# How to calculate the value of a smoothing capacitor

#### Voltz

##### New Member
Assuming 10% Ripple is acceptable does anyone know the formula for the value needed for a capacitor for smoothing. And does it make a difference if the capacitance is greater than this, and if so what difference, I've tried searching but have got 2 or 3 different formulas, need help - thanks in advance

#### MikeMl

##### Well-Known Member
Since the capacitor is usually used at the input of a regulator, the current discharging it is constant. If powered by the AC line, assuming full-wave rectification, the capacitor must supply current to the regulator for 1/2cycle (8.3ms or 10ms, depending if it is 60Hz or 50Hz)

The Charge Q (Coulombs) removed from the capacitor is Q=I*t, where I is current and t is time.

Q also = C*ΔV, where C is the capacitance and ΔV is the voltage drop as the current flows out.

So C*ΔV = Q = I*t

Rearranging gives C = I*t/ΔV.

For a 2A power supply, 60Hz, full-wave, where you can tolerate a 3V sag in the filter capacitor voltage without the regulator dropping out of regulation,

C = 2 * 0.008/3 = 0.0053F = 5300uF

Its not ripple that is important; it is how low does the voltage sag in order not to violate the dropout spec for the regualtor

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#### RCinFLA

##### Well-Known Member
The info you provide is insufficient.

Assuming you are doing 50 Hz full wave rectification you will have 10 millisecs between sinewave peaks. The voltage decay from peak to peak is based on I = C dV/dt.

For example if you want to support 3 amp load and can tolerate a 0.5 volt drop between peaks then capacitance required is 3 amps = C * (0.5v/10 msec) or C = 60,000 uF.

All the capacitor charge current is during the short interval of sinewave peak. A large cap just shortens this interval and increase the peak current on rectifier diodes.

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#### Voltz

##### New Member
Since the capacitor is usually used at the input of a regulator, the current discharging it is constant. If powered by the AC line, assuming full-wave rectification, the capacitor must supply current to the regulator for 1/2cycle (8.3ms or 10ms, depending if it is 60Hz or 50Hz)

The Charge Q (Coulombs) removed from the capacitor is Q=I*t, where I is current and t is time.

Q also = C*ΔV, where C is the capacitance and ΔV is the voltage drop as the current flows out.

So C*ΔV = Q = I*t

Rearranging gives C = I*t/ΔV.

For a 2A power supply, 60Hz, full-wave, where you can tolerate a 3V sag in the filter capacitor voltage without the regulator dropping out of regulation,

C = 2 * 0.008/3 = 0.0053F = 5300uF

Its not ripple that is important; it is how low does the voltage sag in order not to violate the dropout spec for the regualtor
3V is waaay too much, I'm talking about a 240VAC (UK Mains) Transformed down to 24DC Variable,
240VAC -> Transformer (24VAC) -> Rectifier (24DC /w ripple) -> Smoothing (24DC - 23.5V is acceptable) so I'm looking for a maximum ripple of 0.5V ~2% Ripple

So basically using the post below the time is 60Hz then the capacitor has to discharge for half of a cycle which would be ~ 8.3ms, so 0.5V for 8.3ms - is that enough information

#### MikeMl

##### Well-Known Member
Since you are in the UK, the time is 10ms. Since you didn't specify the load current, you will need 0.01/0.5 = 0.2F per AMP of Load current.

#### Voltz

##### New Member
Since you are in the UK, the time is 10ms. Since you didn't specify the load current, you will need 0.01/0.5 = 0.2F per AMP of Load current.
Thanks for that, so what is a simple definitive formula for calculating the smoothing capacitor value because so far I've got the same problem as before - I have two different formulae

EDIT: Interestingly I've just re-read the post and you've both arrived at the same value for my cap, at 3A it will need to be 600mF - so can anyone give me a simple single formula for future use in this sort of thing

EDIT: Actually one of you is saying 600mF another 60mF

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#### audioguru

##### Well-Known Member
I use this simple graph:

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#### Voltz

##### New Member
I use this simple graph:
Thank's Ill save that but I would really appreciate a definitive formula - please?
C = (5*I)/V*t? or something where I is current and V is DC voltage and t is time? Formula!

#### MikeMl

##### Well-Known Member
Thank's Ill save that but I would really appreciate a definitive formula - please?
C = (5*I)/V*t? or something where I is current and V is DC voltage and t is time? Formula!
I gave it to you above: C = I*t/ΔV.

#### Voltz

##### New Member
I gave it to you above: C = I*t/ΔV.
So lets see
example, C = 3A*0.0083s/0.24 = 0.10375 which is 104mF

For a 3A 24VDC line with 1% ripple then the capacitor I need is ~ 104mF?

EDIT: Just realised that the output will be regulated anyway so I think about 10% ripple will be acceptable
so 3A*0.0083s/2.4 = 0.010375 ~ 10mF

Thank you all

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#### AlainB

##### Member
According to my spreadsheet, it would be more like 104167 uF (microfarads).

Alain

Edit on your edit: at 10% ripple it would be 10417 uF. That is if your voltage is at 60 Hz. At 50 Hz, it would be 12500 uF.

For 10% ripple at 60 Hz, a good approximation is C = ( 80,000 * I ) / V.

These calculations are comming from the web mainly for steppers motors power supply.

The file is an Excel file. Rename it to Ripple calculations.xls.

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#### Hero999

##### Banned
You haven't provided enough information.

The ripple will depend on the following:

• Series resistance of the supply, a higher series resistance causes more ripple.
• Supply voltage, the ripple doesn't scale linearly with supply voltage, for example with a constant current load, it's independent of supply voltage, so the capacitor for a given % ripple can't be calculated without the supply voltage.
• A purely resistive load produces the least amount of ripple.
• A constant current source (linear regulator) produces more ripple as the current remains the same as the voltage falls.
• A constant power source (an SMPS) causes even more ripple because the current increases as the voltage falls.

10% ripple might be good enough or it might be too much. Unless it's a purely resistive load, in which case 10% is a good rule of thumb, you need to consider the minimum operating voltage of the load and calculate the capacitor size accordingly.

For a transformer with a low resistance and a constant current load the general rule of thumb is 10,000µF/A for 1V of ripple, 5000µF/A for 2V of ripple, the higher the voltage the smaller capacitor, the basic formula is given below:

C = 10×I/V

Where:
I = Current drawn (mA)
C = Capacitance (µF)
V = Maximum ripple voltage.

This formula only apples to constant current loads and full wave rectification of 50/60Hz, for half wave just double it. The formula tends to oversize the capacitor slightly which is acceptable.

#### Voltz

##### New Member
Ahh there see now I have 2 formulas again

C = (I x t)/ΔV and C = (10xI)ΔV ... and two values totally different
- 125mF and 13mF

- an extra problem is that I can't seem to find a maplins capacitor rated this high suitable for use

#### audioguru

##### Well-Known Member
Capacitor values are uf, not mF.

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#### Voltz

##### New Member
Capacitor values are uf, not mF.
Well you know what I mean 125,000 uF and 13,000 uF
So which formula?

#### MikeMl

##### Well-Known Member
Well you know what I mean 125,000 uF and 13,000 uF
So which formula?
Look at this and figure it out for yourself.

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#### Hero999

##### Banned
Did you read my previous post?

The formula varies depending on the nature of the load.

I gave the one for a constant current load, I think the other one is for a resistive load but I haven't checked it.

#### bychon

##### New Member
radical2 x Capacitance x Eripple (p-p) x Frequency = amps.
√2 C Er F = I

#### Hero999

##### Banned
It turns out, I was wrong earlier.

It doesn't make much difference if its a resistive load.

I thought the difference would be more pronounced with a smaller capacitor so I used 2000µF. I set up the circuit as per Mike's simulation and chose a load resistor with a value to give an RMS current of 2A (14Ω) and the ripple was similar to Mike's simulation.

It appears that the higher current drawn at higher voltages cancels the lower current drawn at lower voltages.

I wonder if it's the same with an SMPS?

Probably.

Still for non-constant current loads, it's good practise to use the highest current value, so for an SMPS that will be at the lowest voltage and the resistive load it'll be at the highest voltage.