# Wilson current mirror problem

Thread Starter

#### OggiSerbia

Joined Dec 28, 2017
34
I have an easy Wilson current mirror problem to solve. I tried to do it but I am not sure if it is correct.
Here is the circuit:

The problem is to find (Ir/I), i.e. Currence Ir over currence I. It is assumed that all the transistors are equal: beta (currence amplification) is the same.
Here is my solution:

Image No.1:

Image No.2:

Image No.3:

Thread Starter

#### OggiSerbia

Joined Dec 28, 2017
34
I'm sorry. My bad. Than... Ignore "Wilson".

#### crutschow

Joined Mar 14, 2008
23,742
It is a Wilson mirror buy why does your circuit have two transistors (Q3 and Q4) in parallel?

Thread Starter

#### OggiSerbia

Joined Dec 28, 2017
34
It is a Wilson mirror buy why does your circuit have two transistors (Q3 and Q4) in parallel?
I really don't know. It was given by the professor just like that... :/

#### danadak

Joined Mar 10, 2018
3,682
Q3, Q4 scaling the output current by 2 ? Assuming they are interdigitated match
pair geometry. Or splitting current to reduce T effects ?

Or eliminating CM effects of source I source ? Or eliminating output supply
effect on output current due to finite output sink. To raise effective complience
of mirror and increase finite dynamic R of output transistor ? The analysis
should lead to answer.

Guess on my part.

Regards, Dana.

Last edited:

#### crutschow

Joined Mar 14, 2008
23,742
I really don't know. It was given by the professor just like that.
Perhaps he put it in there just to make the calculations a little more difficult.
But someone else will have to check your math as I normally stay away from math as much as possible.

Thread Starter

#### OggiSerbia

Joined Dec 28, 2017
34
Perhaps he put it in there just to make the calculations a little more difficult.
But someone else will have to check your math as I normally stay away from math as much as possible.

Hahahaha I also guess it's the reason. OK. Thank you We will wait.

#### Jony130

Joined Feb 17, 2009
5,089
Your math looks good. So, don't worry.

Thread Starter

#### OggiSerbia

Joined Dec 28, 2017
34
Your math looks good. So, don't worry.
Thank you very much. I was worried more about setting up the problem but if everything is OK, then great. O

#### MrAl

Joined Jun 17, 2014
6,785
I have an easy Wilson current mirror problem to solve. I tried to do it but I am not sure if it is correct.
Here is the circuit:
View attachment 155338

The problem is to find (Ir/I), i.e. Currence Ir over currence I. It is assumed that all the transistors are equal: beta (currence amplification) is the same.
Here is my solution:

Image No.1:
View attachment 155339
Image No.2:
View attachment 155341
Image No.3:
View attachment 155340

Hello there,

I used a different approach by substituting an equivalent topology so i could write the equations more directly and got the same result. So you must have done something right

Last edited:

#### atferrari

Joined Jan 6, 2004
3,573
Hello there,

I used a different approach by substituting an equivalent topology so i could write the equations more directly and got the same result. So you must have done something right
I liked that!!

Thread Starter

#### OggiSerbia

Joined Dec 28, 2017
34
Hello there,

I used a different approach by substituting an equivalent topology so i could write the equations more directly and got the same result. So you must have done something right

View attachment 155404
Great. I'm very happy. Have a nice day!

#### MrAl

Joined Jun 17, 2014
6,785
I liked that!!
Hi,

I also updated the schematic to show the transistor Betas now.

Also, i should have mentioned that for Beta 97 or better the ratio Ir/I is within 1 percent of 1:1.
For Beta 197 or better the ratio Ir/I is within 1/2 percent of 1:1.
So the higher the Beta, the closer the ratio is to 1:1.

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