Methods for Increase / Decrease the Delay in Clock / Data path.
(Effect of Transistor's Size On the Slew)
Static Timing analysis is divided into several parts:
- Part1 -> Timing Paths
- Part2 -> Time Borrowing
- Part3a -> Basic Concept Of Setup and Hold
- Part3b -> Basic Concept of Setup and Hold Violation
- Part3c -> Practical Examples for Setup and Hold Time / Violation
- Part4a -> Delay - Timing Path Delay
- Part4b -> Delay - Interconnect Delay Models
- Part4c -> Delay - Wire Load Model
- Part5a -> Maximum Clock Frequency
- Part5b -> Examples to calculate the “Maximum Clock Frequency” for different circuits.
- Part 6a -> How to solve Setup and Hold Violation (basic example)
- Part 6b -> Continue of How to solve Setup and Hold Violation (Advance examples)
- Part 6c -> Continue of How to solve Setup and Hold Violation (more advance examples)
- Part 7a -> Methods for Increase/Decrease the Delay of Circuit (Effect of Wire Length On the Slew)
- Part 7b -> Methods for Increase/Decrease the Delay of Circuit (Effect of Size of the Transistor On the Slew)
- Part 7c -> Methods for Increase/Decrease the Delay of Circuit (Effect of Threshold voltage On the Slew)
- Part 8 -> 10 ways to fix Setup and Hold Violation.
In the Last post we have discussed - how the wire length effects the slew? Now lets discuss about the effect of size of the transistors. Also before that let's discuss few basics also.
Size of transistor:
There are 2 parameters – Width and Length, by which you can decide the size of the transistor. For a particular technology – Channel - length is almost constant. So it means Width is going to decide the size of the transistor. Below figure will refresh your memory - about which, parameter I am talking.
If you want to increase the width of the transistor, then you have 2 options. One – Just increase the Width directly, Second -connect multiple transistors in parallel in such a way that their effective impact remains same. For example – if you want to manufacture a transistor with a width of 20um and a length of 0.2um then it’s similar (not exactly the same) to having four transistors connected in parallel, each with a width of 5um and a length of 0.2um. Here I am not going to discuss the difference in both the way of representation of Layout. If you are interested then you can check any basic book of CMOS design. Below figure will refresh your memory (Note: Below figure I have copied from www.eda-utilities.com )
Now since we are talking about the transition time /transition delay /slew, we know that it depend on the capacitance and resistance. So before we start to discuss how width (means size of the transistor) impact on the transition delay, we should know what all are the capacitance associated with the transistor. Below diagram help you in that. (Note: Below figure I have copied from www.eda-utilities.com )
How the capacitance are calculated (means whole derivation and explanation), I will discuss some other time, right now I am writing/copying the value of these capacitance directly.
Note: the reference of above formulas is from the book written by “J.P.uyemura” - Cmos Logic Circuit Design Edition -2002.
Now from the above, you can see that Gate Capacitance (this gate capacitance has 3 component – Gate to Base, Gate to Source and Gate to Drain) has dependence on the Width of the Channel (W). So it means, if you increase the width, Gate Capacitance will increase and Vice-Versa.
Source and Drain Capacitance has a multiplying factor As and Ad (which is equavilant to WxLs or WxLd). It means source and drain capacitance also increases with Width of the Channel and Vice-Versa.
Now let’s talk about the Resistance. Below Resistance formula is with respect to NMOS. You can derive similar formula for PMOS also (Just replace subscript “n” with “p” J ).
Now here the Resistance is inversely proportional to Width of the Transistor.
Effect of Device Size on the Slew (transition time) and Propogation Delay.:
I can’t write in a single line the effect of size of transistor on the slew because it’s not straight forward (I know you might have doubt on my statementL). There are some other factors which we have to consider. I hope, below paragraph helps you to understand the same.
Consider the above circuit. Gate “A” is the Driving Gate and Gate “B” is the Driven Gate. If we will expand this with the actual capacitance, it will be something similar to…
- Capacitance Cgd12 is the Gate Capacitance of Driving Gate A due to overlap in M1 and M2.
- Cdb1 and Cdb2 are the diffusion capacitances due to the reverse-biased pn-junction
- Cw is the wiring capacitance (pp, fringe, and interwire) that depends on the length and width of the connecting wire. It is a function of the fanout of the gate and the distance to those gates.
- Cg3 and Cg4 are the gate capacitance of the fanout gate (Driven gate).
If we increase the size of the transistor (Width of the Transistor) it’s current carrying capability increase. Means “larger is the size of a transistor, the larger is the driving capability (the ability to source or sink current) of a transistor”. Thus a larger transistor would normally make its output transition faster (when output load is constant). The output load of a driving gate consists of the source/drain capacitance of the driving gate, the routing capacitance of wire, and the gate capacitance of the driven gate.
The larger is the output load, the longer is the time to charge or discharge it. This would increase the transition (rise or fall) time and propagation delay.
Let me summarize few important points.
- On increasing the Size of Gate A –
- On Resistance Decreases (R - inversely proportional to W)
- Means large Driving capability (Ability to source or sink current)
- Decrease the time to charge the output load (capacitance) (Consists of source/drain capacitance of the driving gate, the routing capacitance of wire, and the gate capacitance of the driven gate) **
- Means - Output Transition time of Gate A and Input Transition time for Gate B decreases.
I am sure you have noticed that I have marked point 3 with ** because there are terms and conditions. :)
On increasing the Size of the Gate A – Source/Drain Capacitance also increases which are the part of output load of Gate A. Means it’s going to increase the output load. That means as I have mentioned in my point no 3 – that can be possible only when S/D Capacitance of Driving gate are not dominating the rest of the Capacitance. Which is only possible when either “Net capacitance is large” (length of wire is large) or “Size of the driven gate (Gate B) is large” (which increase the Gate capacitance of GateB) or “Both should be true”.
So for Minimizing Propagation Delay, A fast Gate/Cell is required, which is only possible by
- Keeping the output capacitance CL small (it decreases the charging and discharging time). And for this
- Minimize the area of drain pn junctions. (Decrease W)
- Minimize Interconnect capacitance. (Decrease wire/net Length)
- Avoid large fan-out. Means Minimize gate capacitance of Driven Cell. (Decrease W of Driven cell)
- Decrease L (For a particular technology Node It’s almost constant)
- Increase W
- But this increases pn junction area and hence CL.
So if we want to use the size of the transistor as one of the parameter to increase/decrease of the propagation/transition delay, then we should have understanding of the design and also it depends on the property of Driven Cell and Net length also.
Few last points:
- "Delay reduces with increase in input transition and constant load capacitance".
- "Delay increases with increase in output capacitance and constant input transition"
- Because on increasing the output capacitance – charging and discharging time will increase.
So we can say that
The delay of cell directly depends on input transition and output capacitance.
In the next post we will discuss about the effect of Threshold voltage of the Transistor on the "Transition Delay" and "Propagation Delay".