0
\$\begingroup\$

I have designed a circuit in Cadence Virtuoso. As a first step, I used all the transistors without fingers (i.e. each transistor is a single transistor). Once the circuit satisfied all the requirements I wanted (dc gain, location of the poles, slew rate and so on) I decided to use fingers for each transistor in order to realize the stick diagram. To my astonishment, after using fingers, the dc analysis was completely different with respect to the original circuit: the dc voltages at each node were considerably different with respect to the original circuit.

Question: I know that fingering is used to reduce parasitic capacitances. Anyway, since a capacitor is an open circuit in dc, theoretically nothing should change as far as regards the dc analysis. Why then did Cadence give completely different results in dc when I converted my original circuit into the "fingered" version?

Here the circuit without fingers:

enter image description here

It is a two stages op-amp in open loop configuration (frequency compensation network has been omitted). If now I put 9 fingers for PM21, I get the following dc result:

enter image description here

As you can see, the output dc voltage has changed!

\$\endgroup\$
3
  • 1
    \$\begingroup\$ compare the effective W/Ls; should be the same \$\endgroup\$
    – analogsystemsrf
    Commented Oct 2, 2019 at 3:42
  • 1
    \$\begingroup\$ Your DC operating point may be to sensitive to devices' characteristic dispersion. Have you done sensitivities analysis? Do you have enough DC negative feedback? \$\endgroup\$
    – carloc
    Commented Oct 2, 2019 at 6:02
  • \$\begingroup\$ Your circuit's operating point is very sensitive to device parameters. That sensitivity has to be quantified and then addressed, e.g. by adding negative feedback, changing the topology of the circuit, etc. \$\endgroup\$
    – Kuba hasn't forgotten Monica
    Commented Apr 6, 2022 at 13:38

1 Answer 1

Reset to default
2
\$\begingroup\$

When you change the number of fingers indeed, in first order, the DC solution should not change. If it does then maybe something is wrong.

Some design kits might take edge effects into account so that a 10/1, fingers = 1 doesn't behave exactly the same as a 10/1, fingers = 2.

Note that in some design kits, if you start with a:

W/L = 10/1 transistor with fingers = 1

then change

fingers = 2

the transistor might become 2 x 10/1 = 20/1, so twice as wide!

In such a design kit you would need to enter:

W/L = 5/1 transistor with fingers = 2

and that would then give you the 10/1 transistor with fold = 2 so effectively two 5/1 transistors in parallel and a shared source (or drain).

Pro tip1: when placing a new transistor, make fingers something different from the default 1 unless it's a very small transistor. For example, when I need a 100/0.5 transistor I would make fingers = 5 or 10.

Pro tip2: create a layout cell view and instanciate a 10/1 transistor there. Then copy it so you have two. Now change fingers = 2 on the copy. Compare the layouts, is this what you expected?

\$\endgroup\$
3
  • \$\begingroup\$ Thank you for your answer. When I put the number of fingers, Cadence automatically puts the finger width in order to match with the width of the transistor: PM21 has W=90um, when I put 9 fingers W does not change and the finger width becomes 10um. I upload the schematic of the circuit. \$\endgroup\$
    – Stefanino
    Commented Oct 2, 2019 at 9:18
  • 2
    \$\begingroup\$ Indeed your DC solution changes! Do note that the change is small but after that changed voltage there's a large amount of gain so the output does change a lot but that's a consequence of a small change that is amplified. My guess is that the models of the transistors take some edge effects into account and that then causes the number of fingers to make a small difference. \$\endgroup\$
    – Bimpelrekkie
    Commented Oct 2, 2019 at 9:30
  • \$\begingroup\$ I guess that then it is better starting directly using fingers in the project in order to avoid these problems... \$\endgroup\$
    – Stefanino
    Commented Oct 2, 2019 at 9:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.