Hi all, a couple questions that I'm hoping to get clarified. I'm a total novice who's done a little reading so bare with me.
When upgrading or choosing electrolytic power caps, I see a lot of variety in the capacitance recommended, or actually installed on stock amp units. For instance, on a few units of a SMSL pre-built amp, 16v caps whether single or multiples in parallel, there's a total of 4400uF one (2 in parallel), on another it's 5600uF (single), and on another its 9400uF (2 in parallel). This is on a stmicro tda7492p board. Here are some pictures:
1. Assuming the same quality low impedence-ESR line of caps, why choose one amount of capacitance over another? Taking for granted there is a lower limit, isn't there also an upper limit - with capacitance discharge at larger values starting to affect rate of discharge and thus affecting peak signal timing/distortion? Maybe I'm completely misunderstanding this.
2. Are there any acoustic properties at a set input level that would be affected by the capacitance you choose? I've seen reference to larger capacitance for more bass response or certain size of speaker driver...
3. Are there ways to calculate what a recommended range would be?
4. When combining large wet and small polymer caps (popular oscons), if I understand it this is done to take advantage of the polymer treatment of high frequencies?
5. Since polymers cycle much faster than wet and typically have lower capacitance, is there a calculation or ratio to determine equivalent capacitance between the 2, like 220uf polymer = 660uf wet? Or this a bad comparison because 1uF = 1uF; its the performance characteristics that are different. Is this important in a mixed parallel arrangement, or what if you want to replace a wet with a polymer?
Thanks, lots of great info on this forum!
When upgrading or choosing electrolytic power caps, I see a lot of variety in the capacitance recommended, or actually installed on stock amp units. For instance, on a few units of a SMSL pre-built amp, 16v caps whether single or multiples in parallel, there's a total of 4400uF one (2 in parallel), on another it's 5600uF (single), and on another its 9400uF (2 in parallel). This is on a stmicro tda7492p board. Here are some pictures:
1. Assuming the same quality low impedence-ESR line of caps, why choose one amount of capacitance over another? Taking for granted there is a lower limit, isn't there also an upper limit - with capacitance discharge at larger values starting to affect rate of discharge and thus affecting peak signal timing/distortion? Maybe I'm completely misunderstanding this.
2. Are there any acoustic properties at a set input level that would be affected by the capacitance you choose? I've seen reference to larger capacitance for more bass response or certain size of speaker driver...
3. Are there ways to calculate what a recommended range would be?
4. When combining large wet and small polymer caps (popular oscons), if I understand it this is done to take advantage of the polymer treatment of high frequencies?
5. Since polymers cycle much faster than wet and typically have lower capacitance, is there a calculation or ratio to determine equivalent capacitance between the 2, like 220uf polymer = 660uf wet? Or this a bad comparison because 1uF = 1uF; its the performance characteristics that are different. Is this important in a mixed parallel arrangement, or what if you want to replace a wet with a polymer?
Thanks, lots of great info on this forum!
I suggest reading up on caps a bit more to get a better understanding. Here's a read that's been helpful to me:
Improve sound quality
Then read up on ripple current primer here:
Linear Power Supply Design
as that explains some reasons for choosing multiple caps vs. a couple.
In my very limited experience with Class A/B amps, I've found replacing film caps in the audio path or sometimes on the actual amp board can net the biggest sound improvement, or enough of an improvement it actually sounds better to me.
Power supply caps when recapping an amplifier, sometimes you can increase capacitance there by a small amount, but I wouldn't mess with capacitance changes anywhere else.
Voltage ratings you need to at least match, though you can increase cap voltage ratings if you have the room. Voltage ratings on caps are what they can safely handle, not what the circuit will all of a sudden have.
Film caps for example you might have a cap with a 50v rating, but everything aftermarket is 100v to 600v at the same capacitance. As long as it fits, you can use it. Sometimes if the cap manufacturer has a cap that's typically favored due to the way it affects the audio signal; you might only find it in higher voltage ratings, but always check the sizes as a 600v 1uf cap can be rather large despite the small capacitance value. You also might find the legs to be a thicker diameter than the PCB hole. In that case, you can very carefully enlarge the hole so long as you don't lift the traces, or get creative with the solder.
If you have an old EQ for example, and can trace each slider to a corresponding cap(s) on the tone board, you can try a very small capacitance change on one channel i.e. 1khz slider on right channel and leave left channel alone, then use headphones (for best results) and play a sinewave in the 1khz range or listen to various music, etc. and see if the change is desirable or not. I tweaked an old EQ I have and considerably removed harshness in the higher frequencies by experimenting with one cap (per channel) and a very minute capacitance change really changed the sound there.
Also, since that amp you have pictured is so simple, you might consider installing higher quality resistors with really tight tolerances. Will it drastically improve the sound? Who knows. But for $5 and little time, it might be worth finding out. Though I strongly suggest reading up on resistors before touching them as they need to be able to handle the wattage, etc.
Improve sound quality
Then read up on ripple current primer here:
Linear Power Supply Design
as that explains some reasons for choosing multiple caps vs. a couple.
In my very limited experience with Class A/B amps, I've found replacing film caps in the audio path or sometimes on the actual amp board can net the biggest sound improvement, or enough of an improvement it actually sounds better to me.
Power supply caps when recapping an amplifier, sometimes you can increase capacitance there by a small amount, but I wouldn't mess with capacitance changes anywhere else.
Voltage ratings you need to at least match, though you can increase cap voltage ratings if you have the room. Voltage ratings on caps are what they can safely handle, not what the circuit will all of a sudden have.
Film caps for example you might have a cap with a 50v rating, but everything aftermarket is 100v to 600v at the same capacitance. As long as it fits, you can use it. Sometimes if the cap manufacturer has a cap that's typically favored due to the way it affects the audio signal; you might only find it in higher voltage ratings, but always check the sizes as a 600v 1uf cap can be rather large despite the small capacitance value. You also might find the legs to be a thicker diameter than the PCB hole. In that case, you can very carefully enlarge the hole so long as you don't lift the traces, or get creative with the solder.
If you have an old EQ for example, and can trace each slider to a corresponding cap(s) on the tone board, you can try a very small capacitance change on one channel i.e. 1khz slider on right channel and leave left channel alone, then use headphones (for best results) and play a sinewave in the 1khz range or listen to various music, etc. and see if the change is desirable or not. I tweaked an old EQ I have and considerably removed harshness in the higher frequencies by experimenting with one cap (per channel) and a very minute capacitance change really changed the sound there.
Also, since that amp you have pictured is so simple, you might consider installing higher quality resistors with really tight tolerances. Will it drastically improve the sound? Who knows. But for $5 and little time, it might be worth finding out. Though I strongly suggest reading up on resistors before touching them as they need to be able to handle the wattage, etc.
When using an SMPS together with a Class-D, two things are important, ESR and ripple-current. The capacitance value isn't that important at all. So having multible capacitors in parallel lowers total ESR and multiplies ripple-current handling.
When calculating the needed capacitance, they are mainly choosen for HF ripple.
You may do the math to see, that it wouldn't make a difference if choosen 470uF over i.e. 10000uF. It's still way to less to help in any way on your bass.
Yes, lets do the math for your TDA7492 based amp. We will take the data from datasheet.
Po = 50W
RL = 6R
Vpeak = sqrt(Po * Rl) = sqrt(50W * 6R) = 17.32V
Ipeak = Vpeak / RL = 17.32V / 6R = 2.88A
Cbulk(min) = (Ipeak * Td * Dmax) / (Vripple)
with Td = 1/fsw (switching frequency) - lets take worst case minimum fs from DS = 250kHz
Dmax = maximum duty cycle, unknown so we assume 0.95
Vripple = dV = maximum voltage drop allowed at Ipeak, we assume 0.1V (even 1V wouldnt be an issue)
So:
Cbulk(min) = (2.88A * 1/250kHz * 0.95) / (0.1V) = 110uF
As we have two channels, we'll need at minimum 2x110uF = 220uF.
ESRmax = Vripple/Ipeak = 0.1V / 2.88A = 34.7mR
So either we have 2 caps with C = 110uF, ESR = 34.7mR or 1 cap with C = 220uF, ESR = 17.35mR.
This will be the minimum specs your bulk decoupling should have. When going for 400kHz its even less.
As the caps derate/ages with time, you'll choose your capacitance way higher, like twice so a 470uF <= 17.35mR would do.
The point for having those "big" capacitance values actually in place is, that the cap needs to handle Ipeak as ripple current while having that low ESR. As bigger capacitance usually have higher current handling and lower ESR those are choosen. But they wouldn't need that much capacitance.
Mainly low ESR + higher current handling at 100+kHz.
See math above.
Edit:
To answer the question for "improved" bass response, lets have a look at an example. Picking some random techno/electronic music, a bass note is ~100ms. We now want to support the first half with capacitance.
There is:
C = (dI * dT) / dV
dI = current jump at full power pass, from the math above 2*2.88A (2 channel)
dT = length of current we need to support, 100ms/2 = 50ms
dV = voltage "sag" at current choosen, from the math above 0.1V
C = (5.76A * 50ms) / 0.1V = 2.88F = 2,880,000uF
Impractical?!
So if we relax the situation by choosing bass support for only the first 1/10 with a voltage drop of 10% of the supply voltage (Vsupply = 24V) we will still need:C = (5.76A * 10ms) / 2.4V = 24000uF
Well, would do but whats for the rest of the 90ms? The supply voltage would drop down to Vsupply - Ipeak * Rsupply. If we assume a conservative resistance for the connection from the supply to the first cap of 200mR, There is: 24V - 5.76A * 200mR = 22.84V which is still higher that 24V - 10% drop = 21.6V. So this makes no practical sense. (to me)
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Datasheet has continuous sinewave power output imo, Vpeak in your example would be 21.9 V and Ipeak 3.65A ?
But then that is 10% THD number, more accepted would be 1% 40 watt number. And then some have mentioned the amplifier uses 2 amplifiers per channel, BTL, and claim that gives different numbers when calculating too, I don't know.
But then that is 10% THD number, more accepted would be 1% 40 watt number. And then some have mentioned the amplifier uses 2 amplifiers per channel, BTL, and claim that gives different numbers when calculating too, I don't know.
? Feel free to do the math again. The calculations should be done for worst case. If you calculate for just 1% THD, performance will be worse for power demands with THD > 1%.
If it works for 10%, it will surely work for 1% as well as power demands are more relaxed then.The above calculations are valid for BTL.
If it works for 10%, it will surely work for 1% as well as power demands are more relaxed then.The above calculations are valid for BTL.
The amp features four half bridges, which forms two full bridges (2xBTL). The calculation above is for one BTL output up to the point where current is doubled for both outputs.
The Vpeak/Ppeak statements are at given output. (continuous sine wave)
The reverse math can easily be done.
Ppeak = (Vcc * 1/sqrt2)^2 / RL
This formula doesn't include efficiency - same as the rest of the above calculations.
If you consider the full power path, the current is even less. It's not Rload (RDCspeaker) alone, it's
Rload + 2*Rds_on + 2*Rdc_inductor + Rdc_connections + (ESR_bulk)
So in the end, the formulas are an estimation for the needs of decoupling.
The Vpeak/Ppeak statements are at given output. (continuous sine wave)
The reverse math can easily be done.
Ppeak = (Vcc * 1/sqrt2)^2 / RL
This formula doesn't include efficiency - same as the rest of the above calculations.
If you consider the full power path, the current is even less. It's not Rload (RDCspeaker) alone, it's
Rload + 2*Rds_on + 2*Rdc_inductor + Rdc_connections + (ESR_bulk)
So in the end, the formulas are an estimation for the needs of decoupling.
I edited the above post, it's a bit confusing.
This comes from the given statement of Outputpower at given Load resistance.
Rms Power is 1/sqrt(2) * Ppeak for Sinus signals. You can't just multiply rms power by 2 to get Ppeak power.
The calculations are correct.
Edit:
This can easily be seen if we write:
Vpeak = sqrt(Po * RL)
like this:
Po = Vpeak^2 / RL
It's the same.
Of course Ppeak can be higher than the 50W taken from the datasheet if you use your given PVCC for the calculation like so:
Po = (Vpeak * n)^2 / (RL + 2*Rds_on)
Hope this makes it more clear.
Regards
This comes from the given statement of Outputpower at given Load resistance.
Rms Power is 1/sqrt(2) * Ppeak for Sinus signals. You can't just multiply rms power by 2 to get Ppeak power.
The calculations are correct.
Edit:
This can easily be seen if we write:
Vpeak = sqrt(Po * RL)
like this:
Po = Vpeak^2 / RL
It's the same.
Of course Ppeak can be higher than the 50W taken from the datasheet if you use your given PVCC for the calculation like so:
Po = (Vpeak * n)^2 / (RL + 2*Rds_on)
Hope this makes it more clear.
Regards
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One night later, I understood what you meant. Yes I used the the Average/RMS power in the calculation, so a factor of 1.41 is missing for Vpeak.
Absolutely.
Absolutely.
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So once again:
Po = 50W
Ppeak = 2*Po = 100W
RL = 6R
Vpeak = sqrt(Ppeak * Rl) = sqrt(100W * 6R = 24.49V
So VCC should be 26V to do so.
Ipeak = Vpeak / RL = 24.49V / 6R = 4.08A
Cbulk(min) = (4.08A * 1/250kHz * 0.95) / (0.1V) = 155uF
As we have two channels, we'll need at minimum 2x155uF = 310uF.
ESRmax = Vripple/Ipeak = 0.1V / 4.08A = 24.5mR
So either we have 2 caps with C = 1550uF, ESR = 24.5mR or 1 cap with C =310uF, ESR = 12.25mR.
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Not that much change, but corrent now. for "true" peak power. Getting caps with ripple handling of 4.08A each is the point if you want to do 50W continous (stereo) - which normally wont occur in practice.
Po = 50W
Ppeak = 2*Po = 100W
RL = 6R
Vpeak = sqrt(Ppeak * Rl) = sqrt(100W * 6R = 24.49V
So VCC should be 26V to do so.
Ipeak = Vpeak / RL = 24.49V / 6R = 4.08A
Cbulk(min) = (4.08A * 1/250kHz * 0.95) / (0.1V) = 155uF
As we have two channels, we'll need at minimum 2x155uF = 310uF.
ESRmax = Vripple/Ipeak = 0.1V / 4.08A = 24.5mR
So either we have 2 caps with C = 1550uF, ESR = 24.5mR or 1 cap with C =310uF, ESR = 12.25mR.
----------------------------------
Not that much change, but corrent now. for "true" peak power. Getting caps with ripple handling of 4.08A each is the point if you want to do 50W continous (stereo) - which normally wont occur in practice.
Great info. This conversation has gotten over my head, but I asked for it. 
DoctorMord did you mean 3100uF in the updated calculation for a single cap?
Correct me if I'm wrong, but are you guys running calculations on the TDA7492, not the TDA7492P ? The P (pictured in op) has a lower max output of 25W/Ch:
http://www.st.com/st-web-ui/static/active/en/resource/technical/document/datasheet/CD00211320.pdf
So either we have 2 caps with C = 1550uF, ESR = 24.5mR or 1 cap with C =310uF, ESR = 12.25mR.
DoctorMord did you mean 3100uF in the updated calculation for a single cap?
Correct me if I'm wrong, but are you guys running calculations on the TDA7492, not the TDA7492P ? The P (pictured in op) has a lower max output of 25W/Ch:
http://www.st.com/st-web-ui/static/active/en/resource/technical/document/datasheet/CD00211320.pdf
Oh sorry, tipo. It's 155uF*2 or 310uF.
The calculations were made for the non-P version as I didn't know which version you got. So I searched by Google pictures with the pcb-revision noted on your pictures.
From the last calculations you should be able to do the math again easily.
The calculations were made for the non-P version as I didn't know which version you got. So I searched by Google pictures with the pcb-revision noted on your pictures.
From the last calculations you should be able to do the math again easily.
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P is only marketed at lower voltage and is a little cooler maybe, or hopefully, at same voltage it seems same chip
For capacitors you go for lowest ESR, the pcbtrace from cap to chippin could have same value as ESR, the solderjoint has resistance too. Higher capacitance happens to have lower ESR, for the wet ones.
And there are other ways to calculate, browse here on diyaudio and you'll find many different, this is one here, no idea if accurate or relevant.
For capacitors you go for lowest ESR, the pcbtrace from cap to chippin could have same value as ESR, the solderjoint has resistance too. Higher capacitance happens to have lower ESR, for the wet ones.
And there are other ways to calculate, browse here on diyaudio and you'll find many different, this is one here, no idea if accurate or relevant.
Why should it be irrelevant? The accuracy is limited by the points missed for additional resistances mentioned earlier. (RdcL, Rds_on, Rtrace etc.)
The formulas are straight forward from TI application notes, so I suspect them to be correct.
How do you calculate the (bulk) capacitance for yourself to be accurate and relevant?
It isn't so hard, you have an ampboard you want to fit in its little enclosure and which keeps looking neat. You see inaccurate calculations that in part finally lead to 310uF a channel after some sleepy posts and realize resistance might be bottleneck. You also read another part in same post by doctormord about bass, were it seems calculations were dismissed just because the outcome was impractical. So pratically you take the highest value sleek wet that fits and add an Oscon, you did your best to meet or keep impedance low and you know you are covering a little part of impractically large capacitance needed when looking at it differently. No calculus for me today
Or you buy bigger enclosure and start trying to do the impractical, I wouldn't, yes this is an 7492
Or you buy bigger enclosure and start trying to do the impractical, I wouldn't, yes this is an 7492
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