0
\$\begingroup\$

In all split power supplies there will be at least two capacitors, one in between +V to ground and another in between -V to the ground. In an 300W (RMS output power) amplifier there are two 10 000uF 63V capacitor in each voltage lines.

I have one unit of 68 000uF 125V capacitor which I am planning to install straight between the +V and -V in that power supply.

This will improve or degrade the performance of the amplifier? This connection is obviously safe in electrical perspective, but such connection, at least in my knowledge is never been used in any power supply design.

Pls advice.

\$\endgroup\$
4
  • \$\begingroup\$ Why would you do this when you're not sure it will help? \$\endgroup\$ – Olin Lathrop Nov 28 '16 at 14:56
  • \$\begingroup\$ Mathematically, the third cap should add another 34000uF to each lines. Instead of adding two small caps ( which I need to buy), I'm thinking to make use of what I already have. More capacitance is better for audio amps... The rectifier is safe since I'm already using the soft start relay at AC side. \$\endgroup\$ – soosai steven Nov 28 '16 at 14:59
  • \$\begingroup\$ Why this question get down voted? \$\endgroup\$ – soosai steven Nov 30 '16 at 22:10
  • \$\begingroup\$ I didn't, so I don't now for sure. Possibly it was because your question is rather ambiguous. It's not clear whether you are trying to address a particular problem in a existing amp, want to add the cap "just because" to a working amp, or are designing a new one. \$\endgroup\$ – Olin Lathrop Dec 1 '16 at 11:38
4
\$\begingroup\$

@soosai steven since your speaker is grounded and amplifier is push pull for each polarity of signal, current is best supplied from each supply and ground to speaker to reduce impedance of source power, not between V+&V- . Understand that ESR of large caps may be big and consider you want 8ohm * C = 10*10ms for max out at 8* ripple. But ESR of caps can be big and needs careful selection of quality and additional shunt caps.

I will try to explain but it may be complicated.

Check cap dissipation factor D.F. or measure at 100Hz. If you have a woofer for 25 Hz then D.F. at 25Hz is 4 x worse. compute ESR and determine impedance ratio for woofer.

What dampening factor do you want for woofer? This makes bass clear or muddy from back EMF of cone mass.

Normally 50 is weak, 100 is ok , 1000 for the best PA's for best bass clarity punch like a great bass drum with dampening blanket and tuned port. Thus ESR must be << 8 Ohm/dampening ratio or 80 mOhm for df=100.

This Dissipation Factor for you, 100Hz (for me 120 Hz) ripple current heat ( Ipk ^2*ESR) with 10x speaker current at<10% duty cycle for 10% ripple voltage at full load. You may want better than this 10% to prevent 100Hz distortion.

The dampening factor is for bass step response (mechanical ringing from coil back EMF) are related both to ESR of cap and output impedance of power amp. The power supply ripple is in series with your PA and speaker and ripple reduction depends of feedback. Too much open loop gain, it will oscillate, too little and PSRR suffers. This is a design tradeoff. Even with this, a 20kHz snubber on output is essential.

It gets more complicated to explain but PA's are low voltage gain but very high current gain and supply ripple V sensitivity is poor due gain feedback and low open loop V gain , unlike preamp.

Thus impedance of cap must be very low 1/(2pifC) for lowest bass frequency 1 to 2% of speaker and ESR must be less. Since I know you can compute these, I'll let you decide. When I built my own amp in 1973 , I used 100k uF 63V, caps that were "computer rated " for mainframes with 100A ripple current rating. Then I added 470uF solid tantalum caps.

68kuF will sound better with two, one for each rail, but verify DF or ESR.

If you don't have a scope or spectrum analyzer , use Audacity to sample a scaled signal into PC aux input port, using sweep gen and Spectrum Analysis (free) or measure ESR of caps and compute your distortion and heat loss. Caps are thermally insulated, so the RMS current rating must have good margin for low T rise and long life . You can also use Simulators and add ESR to see the effect.

good luck.

\$\endgroup\$
4
  • \$\begingroup\$ Comprehensive answer..!! Thank you. Based on this, assuming the 3rd cap has lowest ESR, then overall performance of the amp should be improve with the 3rd cap right? \$\endgroup\$ – soosai steven Nov 28 '16 at 18:15
  • \$\begingroup\$ verify with LRC meter , scope or some other method for checking d.f., ESR and ripple but ensure diodes and soft start can handle extra charge up time. as the final Joules in Cap will be the startup Joules in bridge and soft start NTC \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 28 '16 at 18:30
  • \$\begingroup\$ He really wants to add that cap, no matter what.. \$\endgroup\$ – Wesley Lee Nov 29 '16 at 14:19
  • \$\begingroup\$ also he really needs both, but probably can only fit one, unfortunately across V+~V- ends up in series with the small ones on each where the current flows \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 29 '16 at 14:20
6
\$\begingroup\$

Short answer: Don't do it.

It could help, but it could also possibly hurt. Noise on one supply would now couple to the other. How much that happens, and whether it matters is hard to say.

However, the larger point is that someone that knows a lot more than you about circuit design and audio amplifiers designed it the way they did. Leave it alone.

\$\endgroup\$
2
  • \$\begingroup\$ Its a diy amp which I build by my self. \$\endgroup\$ – soosai steven Nov 28 '16 at 15:01
  • 2
    \$\begingroup\$ @soosaisteven: in that case read the statement as "none of the many people that design amps do it that way" \$\endgroup\$ – PlasmaHH Nov 28 '16 at 18:47
1
\$\begingroup\$

It won't work. Consider the current flow through the capacitor and speaker for a large +ve signal. With 2 capacitors ('A' on the +ve rail, 'B' on the -ve rail) and a large +ve output voltage, current flows from the +ve terminal of capacitor A, through the +ve output transistor, through the loudspeaker load to ground, then back through the -ve terminal of capacitor A - current always flows round a closed path. Result: capacitor A provides the current to the load on +ve signals. Ditto capacitor B and -ve signals. Now consider a single capacitor 'A' and the same +ve output signal. Current flows out of the +ve terminal of the capacitor, through the +ve output transistor, through the loudspeaker load to ground. The only way the current can complete the loop is through the secondary winding of the transformer, through the bridge rectifier, to the -ve terminal of the capacitor. In practice, what will happen is that the +ve rail of your power supply will drop towards 0 volts, while the -ve rail goes even more negative - on +ve output signals. And vice versa on -ve output signals. Result: you see little signal across the speaker, but a lot of signal appears between the capacitor terminal and 0 volts.

\$\endgroup\$
5
  • \$\begingroup\$ The third cap is installed to complement the existing first two caps. As such under load,, when any of the first two caps discharge,, 3rd cap should attempt to compensate for the total voltage drop,,right? \$\endgroup\$ – soosai steven Nov 28 '16 at 18:12
  • \$\begingroup\$ @Ian FYI this site does have a circuit tool that you can use to draw circuits in your explanations if you wish \$\endgroup\$ – Voltage Spike Nov 28 '16 at 19:06
  • 1
    \$\begingroup\$ Schematic drawing tool not available for stack exchange users on android platform... Not sure if IOS supports it or not. \$\endgroup\$ – soosai steven Nov 28 '16 at 19:17
  • \$\begingroup\$ @laptop2d - thanks, but I only wandered onto this site because of a mis-typed Google query... \$\endgroup\$ – Ian Nov 29 '16 at 13:22
  • \$\begingroup\$ @soosai steven - sorry, I misunderstood your original query. I see comments have a limited length. I'll post a suggestion as a new answer. \$\endgroup\$ – Ian Nov 29 '16 at 13:28
1
\$\begingroup\$

You can easily estimate the effect of adding the additional capacitor by considering an AC analysis of the circuit.


Consider the +ve side of the amplifier output circuit as a signal source in series with an impedance, that injects 'noise' back into the +ve power rail. The impedance forms a voltage divider with the 10 000 uF cap., producing an amount 'X' of noise on the +ve power rail. Now you add the 68 000 uF cap across the +ve, -ve rails. From an AC point of view, you have just put 2 caps in series (10 000 plus 68 000) in parallel with the +ve rail 10 000 cap. This will roughly halve the impedance on the +ve rail, halving the 'noise' from the output stage current draw. Great! You've effectively halved the voltage drop on the +ve rail. Unfortunately, this is at the expense of injecting almost the same amount of noise onto the -ve rail (the 10 000 and 68 000 caps again form a voltage divider). If I haven't been clear enough, I'll have a go with the tool that laptop2d mentions, but it won't be for a day or two. Ian.

\$\endgroup\$
1
  • \$\begingroup\$ Noise from positive side when leak into negative side thru 3rd cap, will it be additive or subtractive to the overall performance of the amp? I think it will be beneficial to the performance in certain frequency of audio range. My amp in concern is of sub woofer with cut off at 100Hz. \$\endgroup\$ – soosai steven Nov 29 '16 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.