Capacitor smoothing clarification

[shaneshane1]

Just wanted to know if my calculations were correct for voltage smoothing!

Example: If i wanted to smooth out a 1000Hz,20mA signal from a 12V DC power supply, would this be correct?

(5 X 0.020A) / (12V x 1000Hz) = 0.1 / 12000 = 0.000008F so a 8uF Cap is correct, no???

If all is well and good with the math, the only thing i am confused with is the first number 5! is that for the amount of time constants?

This is just an example question for learning purposes , and i don't really plan to smooth a 1000Hz signal.

Thanks!!!

 

[aibelectronics]

You've not said what formula are you using?

One method tries to minimise Xc (reactance) as much as possible. As you should know a Capacitor is generally a short to high frequency signals. The bigger a Capacitor we get the smaller the reactance to hf signals (ripple)...

 

[3iMaJ]

Just wanted to know if my calculations were correct for voltage smoothing!

Example: If i wanted to smooth out a 1000Hz,20mA signal from a 12V DC power supply, would this be correct?

(5 X 0.020A) / (12V x 1000Hz) = 0.1 / 12000 = 0.000008F so a 8uF Cap is correct, no???

If all is well and good with the math, the only thing i am confused with is the first number 5! is that for the amount of time constants?

This is just an example question for learning purposes , and i don't really plan to smooth a 1000Hz signal.

Thanks!!!

You'll have to restate the problem I don't understand what you're trying to do.

 

[shaneshane1]

You'll have to restate the problem I don't understand what you're trying to do.

Really???

I will explain again! i am asking if a 8uF cap is correct for smoothing out a 1000Hz signal with an output of 20mA

So i end up with just a normal output with no signal (or close to no signal)

Here is the link were i found the math

 

[stevez]

"Smoothing" is certainly a good description for the role of the capacitor however smoothing is also not specific - so without more information there is no way to determine if your selection is correct or not.

Your link provides more information and if you followed the author's instructions ( I didn't check your work against the author's work) you'd expect to see 10% ripple voltage. For some applications that's pretty smooth - by other measures it is not - all depends on the application.

Keep in mind that the formula provided, if it is correct, offers only an approximation - and is valid for small amounts of ripple voltage.

 

[shaneshane1]

Keep in mind that the formula provided, if it is correct, offers only an approximation - and is valid for small amounts of ripple voltage.

Thanks!!! I was just checking that i was on the right track, and it seems that i am, I am learning more about capacitors at the moment, i have been putting this subject off for some time now

I am still confused as to the first number in the formula

"5"

Is that the time constance???

 

[stevez]

I do have references (books) that do a nice job explaining the formula - and my guess is that you can find good explanations by googling a bit. Look under "power supplies" - "capacitor power supply filter" as a suggestion. The '5' is a constant that is likely related to frequency if not other factors.

Keep in mind that the formula you used provides a rough approximation and is useful in a very limted way. Unfortunately many authors provide simplified estimation methods but stop short of explaining the limitations of the methods.

 

[shaneshane1]

Keep in mind that the formula you used provides a rough approximation and is useful in a very limted way. Unfortunately many authors provide simplified estimation methods but stop short of explaining the limitations of the methods.

Thanks!!! its still a very positive step in the right direction for me and it will prove to be more useful than nothing at all!

 

[UTMonkey]

I am really rusty so forgive me but isnt "5" as in 5RC considered the time to fully charge a capacitor?

 

[shaneshane1]

I am really rusty so forgive me but isnt "5" as in 5RC considered the time to fully charge a capacitor?

yes, thats why i assume that the "5" is the time constants!

 

[JimB]

Taking this thread back to the beginning, there is a simple formula for calculating the minimum value of smoothing capacitance required to give a required value on a supply.

For a simple (rectifier - capacitor - resistive load) type power supply

Capacitance = (load current x period)/ripple voltage

Where:
Capacitance is in uF
Load Current is in mA
Ripple Voltage is in Volts (peak to peak)
Period is the period of the ripple frequency in milli - seconds

So if we have a load of 500mA, the ripple must be less that 100mV p-p, and the supply frequency is 50hz.

The ripple period will be 10ms for a full wave rectifier.

The smoothing capacitor must be:

C = (500 x 10)/0.1 = 50,000uF

A rather large capacitor, in a practical circuit, you may decide that you can live with a bit more ripple, use a smaller capacitor and a voltage regulator.

JimB

Edit:
On reading some of shaneshanes posts, the expression "smoothing out" appears a couple of times.
The problem is that you have to define how smooth is "smooth".
Without specifying the ripple voltage, the amount of capacitance to use is anyones guess.

 

[shaneshane1]

Thanks JimB, i will use that formula from now on, thanks again!!!

 

[shaneshane1]

Taking this thread back to the beginning, there is a simple formula for calculating the minimum value of smoothing capacitance required to give a required value on a supply.

For a simple (rectifier - capacitor - resistive load) type power supply

Capacitance = (load current x period)/ripple voltage

Where:
Capacitance is in uF
Load Current is in mA
Ripple Voltage is in Volts (peak to peak)
Period is the period of the ripple frequency in milli - seconds

So if we have a load of 500mA, the ripple must be less that 100mV p-p, and the supply frequency is 50hz.

The ripple period will be 10ms for a full wave rectifier.

The smoothing capacitor must be:

C = (500 x 10)/0.1 = 50,000uF

A rather large capacitor, in a practical circuit, you may decide that you can live with a bit more ripple, use a smaller capacitor and a voltage regulator.

JimB

Edit:
On reading some of shaneshanes posts, the expression "smoothing out" appears a couple of times.
The problem is that you have to define how smooth is "smooth".
Without specifying the ripple voltage, the amount of capacitance to use is anyones guess.

I was just looking over the formula you posted and it seems that i get the same results with this formula!

C = (5 X load current in Amps) / Hz

It works and its really easy to remember, but i don't get what the 5 is?

I don't really care that much but it would still be good to know.

 

[JimB]

I was just looking over the formula you posted and it seems that i get the same results with this formula!

C = (5 X load current in Amps) / Hz

It works and its really easy to remember, but i don't get what the 5 is?

I don't really care that much but it would still be good to know.

Using your formula and my figures:

C = (5 x 0.1) / 50 = 0.01 presumably in Farads = 10,000uf

10,000uf is not 50,000uf (my capacitance)

Your formula makes no account of the allowable ripple, it is flawed.

JimB

 

[shaneshane1]

Using your formula and my figures:

C = (5 x 0.1) / 50 = 0.01 presumably in Farads = 10,000uf

10,000uf is not 50,000uf (my capacitance)

Your formula makes no account of the allowable ripple, it is flawed.

JimB

Yes it does, you put 0.1 which is 100mA and is 10,000uF

your example before was 500mA (0.500A) and is 50,000uF

they both are correct and mine is not flawed, look at it again!!!

 

[Hero999]

As a general rule of thumb, I always 10,000:mu:F per Amp for a ripple of 1V so multiplying the capacitance by 10 for 100mV makes sense to me.

 

[shaneshane1]

As a general rule of thumb, I always 10,000:mu:F per Amp for a ripple of 1V so multiplying the capacitance by 10 for 100mV makes sense to me.

So does do you think my formula gives a reasonable answer for a cap (with a 10% ripple voltage)?

 

[sniper007]

Maybe help this one

https://www.bancrofts.essex.sch.uk/pdfs/electronics_Vrip.pdf

 

[JimB]

Yes it does, you put 0.1 which is 100mA and is 10,000uF

your example before was 500mA (0.500A) and is 50,000uF

they both are correct and mine is not flawed, look at it again!!!

Mea Culpa, which I believe is Latin for "I have screwed-up"

You are quite correct, I used the wrong value of current.

However, your formula takes no account of the allowable ripple voltage, it is just "rounded-up" in the figure 5.

Also note that the equation given in the reference from Sniper007 is basically the same as mine, just the figures are transposed.

JimB

 

[arakaru]

i think.............

ripple voltage (Vr) gives by the equation for a full wave rectified one as

Vr = Io/C(2f)

f - frequency of waveform before rectifying

but normally says For many circuits a ripple which is 10% of the supply voltage is satisfactory. so if supply voltage is Vs

then Vr = Vs/10

So combine above two equations the required capacitance gives as C =5 × Io / Vs × f