ptat current source behaviour

[geozog86]

Hello!
I have a theoretical question:
I am using a PTAT current source, and i found the formula you can see next to it. The problem is that in a transient simulation i get a voltage across the resistor of 100mV, while my fingers are f3=8, f1=f2=f4=2, so i should expect a voltage of ln(4)*26mV=less than 40mV. I swept the situation by changing the value of the resistor and strange thing, for lower resistor, i saw higher current and more voltage across the R, and from size on it "stabilizes" in 100mV!!!
What am i not taking into account in my way of thinking?

Thx everyone!

 

[saiaditya]

hi..i don't think that the equation is correct...the voltage across the resistor will be delta_vgs = vgs1-vgs3 and thus the current will be delta_vgs/R...thus for smaller R u get larger currents

 

[geozog86]

i would agree with you...but the formula is derived exactly starting from this formula v_R=delta_vgs and ends up in this ln form .... theory on ptat current source confuses me obviously

 

[saiaditya]

i am not sure how m2, m4 can come into the picture...assuming strong inversion...delta_vgs = vgs1-vgs3 = sqrt(2*I/k) - sqrt(2*I/4*k) (neglecting body effect...factor of 4 because f3/f1 =4)

 

[leo_o2]

It depends on what region transistors work. The circuit can be designed for satuation region or subthreshold region. When they work under subthreshold region, the voltage drop will be close to the formula. On the other hand, satuation region will have different story.

 

[palmeiras]

I agree with leo_o2.
Take a look in your circuit and check the operation point of M1 and M3, they should operate in weak inversion. Transistors M2 and M4, can operate in strong region in order to improve the matching.
The above formula is obtained when the current is an exponential function of VBE, what is only valid for weak inversion operation.

Best regards.