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In the transistor fingering diagram (b) below, shouldn't the two transistor fingers be connected at the right side as well in order for the fingers to actually appear in parallel? As it is right now it seems clear that each finger individually has half the distributed gate resistance that a non-fingered transistor would, but I don't see how they appear in parallel if they aren't connected at the right side so the gate fingers are just dangling.

(Diagram is from page 232 of Razavi.)

enter image description here

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2 Answers 2

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shouldn't the two transistor fingers be connected at the right side as well in order for the fingers to actually appear in parallel?

Aren't the two fingers already connected? I think they are.

All the gray area is conductive polysilicon and I see only one gray area in each transistor.

The only reason to also connect the right side of the gate in (b) is if you wanted a lower gate resistance. Then, like in (a) also add a metal connection to connect both sides.

the gate fingers are just dangling.

They're not really "dangling", it is OK to make a layout like this (I have done that thousands of times) and they are connected at the contact on the far left of (b).

In my opinion, in (a) and (B) the gate resistance is similar. If you really want a low gate resistance then the gate must be connected on both sides using metal, like is done in (a).

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  • \$\begingroup\$ So to clarify, if I modify (b) so that there's a gate connection on either end, does the distributed gate resistance decrease? Figure for reference: i.imgur.com/x2jNVeO.png \$\endgroup\$
    – HallEffect
    Mar 2, 2021 at 23:23
  • \$\begingroup\$ How connecting the right side of the gate in (b) helps reduce its resistance? As mentioned in the book, the resistance is already RG/4 where the right ends of the two fingers are not connected. What would resistance of the gate be if you connect the right ends? \$\endgroup\$
    – emnha
    Sep 27, 2021 at 9:15
  • \$\begingroup\$ @anhnha I have a feeling that you're not taking into account that you have to see that MOSFET's gate as a distributed RC network. If you see the gate just as one resistor or distributed resistors (but no capacitors) then indeed, connecting the gate at both sides does not make sense. I suggest you read: researchgate.net/figure/… then consider what it means if you want to switch a MOSFET on/off as fast as possible, then maybe what I have written makes more sense to you. \$\endgroup\$ Sep 27, 2021 at 11:10
  • \$\begingroup\$ I did think about distributed RC. Consider the folding gate as in the figure b, now you connect the right ends of these two fingers together BUT you don't add a contact to the right end. You only have contact on the left end. So would it make any difference with the case where you don't connect the right ends? It's like this: ibb.co/0Qv3RSP \$\endgroup\$
    – emnha
    Sep 27, 2021 at 12:15
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If you did connect the right ends of the fingers together, would it make any difference to the voltage across each finger, or to the current flowing through each finger. No, it wouldn't. Electrically these two fingers behave exactly as though they were connected together at both ends.

Note that the original text did not use the word "parallel" to describe this arrangement.

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    \$\begingroup\$ Electrically these two fingers behave exactly as though they were connected together at both ends I disagree! :-) OK, for DC I agree, it makes no difference whatsoever. However for AC and particularly very high frequencies, it can make a difference as the gate is actually a distributed RC (R = sheet resistance of the gate's polysilicon, C = Gate to channel capacitance) filter that can influence the AC behavior of the transistor. \$\endgroup\$ Mar 3, 2021 at 8:52
  • \$\begingroup\$ Long ago I designed a gmC bandpass filter where the the gate resistance influenced the transfer curve of my filter. Doing everything to get the gate's resistance down (including connecting the gates at both sides) solved that. \$\endgroup\$ Mar 3, 2021 at 8:52
  • \$\begingroup\$ @Bimpelrekkie Yes I think this is my question. Since clearly there is plenty of noise at non-DC, I think the figure (b) in the OP should have a gate connection on the right side in order to have a total distributed gate resistance of Rg/4. As it is right now, I think it's two separate resistors of value Rg/2. \$\endgroup\$
    – HallEffect
    Mar 4, 2021 at 2:47
  • \$\begingroup\$ @Bimpelrekkie I understand that the gate poly needs to be treated like a distributed RC, but if we assume that the gate material is uniform wouldn't the RC effect be exactly the same in both fingers? So that the voltage at any instant in time at the right ends of the fingers would be identical? \$\endgroup\$ Mar 4, 2021 at 19:33
  • \$\begingroup\$ The gate material is uniform (well, it should be). I see the gate as a distributed RC network (see: literature.cdn.keysight.com/litweb/pdf/ads2008/ccdist/ads2008/… ). Now consider the time it takes for voltage step (low to high) to travel from left to right. I hope you agree that there will be some delay, let's say 1 ns from left to right. That means it takes 1ns for the complete gate to be at the high voltage level. Now consider that we connect both ends, then it will take only 0.5 ns as the "slowest" point will be in the middle of the gate. \$\endgroup\$ Mar 4, 2021 at 19:52

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