate like that until C does not change significantly.
Or, using an approximate expression for ESR,
ESR = 0.02 / (C x VR), where VR = Voltage Rating of capacitor, and substituting it into the first equation, you could use:
C ≥ imax ((0.04πf / VR) + 1) / (πf∙Δv) [approximate, for electrolytic cap, only]
If we assume that the charging pulses would always be sufficient to recharge the capacitors as much as needed, then the two equations above would probably become the two equations below (which are not fully simplified; sorry):
Capacitance Needed for a Sine Down to Bass Frequency = f, at Max Rated Power:
(1) C ≥ ( (imax ∙ f ) / ( 2 ∙ fmains )) / ( π ∙ f ∙ (Δv - ( ESR ∙ imax )))
or
(2) C ≥ (( imax ∙ f ) / ( 2 ∙ fmains ))∙(( ( 0.04 ∙ π ∙ f ) / VR ) + 1) / ( π ∙ f ∙ Δv ) [approximate, for electrolytic cap, only]
where VR = Voltage Rating of capacitor.
-----
But, the above is only for a single sine at a time. For it to be truly "bulletproof", for any signal shape, including DC at the maximum rated peak sine voltage and current, you would want to use:
C ≥ (imax∙Δt ) / (Δv - (ESR∙imax))
which gives C as the capacitance that will allow only up to some desired maximum rail-voltage dip = Δv, when a specified maximum rated current imax is pulled out of the cap for a specified time Δt.
In this case, for a full wave bridge rectifier, we would use the time between charging pulses as Δt, i.e. Δt = 1/2f, where f is the AC mains frequency (50 Hz or 60 Hz), giving:
Capacitance Needed for Worst-Case Signal Shape, at Max Rated Power:
(3) C ≥ ( 1 /( 2 ∙ fmains )) ∙ imax / (Δv - ( ESR ∙ imax ))
Or, using the same approximation for ESR as was used before:
(4) C ≥ ( imax / Δv ) ∙ ( ( 0.5 / fmains ) + ( 0.02 / VR )) (approximation, for electrolytic capacitors ONLY)
where VR = Voltage Rating of capacitor.
Note that if the ESR is not "much lower" than the desired Δv / imax (which is also the maximum impedance we want the load to see), then the required capacitance value would become excessive.
Cheers,
Tom
C ≥ imax / ( πf∙(Δv - (ESR∙imax)))
which was derived in the post at http://www.diyaudio.com/forums/power-supplies/216409-power-supply-resevoir-size-169.html#post3320547 .
and gives the capacitance value, C, that would be required in order to supply the current for one half-cycle of a sine signal of frequency f (in Hz), with a 0-to-peak amplitude of imax Amperes, while causing the voltage across the capacitor to dip by no more than your choice of Δv Volts. Note that no re-charging pulses are assumed, so far.
imax = √(2 ∙ P_rated_rms / R_load) (that's a square root; the √ symbol is a bit hard to read)
With the C equation above, you could first let ESR = 0, then calculate C, then find the ESR for that C at frequency f, then recalculate C with the ESR you found, and you might need to iterate like that until C does not change significantly.
Or, using an approximate expression for ESR,
ESR = 0.02 / (C x VR), where VR = Voltage Rating of capacitor, and substituting it into the first equation, you could use:
C ≥ imax ((0.04πf / VR) + 1) / (πf∙Δv) [approximate, for electrolytic cap, only]
If we assume that the charging pulses would always be sufficient to recharge the capacitors as much as needed, then the two equations above would probably become the two equations below (which are not fully simplified; sorry):
Capacitance Needed for a Sine Down to Bass Frequency = f, at Max Rated Power:
(1) C ≥ ( (imax ∙ f ) / ( 2 ∙ fmains )) / ( π ∙ f ∙ (Δv - ( ESR ∙ imax )))
or
(2) C ≥ (( imax ∙ f ) / ( 2 ∙ fmains ))∙(( ( 0.04 ∙ π ∙ f ) / VR ) + 1) / ( π ∙ f ∙ Δv ) [approximate, for electrolytic cap, only]
where VR = Voltage Rating of capacitor.
-----
But, the above is only for a single sine at a time. For it to be truly "bulletproof", for any signal shape, including DC at the maximum rated peak sine voltage and current, you would want to use:
C ≥ (imax∙Δt ) / (Δv - (ESR∙imax))
which gives C as the capacitance that will allow only up to some desired maximum rail-voltage dip = Δv, when a specified maximum rated current imax is pulled out of the cap for a specified time Δt.
In this case, for a full wave bridge rectifier, we would use the time between charging pulses as Δt, i.e. Δt = 1/2f, where f is the AC mains frequency (50 Hz or 60 Hz), giving:
Capacitance Needed for Worst-Case Signal Shape, at Max Rated Power:
(3) C ≥ ( 1 /( 2 ∙ fmains )) ∙ imax / (Δv - ( ESR ∙ imax ))
Or, using the same approximation for ESR as was used before:
(4) C ≥ ( imax / Δv ) ∙ ( ( 0.5 / fmains ) + ( 0.02 / VR )) (approximation, for electrolytic capacitors ONLY)
where VR = Voltage Rating of capacitor.
Note that if the ESR is not "much lower" than the desired Δv / imax (which is also the maximum impedance we want the load to see), then the required capacitance value would become excessive.
Cheers,
Tom
Haha ????
I spent a large portion of my life as an Auto mechanic, breathing in hydrocarbons by day and drinking beer by night. I have decided to take a different direction in my life but am still back at Ohms law!
I was hoping for an answer like "keep Cap voltage close to rail voltage or have tons of capacity". Same with the Capacitance. I have two Capacitors after my rectifier and I have heard of using everything from 4700 uF to 10000uF. I know it makes a difference but I do not know what difference.
I dream of the day that I can understand and implement the formulas that you have provided. At 44 years old, I am getting a late start and hope that it is possible for me.
Thanks
I spent a large portion of my life as an Auto mechanic, breathing in hydrocarbons by day and drinking beer by night. I have decided to take a different direction in my life but am still back at Ohms law!
I was hoping for an answer like "keep Cap voltage close to rail voltage or have tons of capacity". Same with the Capacitance. I have two Capacitors after my rectifier and I have heard of using everything from 4700 uF to 10000uF. I know it makes a difference but I do not know what difference.
I dream of the day that I can understand and implement the formulas that you have provided. At 44 years old, I am getting a late start and hope that it is possible for me.
Thanks
Bleeder resistors should do nothing but bleed (and provide some voltage stabilization). They shouldn't affect the sound from your amp. To allow the reservoir caps to discharge when the amp is turned off (and therefore semiconductor junctions aren't biased on in the circuit ahead and rectifiers behind) the bleeder resistor allows a path for current. The alternative is a slow self-discharge via cap leakage; I don't know how that might affect component reliability, if at all.
I recently read Don Lancaster's Hardware Hacker column for October 1989, Picking Filter Capacitors. His "magic rule" is,
I recently read Don Lancaster's Hardware Hacker column for October 1989, Picking Filter Capacitors. His "magic rule" is,
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I generally just put a LED on each DC power rail which suits the purpose of telling me that both power rails are working with the added benefit of discharging the capacitors after it has been turned off ......
Just for a reference...in my amp supply I've used two 820 ohm (5 Watt) to "bleed" a total capacitance of about 33000 uF @ 56 Volt per rail (the discharge current is about 70 mA).
Regards,
Roberto
Regards,
Roberto
Sorry. I didn't mean for it to be overwhelming. (But also note that I just re-learned much of that math, in the last few years, after letting it rust for about 30 years. So don't give up hope.)
You don't need those equations. That's one of the reasons this forum is here, after all. And if you didn't have people to ask, in places like this, you still wouldn't REALLY need them. But then you'd have to go to extremes, to be sure it was enough. So the equations just give the absolute least capacitance, and also the absolute most capacitance you would ever need, so you can figure out what range to shoot for.
Basically, for a PSU, more capacitance is always better, and multiple smaller capacitors in parallel are better than one large one. It's also usually important to have a relatively large amount of local decouping capacitance, as close as possible to the chip pins, where multiple parallel caps are still better than one large one, but even more so. (Or, better yet, make the PSU output be right at the chip's power supply pins.)
What happens is that when the chipamp's power pins demand more current, to send to the speakers, the inductance (and resistance) of the wiring or PCB traces will have an unwanted voltage induced across them, and will also delay the arrival of the current (slightly). And also, of course, as the current is drawn, the voltage across the PSU caps will start to fall (making the ripple). With more PSU capacitance, the voltage will fall less. And with more local decouping capacitance, the current can be delivered more accurately, especially for the leading edges of transients, and there will be less disturbance of the rail voltages.
To use the equations to calculate the needed C value, the only thing you would have to figure out is how much ripple you want to allow. Lower ripple might make the amplifier have less distortion.
But, before worrying about that, there is a more-basic consideration: The ripple shouldn't be allowed to dip down into the portion of the power supply voltage that's reserved for the signal, and for the amp itself. That would cause rather-blatant distortion, like clipping.
The peak rail voltage minus the ripple amplitude (peak-to-peak) minus about 4 Volts where the amp itself sits leaves what voltage range is left for the peak signal voltage.
But the ripple voltage is really just a side-effect type of thing. What the caps are really for is providing big honking surges of CURRENT, accurately and on demand. Or little ones. The capacitor current IS THE MUSIC. The cap currents make the sound come out of the speakers.
So those equations were derived so we could KNOW what cap values to use, for the PSU to even be ABLE to provide the current needed, for the worst-case scenario (i.e. operating at full rated power), while ALSO not making more than "X" amount of ripple.
Basically, you just plug in all of the numbers on the right side of the equation, turn the crank, and the needed C value pops out.
EXAMPLE:
An amp is rated at 100 Watts RMS power. Speaker is 8 Ohms. How much PSU capacitance is needed?
First, find the maximum peak current needed, through the 8 Ohms load:
imax = √(2 ∙ P_rated_rms / R_load)
In this case, that would be:
imax = √(2 ∙ 100 / 8)
imax = 5 Amps peak
The last two equations (3) and (4) give the worst-case C value, which should be enough for absolutely any possible type of signal, even DC at the level that is normally just the peak level. Equation (4) is easier to use than (3):
C ≥ ( imax / Δv ) ∙ ( ( 0.5 / fmains ) + ( 0.02 / VR ))
Assuming 5 Amps max, 60 Hz AC power, and caps rated at 50V:
C ≥ ( 5 / Δv ) ∙ ( ( 0.5 / 60 ) + ( 0.02 / 50 ))
C ≥ 0.0437 / Δv
which means that if the ripple, Δv, was allowed to be no more than 1 V peak-to-peak, then
C ≥ 0.043700 Farads, or
C ≥ 43700 uF
The example amplifier should never need more than that in its PSU, but less COULD also be OK (assuming 1V ripple is fine).
We can also check what capacitance it would need for a single 20 Hz sine wave at the rated max power, to see how much less capacitance that would require (knowing that THAT C value might not always be enough, since real music is not a sine wave):
The equation is:
(2) C ≥ (( imax ∙ f ) / ( 2 ∙ fmains ))∙(( ( 0.04 ∙ π ∙ f ) / VR ) + 1) / ( π ∙ f ∙ Δv )
Plugging in the numbers from our example makes it:
C ≥ (( 5 ∙ 20 ) / ( 2 ∙ 60 ))∙(( ( 0.04 ∙ 3.14 ∙ 20 ) / 50 ) + 1) / ( 3.14 ∙ 20 ∙ Δv )
You just start at the innermost sets of parentheses and go from there, with a calculator. Eventually, you get:
C ≥ 0.013930 / Δv
C ≥ 13930 uF / Δv
So if we allowed 1 Volt of ripple, max, then C could be anything at or above 13930 uF. Less ripple would require more capacitance and more ripple allowed would require less capacitance. You can also always turn that around and figure out what ripple you would get, for any particular C value.
We COULD then also check to see how much power the amp could actually put out, with a sine wave, before the signal ran out of room and started to distort:
Since RMS power = RMS (sine) voltage squared divided by R (speaker rated Ohms), and any sine voltage peak equals RMS volts times 1.414, we can calculate that the peak sine voltage across the speaker would be 40 Volts, in this example, to make a maximum rated 100 Watts into 8 Ohms. But we have to leave 1 V for ripple and about 4 volts for the amp itself. So if your peak rail voltage wasn't at least 45 Volts, you wouldn't actually be able to get a maximum output of 100 Watts RMS.
To figure out what max power you actually COULD get, you can take the peak rail voltage and subtract "ripple p-p voltage plus about 4 volts", to get the maximum allowable peak signal voltage. Then square that max peak signal voltage and divide it by two times the speaker Ohms, to get the maximum RMS output power.
None of that actually answered the question about how to decide how much ripple to allow. After you have enough capacitance to make the ripple small-enough to still give you a decent maximum power output, you would look at the PSRR (Power Supply Rejection Ratio) of the amp, which tells you how much distortion could be added by variations in the voltage at the power supply pins of the amp. But most of the chipamps are pretty darn good, at least at lower frequencies. If the PSRR is not blatantly bad, then adding more capacitance will still improve the sound, but not by a whole lot, probably. I would probably just go with the higher estimate for the required C, from equation (4), and see if it sounded like it needed more or not. And remember to keep all of your decoupling and PSU cap connections as short and wide as possible.
I made an MS Excel spreadsheet that does most of the above, for you. It's downloadable from the post at:
http://www.diyaudio.com/forums/power-supplies/216409-power-supply-resevoir-size-167.html#post3287619
Cheers,
Tom
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Thanks a million!
Do you recall when mathematics for some of seemed a waste of time due to lack of application? I experienced that with trig in high school. It made perfect sense to me when I was 25 years old designing a Car frame. You have totally clarified this for me by connecting the application of it.
The next step would be for me to practise it.
Do you recall when mathematics for some of seemed a waste of time due to lack of application? I experienced that with trig in high school. It made perfect sense to me when I was 25 years old designing a Car frame. You have totally clarified this for me by connecting the application of it.
The next step would be for me to practise it.
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