Monte Carlo Mismatch and Percentage

[CAMALEAO]

Hi guys,

After running some Monte Carlo simulations in a current mirror, I would like to compute the amount of mismatch in percentage.

To compute the mismatch in percentage I should do it taking into account the (3 sigma * std. deviation) or just the std. deviation?

For example I see people saying that the current mirror has 5% of mismatch. This calculation was done taking into account the (3 sigma * std. deviation) or just the std. deviation?

To compute the mismatch in percentage I should do, for example:

(Mean iout - iref) / Mean iout

or

(Mean iout - iref) / iref?

In either the case the mean iout will depend on the first question, that is:

Mean iout = Mean iout +- std. deviation

or

Mean iout = Mean iout +- (3 sigma * std. deviation)

Practical case for example:

Iref = 1uA
Mean Iout = 1.01uA
Std. Dev. = 10nA

Mismatch % = ( 1.01uA - 1uA ) / 1.01uA
Mismatch % = ( 1.01uA - 1uA ) / 1uA

Mismatch % = ( (1.01uA + 3 * 10nA) - 1uA ) / 1.01uA
Mismatch % = ( (1.01uA + 3 * 10nA) - 1uA ) / 1uA

Regards.

 

[kemiyun]

Both of the definitions you gave are valid. One is absolute error, which is useful if you are designing a current reference, the other one is the error around the mean value, you can use the latter for matched circuits that don't require absolute accuracy but good matching between them.

The interval width (number of standard deviations) depends completely on your choice. If you want almost all of your chips to perform within specs then it should be 3 standard deviations, if you are more relaxed about it then it can be 1 standard deviation. 3 sigma = 99% of the samples will be within range, 2 sigma = 95 or something I forgot and 1 sigma is around 65%. It's your decision to choose which one to use.

 

[CAMALEAO]

Thanks for the reply.

So, this one:

Iref = 1uA
Mean Iout = 1.01uA
Std. Dev. = 10nA

Mismatch % = ( 1.01uA - 1uA ) / 1.01uA

Is better for absolute error when designing current reference.

The other one:

Iref = 1uA
Mean Iout = 1.01uA
Std. Dev. = 10nA

Mismatch % = ( 1.01uA - 1uA ) / 1uA

is the error around the mean value? Better to assess the matching between them.

Is that it?

 

[kemiyun]

Pretty much, but you need to decide what you want, I just gave examples. You need to have your own justification for using whichever you want or whichever you think is the performance indicator in your circuit.

 

[CAMALEAO]

Well, if you are characterizing the mismatch in a current mirror, for a couple of W/L, the second one, this one:

Iref = 1uA
Mean Iout = 1.01uA
Std. Dev. = 10nA

Mismatch % = ( 1.01uA - 1uA ) / 1uA

then is better?