filter capacitor sizing calcs

Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second. The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

Thanks,

Michael

 

[Paul E. Schoen]

<mrdarrett@gmail.com> wrote in message
news:15d20671-6385-4340-afa9-bb23412626d8@e10g2000prf.googlegroups.com...

Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second. The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

Thanks,

Michael

As a rule of thumb, I use 1000 to 2000 uF per amp. But that was for a
nominal 12 VDC supply.

Another way to look at it is using the RC time constant, which allows the
voltage to drop to 63% of peak. Also, you must account for the entire time
the sine wave is below peak, so use 1/120 Sec. So for 4 ohms,

C = (1/120)/4 = ~2000 uF. Three time constants are needed for 5% drop, so
that would be ~6000 uF.

For a 12 ohm load, you get 690 uF. So three times that is 2000 uF, for 1
amp at 12 volts.

For an amplifier power supply, the load is not quite the same as the output
load. It will depend on the efficiency and the frequency you are
amplifying.

You should do an LTSpice simulation and check the voltage waveform as well
as the RMS current from the power supply (transformer) and peak current
through the diodes. You might be surprised at what you find. Also check the
RMS current in the capacitors (with real values for source impedance and
capacitor ESR), and make sure they will handle the ripple current. And
check for overall power dissipation to see what sort of efficiency you get.
Switching supplies start looking pretty good.

Paul

 

[Greg Neill]

<mrdarrett@gmail.com> wrote in message
news:15d20671-6385-4340-afa9-bb23412626d8@e10g2000prf.googlegroups.com

Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second. The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

You will want to know what the peak current draw is
for the amplifier. The speaker impedance (4 ohms)
is not really relevant to the calculation. The
load on the power supply is the amplifier.

 

[John O'Flaherty]

On Apr 1, 11:23 am, mrdarr...@gmail.com wrote:

Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second. The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

You lost a set of parentheses. You should have
(1/240) / (4* ln(Vo / 0.95 Vo)) = 0.02 F, or 20 mF, or 20,000 uF.

But the time should be 1/120, or 8.3 ms, since worst case, and
desiring only a small drop, you have to wait almost from one peak to
the next, so that would be 40,000 uF. Also, note that the 4 ohm rating
for a speaker is an impedance and may not be that high for the lower
frequencies.
Another way to look at it is to figure out your peak current and
maximum allowable voltage drop. Since you don't want much percentage
voltage drop, you can use a linear approximation instead of an
exponential. Then,
Q = C * V --> C = Amps * seconds / volts. If you have a supply of 30
V, and you want no more than 5% drop, or 1.5 V, and you have 7.5 A
(about 4 ohm load), full-wave, you have C = 7.5 A * 8.3 ms / 1.5 V = .
0415 F, or 41,500 uF, close to what was calculated above.
--
John

 

On Apr 1, 12:02 pm, "John O'Flaherty" <quias...@yahoo.com> wrote:

On Apr 1, 11:23 am, mrdarr...@gmail.com wrote:



Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second. The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

You lost a set of parentheses. You should have
(1/240) / (4* ln(Vo / 0.95 Vo)) = 0.02 F, or 20 mF, or 20,000 uF.

Absolutely right! My goodness, I'm losing the ability to do basic
algebra too... I messed up right after ln(Vcap/Vo) = -t/RC.

But the time should be 1/120, or 8.3 ms, since worst case, and
desiring only a small drop, you have to wait almost from one peak to
the next, so that would be 40,000 uF. Also, note that the 4 ohm rating
for a speaker is an impedance and may not be that high for the lower
frequencies.
Another way to look at it is to figure out your peak current and
maximum allowable voltage drop. Since you don't want much percentage
voltage drop, you can use a linear approximation instead of an
exponential. Then,
Q = C * V --> C = Amps * seconds / volts. If you have a supply of 30
V, and you want no more than 5% drop, or 1.5 V, and you have 7.5 A
(about 4 ohm load), full-wave, you have C = 7.5 A * 8.3 ms / 1.5 V = .
0415 F, or 41,500 uF, close to what was calculated above.
--
John

Thanks.

Michael

 

[Phil Allison]

<mrdarrett@gmail.com>

Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

** The formula you use is I = C dv/dt or C = I dt/dv

You merely calculate the drop in voltage in the period when the cap is NOT
being charged by the transformer.

So, setting " dt " at 6 mS and "dv" at 2 volt and I at 4 amps average.

C = 4 x 0.006 / 2

= 0.012 F

= 12,000 uF



....... Phil

 

[Bill Bowden]

On Apr 1, 8:23 am, mrdarr...@gmail.com wrote:

Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second.  The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

Thanks,

Michael

The formula is C=IT/E where I is the current, T is the time, and E is
drop in voltage, and C is the capacitor. If it's 60Hz, full wave
rectified, the time is 8.33 mS. If the voltage is 1 volt and the load
is 4 ohms, the current will be 250mA. and a 5% voltage drop will be
50mV. So it should be somewhere around C=(.25*.0083) / .05 = 41,665
uF, but that seems a little large. Where is the math error?

-Bill

 

[Paul E. Schoen]

"Paul E. Schoen" <pstech@smart.net> wrote in message
news:47f26de6$0$17425$ecde5a14@news.coretel.net...

mrdarrett@gmail.com> wrote in message
news:15d20671-6385-4340-afa9-bb23412626d8@e10g2000prf.googlegroups.com...
Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second. The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

Thanks,

Michael

As a rule of thumb, I use 1000 to 2000 uF per amp. But that was for a
nominal 12 VDC supply.

Another way to look at it is using the RC time constant, which allows the
voltage to drop to 63% of peak. Also, you must account for the entire
time the sine wave is below peak, so use 1/120 Sec. So for 4 ohms,

C = (1/120)/4 = ~2000 uF. Three time constants are needed for 5% drop, so
that would be ~6000 uF.

For a 12 ohm load, you get 690 uF. So three times that is 2000 uF, for 1
amp at 12 volts.

For an amplifier power supply, the load is not quite the same as the
output load. It will depend on the efficiency and the frequency you are
amplifying.

You should do an LTSpice simulation and check the voltage waveform as
well as the RMS current from the power supply (transformer) and peak
current through the diodes. You might be surprised at what you find. Also
check the RMS current in the capacitors (with real values for source
impedance and capacitor ESR), and make sure they will handle the ripple
current. And check for overall power dissipation to see what sort of
efficiency you get. Switching supplies start looking pretty good.

Paul

I was a bit wrong on this. According to an LTSpice simulation, my 6000 uF
capacitor would have a voltage dropoff of about 20%. Just about 30,000 uF
gives a 5% drop in voltage, using a 12 VAC source and a 4 ohm load. The
peak voltage is 15.57, and RMS output is 15.2 volts. The load current is
3.8 amps, the capacitor current is 7.6 A RMS, and the supply current is
8.52 amps RMS.

With Schottky diodes, there is 61.5 watts input and 57.8 watts output, for
94% efficiency. But if the capacitor has just 0.1 ohms ESR, the efficiency
drops to 90%. And silicon diodes drops that down to 81%.

Paul

========================================================================

Version 4
SHEET 1 880 680
WIRE 64 128 -16 128
WIRE 96 128 64 128
WIRE 208 128 160 128
WIRE 96 208 48 208
WIRE 208 208 208 128
WIRE 208 208 160 208
WIRE 272 208 208 208
WIRE 368 208 272 208
WIRE -16 224 -16 128
WIRE 272 224 272 208
WIRE 64 304 64 128
WIRE 96 304 64 304
WIRE 208 304 160 304
WIRE 272 304 272 288
WIRE 272 304 208 304
WIRE 368 304 368 288
WIRE 368 304 272 304
WIRE -16 336 -16 304
WIRE 48 336 48 208
WIRE 48 336 -16 336
WIRE 368 352 368 304
WIRE 48 400 48 336
WIRE 96 400 48 400
WIRE 208 400 208 304
WIRE 208 400 160 400
FLAG 368 352 0
SYMBOL voltage -16 208 R0
WINDOW 3 -197 150 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 -94 108 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 16.9 60 0 0 0 100)
SYMATTR SpiceLine Rser=.01
SYMBOL diode 96 144 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL diode 96 224 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL diode 160 320 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D3
SYMATTR Value MUR460
SYMBOL diode 160 416 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D4
SYMATTR Value MUR460
SYMBOL cap 256 224 R0
WINDOW 3 21 55 Left 0
SYMATTR InstName C1
SYMATTR Value 30000ľ
SYMATTR SpiceLine Rser=.1
SYMBOL res 352 192 R0
SYMATTR InstName R1
SYMATTR Value 4
TEXT -70 418 Left 0 !.tran 1

 

[Paul E. Schoen]

"Bill Bowden" <wrongaddress@att.net> wrote in message
news:ed8ad664-acd8-45f2-898f-a163b1df4722@c19g2000prf.googlegroups.com...
On Apr 1, 8:23 am, mrdarr...@gmail.com wrote:

Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second. The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

Thanks,

Michael

The formula is C=IT/E where I is the current, T is the time, and E is
drop in voltage, and C is the capacitor. If it's 60Hz, full wave
rectified, the time is 8.33 mS. If the voltage is 1 volt and the load
is 4 ohms, the current will be 250mA. and a 5% voltage drop will be
50mV. So it should be somewhere around C=(.25*.0083) / .05 = 41,665
uF, but that seems a little large. Where is the math error?

-Bill

==========================

I did an LTSpice simulation and found that 30,000 uF was just about right.
The time is actually a little less than 0.0083 seconds because the
discharge time starts when the peak of the AC waveform drops below the
actual peak output. There is about 1.5 mSec charging time and 6.8 mSec
discharge time, resulting in a calculated capacitance of 34,000. It is
quite high, but really you don't normally need to hold within 5% of the
peak. My capacitor value of 6000 uF, based on 3 TCs with the 4 ohm
resistor, gives about 20% P-P ripple, which is usually good enough. It's
probably better to add an inductor if you want to get better ripple without
huge capacitors and high peak diode currents.

But I still say a switcher is the way to go for anything over about 10
watts, or if you are concerned about efficiency. Don't allow your power
supply to contribute to Global Warmongering!

Paul

 

[Eeyore]

Bill Bowden wrote:

On Apr 1, 8:23 am, mrdarr...@gmail.com wrote:
Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second. The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

Thanks,

Michael

The formula is C=IT/E where I is the current, T is the time, and E is
drop in voltage, and C is the capacitor. If it's 60Hz, full wave
rectified, the time is 8.33 mS. If the voltage is 1 volt and the load
is 4 ohms, the current will be 250mA. and a 5% voltage drop will be
50mV. So it should be somewhere around C=(.25*.0083) / .05 = 41,665
uF, but that seems a little large. Where is the math error?

Several.

The time isn't 8.33 ms or even 10ms where 50Hz current is the norm. It'll be
around 80-75% of that figure, since for a good 20-25% of the half cycle, the
cap is actually being charged.

Also, your calculation assumes that the 4 ohms is driven straight from the
supply. In audio, the polarity is constantly reversing, so power is drawn
from BOTH supplies, so you can halve the current for starters. Also, with a
sinewave signal, the current drawn will be reduced also.


Graham

 

[Phil Allison]

"Paul E. Schoen Monumental Fuckwit "

But I still say a switcher is the way to go for anything over about 10
watts, or if you are concerned about efficiency.

** Complete BOLLOCKS !!

SMPS are far less reliable and generally LESS efficient than transformer/
rectifier PSUs.

Don't allow your power supply to contribute to Global Warmongering!

** What a MONUMENTAL FUCKWIT !!



....... Phil

 

[Bob Masta]

On Wed, 02 Apr 2008 09:57:32 +0100, Eeyore
<rabbitsfriendsandrelations@hotmail.com> wrote:
<snip>

Also, your calculation assumes that the 4 ohms is driven straight from the
supply. In audio, the polarity is constantly reversing, so power is drawn
from BOTH supplies, so you can halve the current for starters. Also, with a
sinewave signal, the current drawn will be reduced also.

Actually, the speaker is never driven from both supplies at once.
(If both positive and negative output devices were on at the same
time, they'd be shorting both supplies together!) So you don't have
half the current, you have the same (peak) current but it is drawn
from alternate supplies on alternate half-cycles of the signal.

Best regards,


Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

 

[Bill Bowden]

On Apr 2, 12:57 am, Eeyore <rabbitsfriendsandrelati...@hotmail.com>
wrote:

Bill Bowden wrote:
On Apr 1, 8:23 am, mrdarr...@gmail.com wrote:
Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second.  The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

Thanks,

Michael

The formula is C=IT/E where I is the current, T is the time, and E is
drop in voltage, and C is the capacitor. If it's 60Hz, full wave
rectified, the time is 8.33 mS.  If the voltage is 1 volt and the load
is 4 ohms, the current will be 250mA. and a 5% voltage drop will be
50mV. So it should be somewhere around  C=(.25*.0083) / .05  = 41,665
uF, but that seems a little large. Where is the math error?

Several.

The time isn't 8.33 ms or even 10ms where 50Hz current is the norm. It'll be
around 80-75% of that figure, since for a good 20-25% of the half cycle, the
cap is actually being charged.

Also, your calculation assumes that the 4 ohms is driven straight from the
supply. In audio, the polarity is constantly reversing, so power is drawn
from BOTH supplies, so you can halve the current for starters. Also, with a
sinewave signal, the current drawn will be reduced also.

Graham- Hide quoted text -

- Show quoted text -

Yes, it's 72 degrees where the voltage falls 5% so the discharge time
would be 80 percent (72/90= 0.8) of the cycle, or 6.66 mS since the
frequency is 120 Hz. So the new figure would be C=(.25 * .00666) / .05
= 33,000uF. But the fact the amplifier uses both positive and negative
supplies doesn't change that much since a low 100Hz bass beat will be
5mS on each supply.

-Bill

 

[Eeyore]

Bill Bowden wrote:

Eeyore wrote:
Bill Bowden wrote:
On Apr 1, 8:23 am, mrdarr...@gmail.com wrote:
Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60 sec.
One peak will occur in 1/120 of a second. The worst case scenario is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

Thanks,

Michael

The formula is C=IT/E where I is the current, T is the time, and E is
drop in voltage, and C is the capacitor. If it's 60Hz, full wave
rectified, the time is 8.33 mS. If the voltage is 1 volt and the load
is 4 ohms, the current will be 250mA. and a 5% voltage drop will be
50mV. So it should be somewhere around C=(.25*.0083) / .05 = 41,665
uF, but that seems a little large. Where is the math error?

Several.

The time isn't 8.33 ms or even 10ms where 50Hz current is the norm. It'll be
around 80-75% of that figure, since for a good 20-25% of the half cycle, the
cap is actually being charged.

Also, your calculation assumes that the 4 ohms is driven straight from the
supply. In audio, the polarity is constantly reversing, so power is drawn
from BOTH supplies, so you can halve the current for starters. Also, with a
sinewave signal, the current drawn will be reduced also.

Graham


Yes, it's 72 degrees where the voltage falls 5% so the discharge time
would be 80 percent (72/90= 0.8) of the cycle, or 6.66 mS since the
frequency is 120 Hz. So the new figure would be C=(.25 * .00666) / .05
= 33,000uF. But the fact the amplifier uses both positive and negative
supplies doesn't change that much since a low 100Hz bass beat will be
5mS on each supply.

Well ... if you insist on doing a worst case analysis ... ! ;~)

You also forgot that a sinewave won't demand as much current as a square wave
though.

I also reckon the 5% ripple is unrealistic. 10% is a more sensible number to
target. Put all that into the equation and 10,000 uF is more than adequate.

Graham

 

[Paul E. Schoen]

"Eeyore" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:47F427F4.4108E710@hotmail.com...

Bill Bowden wrote:

Eeyore wrote:
Bill Bowden wrote:
On Apr 1, 8:23 am, mrdarr...@gmail.com wrote:
Let's say I've got a bridge rectifier downstream of a power
transformer, and I want to calculate the size of filter capacitor
I'll
need, allowing for a 5% drop in voltage.

From Vcap = Vo exp( -t/RC ),
C = t/R ln(Vo/Vcap)

Let's say that the power supply will need to power an amplifier
feeding a 4-ohm speaker, so R=4 ohms.

At 60 Hz, a bridge rectifier will give two |sin x| peaks in 1/60
sec.
One peak will occur in 1/120 of a second. The worst case scenario
is
half of this - the portion from the top of the peak to 0V, which
occurs in 1/240 sec.

C = (1/240s) / (4 ohm) ln ( Vo / 0.95 Vo )

C = 53 uF.

Anything larger than this is a waste, then... ?

Thanks,

Michael

The formula is C=IT/E where I is the current, T is the time, and E
is
drop in voltage, and C is the capacitor. If it's 60Hz, full wave
rectified, the time is 8.33 mS. If the voltage is 1 volt and the
load
is 4 ohms, the current will be 250mA. and a 5% voltage drop will be
50mV. So it should be somewhere around C=(.25*.0083) / .05 =
41,665
uF, but that seems a little large. Where is the math error?

Several.

The time isn't 8.33 ms or even 10ms where 50Hz current is the norm.
It'll be
around 80-75% of that figure, since for a good 20-25% of the half
cycle, the
cap is actually being charged.

Also, your calculation assumes that the 4 ohms is driven straight from
the
supply. In audio, the polarity is constantly reversing, so power is
drawn
from BOTH supplies, so you can halve the current for starters. Also,
with a
sinewave signal, the current drawn will be reduced also.

Graham


Yes, it's 72 degrees where the voltage falls 5% so the discharge time
would be 80 percent (72/90= 0.8) of the cycle, or 6.66 mS since the
frequency is 120 Hz. So the new figure would be C=(.25 * .00666) / .05
= 33,000uF. But the fact the amplifier uses both positive and negative
supplies doesn't change that much since a low 100Hz bass beat will be
5mS on each supply.

Well ... if you insist on doing a worst case analysis ... ! ;~)

You also forgot that a sinewave won't demand as much current as a square
wave
though.

I also reckon the 5% ripple is unrealistic. 10% is a more sensible number
to
target. Put all that into the equation and 10,000 uF is more than
adequate.

Graham

Yes, and with realistic values of transformer resistance, capacitor ESR,
and diode characteristics, you really can't get much better than that. You
need almost ideal components to get enough charge during that sine wave
peak. Even 15-20% is OK, and you can use reasonable size capacitors.

If load is fairly constant, an inductor can really help, but the output
voltage will be closer to the RMS value of the AC input under load, and can
become unstable (with oscillations) and poorly regulated with light loads.

Paul