# capacitor question.

Status
Not open for further replies.

#### allegro

##### New Member
I have a 2.2V @ 20mA LED, that is driven by a 5V PSU, and therefore is connected to a 140Ω resistor.

I'd like to use to purchase a capacitor, that i'd be able to charge the capacitor to 5V, and that when I connect the capacitor to the LED circuit, as the PSU of this circuit, the LED's volume of illumination will not decline from 90% for 5 seconds.

I treated the LED as 2.2V/20mA=110Ω resistor, and did the next math:
4.5V = 5V * e^[-5sec/(250Ω*C)]
I reached C = 190mF.

Was it ok to treat the LED as a 110Ω resistor?
Is there any other way to reach this answer, reaching it using Q of the capacitor?

• alphacat

#### ericgibbs

##### Well-Known Member
I have a 2.2V @ 20mA LED, that is driven by a 5V PSU, and therefore is connected to a 140Ω resistor.

I'd like to use to purchase a capacitor, that i'd be able to charge the capacitor to 5V, and that when I connect the capacitor to the LED circuit, as the PSU of this circuit, the LED's volume of illumination will not decline from 90% for 5 seconds.

I treated the LED as 2.2V/20mA=110Ω resistor, and did the next math:
4.5V = 5V * e^[-5sec/(250Ω*C)]
I reached C = 190mF.

Was it ok to treat the LED as a 110Ω resistor?
Is there any other way to reach this answer, reaching it using Q of the capacitor?
hi,
Using Q = CV, where Q = I * t

So C= [I *t] /V

Is this what you are asking.?

#### ericgibbs

##### Well-Known Member
hi,
This option will give a 'steady' current thru the LED for 5secs.

Its a big capacitor.! #### Attachments

• 25.3 KB Views: 97

#### allegro

##### New Member
Thank you.

How come we both got different results?
you reached C = I*t/V = 20mA*5sec/5V = 20mF.
Isnt it a problem that in your formula you treat I as being constant during the whole 't' seconds?
When the capacitor discharges, its current doesnt stay I for t seconds, but it declines.

I was wrong when I treated the LED as resistor, since its current doesnt change linearly with changing its voltage.
So my formula was wrong, but i dont see why yours is correct.

Last edited:

#### MikeMl

##### Well-Known Member
I have a 2.2V @ 20mA LED, that is driven by a 5V PSU, and therefore is connected to a 140Ω resistor.

I'd like to use to purchase a capacitor, that i'd be able to charge the capacitor to 5V, and that when I connect the capacitor to the LED circuit, as the PSU of this circuit, the LED's volume of illumination will not decline from 90% for 5 seconds.

I treated the LED as 2.2V/20mA=110Ω resistor, and did the next math:
4.5V = 5V * e^[-5sec/(250Ω*C)]
I reached C = 190mF.

Was it ok to treat the LED as a 110Ω resistor?
Is there any other way to reach this answer, reaching it using Q of the capacitor?
Sure, simulate it using Spice.

You would be much better off starting with a smaller capacitor charged to a much higher voltage.

#### Attachments

• 264.2 KB Views: 71

#### ericgibbs

##### Well-Known Member
Thank you.

How come we both got different results?
you reached C = I*t/V = 20mA*5sec/5V = 20mF.
Isnt it a problem that in your formula you treat I as being constant during the whole 't' seconds?
When the capacitor discharges, its current doesnt stay I for t seconds, but it declines.

I was wrong when I treated the LED as resistor, since its current doesnt change linearly with changing its voltage.
So my formula was wrong, but i dont see why yours is correct.
hi,
How do you get 20uF.??? Cap = [0.02 * 5] /5 = 0.1/5 = 0.02Farad

Also in your calculations you have not allowed for the voltage drop across the led of 2.2V, so your actual 'usable' Voltage is 5 - 2.2 = 2.8V ~ 0.035F.

#### Hero999

##### Banned

You shouldn't use mF because it's too easily confused with micro-Farads, use Farads for anything ≥0.1F and µF for anything ≤99,999µF.

#### allegro

##### New Member
To Eric,
Why using C = I*t/V is correct? the current wont remain permament during the whole t seconds.

To MileML,
If I'll use a smaller capacitor charged to higher voltage i'll apply to much voltage on the LED.
I'm not interested in inserting numbers into Pspice untill i reach the right answer, i want to understand.

#### MikeMl

##### Well-Known Member
To Eric,
...
If I'll use a smaller capacitor charged to higher voltage i'll apply to much voltage on the LED.
No, notice the size of the current limiting resistor that I used. You could start with a 1000V in the capacitor, adjust the resistor, and the forward drop across the LED would still be only ~2.2V!!!

Energy stored in a capacitor is proportional to CV²; that is why a higher voltage lets you use a smaller capacitor for a fixed time...

#### allegro

##### New Member
Thanks for this.
What is the advantage in having a capacitor with larger energy stored in it?

I see indeed that in the graph, by using a smaller capacitor, which takes smaller space, you achieved larger current during the first 5 seconds.

Could you detail about the consequences of the capacitor's energy on the current?

--
For example, if two capacitors, C1 and C2, have been charged with the same amount of charges, Q, but C1 stores larger energy than C2, what is its advantage over C2? how does it come to fruition?

Last edited:

#### allegro

##### New Member
This issue about capacitor energy is really interesting me.

#### ericgibbs

##### Well-Known Member
This issue about capacitor energy is really interesting me.
hi,
It would help if you told us about the 5sec/cap project purpose, rather than the solution. #### allegro

##### New Member
the purpose if for a capcitor to keep the led at above 90% of its 20mA forward current for the first 5 seconds.

I'd like to know why its better to have a small cap with larger energy, rather then larger cap with smaller energy, assuming both RC (R is the needed limitting current resistor) are identical, as Mike showed.

#### MikeMl

##### Well-Known Member
the purpose if for a capcitor to keep the led at above 90% of its 20mA forward current for the first 5 seconds.

I'd like to know why its better to have a small cap with larger energy, rather then larger cap with smaller energy, assuming both RC (R is the needed limitting current resistor) are identical, as Mike showed.
First principles (see wiki)

Another way to explain it:

Assuming a constant 20mA current discharge of a charged capacitor, how long can a given capacitor support that current? An Ampere is a Coulomb per second, or I=Q/t. Rearranging gives Q=I*t.

Also, amount of charge Q in/out of a capacitor when it's voltage changes by ΔV is Q=C*ΔV, where C is the capacitance in Farads.

Which can be written as C*ΔV=Q=I*t. Solving for C=I*Δt/ΔV

So, your problem is to supply 20mA for 5 sec. The bigger you make ΔV, the smaller C will be.

Your underlying problem is one that EricGibbs already showed you the solution to: If you want the current through the LED to be constant during the 5 seconds, then you need a more complicated circuit between the capacitor and the LED, namely a CONSTANT-CURRENT source. That is what Eric's circuit does...

#### Attachments

• 317.1 KB Views: 61
Last edited:

#### colin55

##### Well-Known Member
There are two ways to get the energy you require.
1. A high value electrolytic (capacitance - in uF) and charge it to a low voltage, or:
2. A low value electrolytic (capacitance - in uF) and charge it to a high voltage.
Basically it is much easier to get a low-value electrolytic (capacitance - in uF) and charge it to a high voltage to get the energy you require.
There is a limit to this and the voltage on the capacitor should not be more than 3 times the minimum supply voltage.
If the minimum supply voltage is 6v and the current is 15mA, the dropper resistor will be 270R. If the capacitor is charged to 18v, the current will start at 45mA and drop to 15mA.
From these values you can work out the size of the "super cap" you need.

##### Banned
Or you can use this handy dandy calculator. Courtesy of powerstor. You pick your input voltage, your acceptable cutoff voltage, and required hold up time and current draw and it pops out the value of capacitance you need. Just to warn you to maintain 90% brightness for 5 seconds is going to be a very large capacitor. To do it sanely you should probably use a constant current source instead of a resistor the value of the cap you need is going to be at least one or two orders of magnitude smaller.

To maintain a voltage from 5.0 volts to 2.2 volts with a draw of 20ma's for 5 seconds requires a capacitance of .0357 farrads This would only produce a uniform brightness in the LED with a constant current source.

I had to zip the file, it's not a supported file type in the upload section. It's an Excel worksheet.

#### Attachments

• 9.5 KB Views: 152
Last edited:

#### allegro

##### New Member
First principles (see wiki)

Another way to explain it:

Assuming a constant 20mA current discharge of a charged capacitor, how long can a given capacitor support that current? An Ampere is a Coulomb per second, or I=Q/t. Rearranging gives Q=I*t.

Also, amount of charge Q in/out of a capacitor when it's voltage changes by ΔV is Q=C*ΔV, where C is the capacitance in Farads.

Which can be written as C*ΔV=Q=I*t. Solving for C=I*Δt/ΔV

So, your problem is to supply 20mA for 5 sec. The bigger you make ΔV, the smaller C will be.
I read your post thoroughly and I dont see in your calculation of the capacitor's value any reference to the capacitor's energy.

You said that its better to increase V in order to get smaller C.
I understand what is the benefit in having smaller C - it takes less space on the PCB.

In other post you said that increasing V increases the capacitor's energy.
I dont understand what is the benefit in increasing the energy.

There are two ways to get the energy you require.
How can I know what energy I require?
If I want to have a capacitor to be able to supply 20mA for 5 seconds, there are many options for the capacitor's enegy, since I can pick different pair of (C,V).
My question is what larger energy gives you? why would you want to have large V and small C,

#### ericgibbs

##### Well-Known Member
hi allegro,

Using a hign voltage with a smaller capacitor could give the same energy storage as a larger cap with a lower voltage.

The problem is the release of this stored energy into a fixed load ie: resistor and led.

Initially the current into the load would be high, falling exponentially.
So very bright led going to dim over a 5 second period.

This is where a constant current circuit has some advantage, it controls the release of the stored energy,
this gives the led a steady brightness over the 5sec period.

Last edited:
Status
Not open for further replies. 