# Figuring out capacitor value

#### PsySc0rpi0n

Joined Mar 4, 2014
1,772
Hello.

I have removed an SMD capacitor out of a circuit and put it in series with a resitor to try to get its capacitance value.
Starting from time constant formula ζ = RC.
So, I used a 10kΩ resistor (real value of 9630Ω) and 5.05V on the power suply.
63% of 5.05V is 3.18V, so on the scope I checked when the capacitor reached this voltage and I got 188ms.
Knowing this I got the C out as C = ζ / R <=> C = 188ms / 9630Ω = 19.52μF.
Ok, so the nearer nominal value is 22μF, so I think ~11% of error was OK.
(did a check by doing the same math but for 22μF value - time constant of 211ms - and compared the error and I also got 11%)

But then, I tried the same exercise but with 12.05V.
I got a time for the 63% of voltage of 0.103ms.
I found it weird because the ζ formula doesn't depend on charging voltage.
And just to make sure I wasn't doing anything wrong I went until the end and got around 10.70μF.

So, shouldn't I get the same time constant regardless the charging voltage? Shouldn't be the charging current the one being different?

#### ericgibbs

Joined Jan 29, 2010
18,865
Hi Psy,
Consider what parameter that is 63%., it's the final-applied voltage.
E

#### PsySc0rpi0n

Joined Mar 4, 2014
1,772
Hi Psy,
Consider what parameter that is 63%., it's the final-applied voltage.
E
Isn't the voltage that I applied to the capacitor to charge it up to that voltage?
I applied 2 voltages. One of 5.05V and another of 12.05V. So, the 63% is with respect to this 2 values of voltage, right?

#### ericgibbs

Joined Jan 29, 2010
18,865
hi Psy,
Plots of charge voltages for same R/C but at 5V and 10v
E

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#### PsySc0rpi0n

Joined Mar 4, 2014
1,772
hi Psy,
Plots of charge voltages for same R/C but at 5V and 10v
E
Yes, but this seems to kind of match the math but in my case above, it doesn't seem to match.

The 1st time constant seems to be independent of charge voltage. But not in my case because it gives me different values of 1st time constant and capacitance!

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#### eetech00

Joined Jun 8, 2013
3,958
So, shouldn't I get the same time constant regardless the charging voltage?
No. The charge current must held constant.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,772
No. The charge current must held constant.
ok, so am I not getting the same value for the 1st time constant? I didn't get it yet!

#### WBahn

Joined Mar 31, 2012
30,076
Measuring the time constant this way tends to be pretty inaccurate, because it's hard to get a good start time value.

This is why most scopes have 10%/90% lines on them.

First, use a frequency that is slow enough so that you have a period of at least ten time constants and use a slow sweep speed that you get a couple of full cycles displayed. Then take the vertical adjust out of the cal mode and adjust the gain and offset until the top and bottom of the waveform is at the 0% and 100% lines. Then increase the sweep speed (in cal mode!) and move the waveform horizontally until it crosses the 10% line on the first vertical grid line (not the left edge, but the first one within the screen region). Use a sweep speed that puts the point at which it crosses the 90% line as far to the right as possible. Measure that time. This is the 10/90 rise time.

If you do the math (and you should do it, at least once), the relationship between the time constant and 10/90 rise time (for a first-order step response) is

$$T_{10/90) \; = \; ln(9) \tau \; \approx \; 2.2 \tau$$

#### MisterBill2

Joined Jan 23, 2018
18,584
In addition to the other comments, it is possible that the leakage current of the capacitor varies with the applied voltage, and possibly not in a linear manner.

#### WBahn

Joined Mar 31, 2012
30,076
In addition to the other comments, it is possible that the leakage current of the capacitor varies with the applied voltage, and possibly not in a linear manner.
I wouldn't expect leakage current to be very much for a 22 µF SMD cap. Plus, leakage current that is unaccounted for would result in a higher calculated capacitance.

Applying 12 V to a capacitor of unknown specs can result in interesting, and energetic, results given that some caps of that size have rated voltages of only 2 V.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,772
Measuring the time constant this way tends to be pretty inaccurate, because it's hard to get a good start time value.

This is why most scopes have 10%/90% lines on them.

View attachment 315977

First, use a frequency that is slow enough so that you have a period of at least ten time constants and use a slow sweep speed that you get a couple of full cycles displayed. Then take the vertical adjust out of the cal mode and adjust the gain and offset until the top and bottom of the waveform is at the 0% and 100% lines. Then increase the sweep speed (in cal mode!) and move the waveform horizontally until it crosses the 10% line on the first vertical grid line (not the left edge, but the first one within the screen region). Use a sweep speed that puts the point at which it crosses the 90% line as far to the right as possible. Measure that time. This is the 10/90 rise time.

If you do the math (and you should do it, at least once), the relationship between the time constant and 10/90 rise time (for a first-order step response) is

$$T_{10/90) \; = \; ln(9) \tau \; \approx \; 2.2 \tau$$
I'll check if my scope has those lines and will try to follow your suggestions, but only next monday because my scope is at work!
My scope is a Rigol MSO1104Z.

I wouldn't expect leakage current to be very much for a 22 µF SMD cap. Plus, leakage current that is unaccounted for would result in a higher calculated capacitance.

Applying 12 V to a capacitor of unknown specs can result in interesting, and energetic, results given that some caps of that size have rated voltages of only 2 V.
That capacitor was in a Max745 module. This one here:
https://pt.aliexpress.com/item/32851664522.html

And the cap is this one circled in red.

I was measuring it because the one pointed by the green arrow burned to coal due to some short circuit that happened when I was taking something that was sitting next to the module that was already connected to a battery pack. There was a wire that touched somewhere and made that cap to short and burn.
Now, I was assuming that those 2 caps are of the same capacity and was measuring the one next to the one that burned to replace it by another one of the same capacity!
I can't know the voltage rating of these caps, I think.

#### WBahn

Joined Mar 31, 2012
30,076
I'll check if my scope has those lines and will try to follow your suggestions, but only next monday because my scope is at work!
My scope is a Rigol MSO1104Z.

That capacitor was in a Max745 module. This one here:
https://pt.aliexpress.com/item/32851664522.html

And the cap is this one circled in red.
View attachment 316071

Now, I was assuming that those 2 caps are of the same capacity and was measuring the one next to the one that burned to replace it by another one of the same capacity!
Those caps don't look the same to me -- the one with the green arrow appears narrower. Since it appears that they are in parallel, my guess would be that one of then is one or two orders of magnitude smaller than the other in order to get better suppression of noise at different frequencies.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,772
Measuring the time constant this way tends to be pretty inaccurate, because it's hard to get a good start time value.

This is why most scopes have 10%/90% lines on them.

View attachment 315977

First, use a frequency that is slow enough so that you have a period of at least ten time constants and use a slow sweep speed that you get a couple of full cycles displayed. Then take the vertical adjust out of the cal mode and adjust the gain and offset until the top and bottom of the waveform is at the 0% and 100% lines. Then increase the sweep speed (in cal mode!) and move the waveform horizontally until it crosses the 10% line on the first vertical grid line (not the left edge, but the first one within the screen region). Use a sweep speed that puts the point at which it crosses the 90% line as far to the right as possible. Measure that time. This is the 10/90 rise time.

If you do the math (and you should do it, at least once), the relationship between the time constant and 10/90 rise time (for a first-order step response) is

$$T_{10/90) \; = \; ln(9) \tau \; \approx \; 2.2 \tau$$
Hello.
I'm next to my scope, and unless I have to activate those 10% and 90% lines, I don't see them.
What I used were the cursors of my scope to fin the values.
For instance, for a known 1μF, 63V, charging it through the same 10kΩ (9630Ω) and 5.05V I get this:

Second vertical cursor is on the intersection of the second horizontal cursor at 3.18V with the plot line and gives me 13.20ms.
Doing the same math, I get 1.37μF. It's rated at 1μF so I have about 37% of error.
Even though the error margin is a bit high, I still can say that this is a match.

Doing the same for 12.05V I get more or less the same values:
1.41μF. A little bit worse but still in the same range. I don't know why with the other caps (the SMD ones) I didn't get similar values!

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#### MisterBill2

Joined Jan 23, 2018
18,584
I suspect that capacitance meters use a different approach. Many of them provide the capability of testing at different frequencies. and certainly there are other methods that require much less skill. A common scheme uses an AC voltage at a specific frequency to determine capacitive reactance. That can be quite accurate, and use less complex equipment.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,772
Ok, guys.
I remember I have here one of those devices that test components and tells us what they are and some of the common values.

I tested one of these SMD capacitors and it tells me 116.7μF, ESR= .00Ω and a Vloss = 1.4%.
I tried to use the scope and the time constant and got very different values.

I would like to understand how to do it @WBahn way just to check what I would get but I can't find those lines he refers to on my scope!

Joined Feb 7, 2024
125
I suspect that capacitance meters use a different approach. Many of them provide the capability of testing at different frequencies. and certainly there are other methods that require much less skill. A common scheme uses an AC voltage at a specific frequency to determine capacitive reactance. That can be quite accurate, and use less complex equipment.
They use the charge current to determine capacitance in DMM. Bench type capacitor checker tests leakage current in addition to capacitance.
Here is a schematic of one that would be in a DMM:

Of course this DIY one would be easy to build:

The DIY project article for the above schematic, is here: https://www.diy-electronic-projects.com/p207-Capacitance-Meter

#### Tonyr1084

Joined Sep 24, 2015
7,905
Take a dual trace scope and set one line on the unknown cap and the second line on a known value cap. Set both traces up exactly the same way and check the sine waves. If they match then you know what value you have. If the one under test has a higher frequency then it's smaller than the known value. Swap the known for another known value and retest.

I built a 555 circuit with multiple capacitances and can switch out one and another in, or I can even switch in multiple caps. If I need a frequency above or below what's built in I have an IC socket that I can push a cap into of any value I want or need. Using two traces and matching the sine wave - you will know the value reasonably well. I, too, have wanted to know a good way of determining SMT caps. I may just build such an apparatus. Thinking of taking a quad NAND gate and setting them up as oscillators. One from a known cap value and one from a test circuit. Match them and I know what I have.

#### MrAl

Joined Jun 17, 2014
11,494
Hello.

I have removed an SMD capacitor out of a circuit and put it in series with a resitor to try to get its capacitance value.
Starting from time constant formula ζ = RC.
So, I used a 10kΩ resistor (real value of 9630Ω) and 5.05V on the power suply.
63% of 5.05V is 3.18V, so on the scope I checked when the capacitor reached this voltage and I got 188ms.
Knowing this I got the C out as C = ζ / R <=> C = 188ms / 9630Ω = 19.52μF.
Ok, so the nearer nominal value is 22μF, so I think ~11% of error was OK.
(did a check by doing the same math but for 22μF value - time constant of 211ms - and compared the error and I also got 11%)

But then, I tried the same exercise but with 12.05V.
I got a time for the 63% of voltage of 0.103ms.
I found it weird because the ζ formula doesn't depend on charging voltage.
And just to make sure I wasn't doing anything wrong I went until the end and got around 10.70μF.

So, shouldn't I get the same time constant regardless the charging voltage? Shouldn't be the charging current the one being different?
The charging of a capacitor with a fixed resistor in series is (RC is R*C):
Vc=Vcc*(1-e^(-t/RC))

Setting Vc to some percentage of Vcc:
pc*Vcc=Vcc*(1-e^(-t/RC))

and since Vcc is on both sides we get:
pc=(1-e^(-t/RC))

and solving for t we get:
t=ln(1/(1-pc)*RC

Now setting pc to 0.90 and subtracting that with pc=0.10 we get:
t=ln(9)*RC

which is close to:
t=2.2*RC

To charge to one time constant t=RC we would get:
Vc=Vcc*(1-e^(-1))

Vc=Vcc*0.632

Some of the things you have to consider is the impedance of the voltage source being used, and also the ESR of the capacitor.

Another experiment you can do if you have a decent sine frequency generator is to put a resistor in series with the cap and then power this by the sine source, then the AC voltage across the cap is (Vs is the voltage of the sine source):
Vc=Vs/sqrt(w^2*C^2*R^2+1)

and the phase shift (in radians) measured across the cap is:
Ph=-atan(w*R*C)

This gives you two different measurements you can check.

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#### WBahn

Joined Mar 31, 2012
30,076
Ok, guys.
I remember I have here one of those devices that test components and tells us what they are and some of the common values.

I tested one of these SMD capacitors and it tells me 116.7μF, ESR= .00Ω and a Vloss = 1.4%.
I tried to use the scope and the time constant and got very different values.

I would like to understand how to do it @WBahn way just to check what I would get but I can't find those lines he refers to on my scope!
If you have cursors, then you don't need the lines, just use the cursors to set one line at 10% and one line at 90% the same way that you are using them to set one of your cursors at 63% now.

#### MrAl

Joined Jun 17, 2014
11,494
Or just look at the dang vertical scale and figure out where the 10 percent and 90 percent marks would be located