7.4a | 7.4b | 7.4c |

Maximum Transition Time | Maximum Fanout | Maximum And Minimum Capacitance |

Design Constraints are divided into several parts because it’s really a wide and important topic. I want to discuss this in detail. I have also noticed that lot of information is present in internet but those are bits and pieces. So I am trying my best to cover everything here in a proper way. Let me know in case any of you have any feedback.

- Part 1a -> Basics of Design Constraints and Details of "Maximum Transition Time" (max_transition)
- Part 1b -> Maximum Fanout Constraint. (max_fanout)
- Part 1c -> Maximum (and minimum) capacitance (max_capacitance and min_capacitance)
- Part 1d -> Cell degradation (cell_degradation)

**Note: Rest Of the parts is still in development.**

In the last blog we have discuessed that Design Constraints are of 2 types:

- Design Rule Constraints
- Optimization Constraints

And further Classification of Design Rule Constraints are -- 4 types:

- Maximum transition time
- Maximum fanout.
- Maximum (and minimum) capacitance.
- Cell degradation

Details of Maximum Transition Time - we have discuessed in last blog. In this part we will discuss about the "Maximum Fanout".

__Maximum fanout:__- The
**maximum fan-out**of an output measures its load-driving capability: - It is the greatest number of inputs of gates to which the output can be safely connected.
- Fanout load is a dimensionless number
- This info is present in the .lib file. Please see the below snapshot of .lib with respect to one cell definition.
__“max_fanout” is available only for “output” pins.__- Whether the circuit is violated or not – calculated as per the following e.g
- Way to calculate:
- Sum up all the fanout loads for inputs driven by a pin (driving pin).
- If the sum of the fanout loads is not more than the max_fanout value, the net driven by X is valid or say not violated , else you have to make changes in your design by adding a buffer or anyother way.

As in the above figure : the max_fanout of X defined aas per the l.lib file is 20. Now if few loads are connected as the output at X then first calculate total fan loads.

So Total Fanout Load is 1.0 +1.0 +3.0 +2.0 = 7.0

Since max_fanout of X > 7.0 - so no violation.

Now let assume that you have set the fanload of OUT1 18.0 ( by using set_fanout_load 18.0)

then Total Fanout Load is 1.0+1.0+18.0+2.0 = 22.0

Since max_fanout of X < 22.0 - so its violation.

Similarly It may be that in you reset the max_fanout of X to 6.0 (by using set_max_fanout 6.0 )

then Toatal Fanout Load is 1.0 +1.0 +3.0 +2.0 = 7.0

Since max_fanout of X < 7.0 - so its violation.

*Now same question : What's the need/importance/significance of this parameter (max_fanout) in the design?*Typically manufacturer defines the maximum input current for a particular cell at each logic level. So fanout of a cell A is the max no of input that can be connected to an output (of A) before the current requirement by any of the input exceeds the current that can be delivered by the output while maintain the correct logic level. ( very big sentance :) Read it 4-5 times.. :) ). Actually these are very basics- But its very important to understand this here. :) .. (So I have copied this from Ref1.)

*The fan-out depends on the amount of electric current a gate can source or sink while driving other gates*.

- When an output pin is HIGH, the I
_{IH}requirements for all receivers must add-up to be ≤ to the driver's I_{OH}. - When an output pin is LOW, the I
_{IL}requirements for all receivers must add-up to be ≤ to the driver's I_{OL}.

Given that an output of any logic device can go either HIGH or LOW (High-impedance is not relevant), and hence exhibit either I

_{OH}or I_{OL}, respectively, the Fan-Out is the minimum of two ratios:**Fan-Out = min ( I**

_{OH}/I_{IH}, I_{OL}/I_{IL })For example, Input and output currents are the following.

**:***Recall that negative current values indicate current flowing out of the gate while positive current values indicate current flowing into the gate*- I
_{OH}= -400 µA (i.e., output can source a maximum of 400µA) - I
_{OL}= 16 µA (i.e., output can sink a maximum of 16µA) - I
_{IH}= 40 µA (i.e., input can sink a maximum of 40µA) - I
_{IL}= -1.6 µA (i.e., input can source a maximum of 1.6µA)

Therefore the fan-out is min ( 400/40, 16/1.6) = min (10, 10) = 10. In other words,

**of same type without getting out of its guaranteed range of operation. If more than 10 gates were connected, the output voltage levels will degrade and the gate will slow down.***each gate can drive 10 other gates*What if there were different IC families, where the output was coming from one family while the other inputs related to different logic families? In this case, add I

_{IH}for all inputs connected to an output. The sum must be less than the output's I_{OH}. Then add I_{IL}for all inputs connected to an output. The sum must be less than the output's I_{OL}.The following diagram illustrates the principle of

**Fan-Out**for both HIGH and LOW outputs:When the NOR gate output is HIGH, the output bin behaves as a current source since I

_{OH}flows out of the driver gate and into the set of driven gates. The current I_{OH}equals the sum of all input currents indicated by I_{IH}, flowing into the driven gates. In other words, I_{OH}= ∑I_{IH}.When the NOR gate output is LOW, the output bin behaves as a current sink since I

_{OL}flows into the gate and out of the driven gates. The current I_{OL}equals the sum of all input currents indicated by I_{IL}, flowing out of the driven gates. In other words, I_{OL}= ∑I_{IL}.For example, I

_{OL (MAX)}= 16mA and I_{IL (MAX)}= -1.6mA (negative current values indicate current flowing out of the gate). Therefore, fan-out is 16/1.6=10:

__Snapshot of .Lib File (Liberty File)__cell (<cellname>) {

cell_leakage_power : 3.748077e-03;

threshold_voltage_group : "si38p" ;

area : "8.775" ;

….

abc_cell () {

…

pin (Z) {

direction : "output";

}

**pin (CP) {**

**direction : "input";**

}

**pin (D) {**

**direction : "input";**

}

….

}

pin (Z) {

direction : "output";

related_bias_pin : "VDDB VSSB";

**max_capacitance : 0.334971**;

**max_fanout : 20**;

…

timing () {

**cell_degradation (constraint)**{

index_1 ("1.0, 1.5, 2.0") ;

values ("1.0, 1.5, 2.0") ;

}

....

}

}

In the next blog (part) we will discuss rest of the Design rule Constraints.

Thanks a lot for sharing this kind of knowledge sir, excellent . I had clear a lot of doubts from this information.

ReplyDelete" maximum input current for a particular cell at each logic level."

ReplyDeletethis should be either minimum input current or maximum output current.

No Dear .. even max input current is also required

Deletehow to design a cmos differential amplifier (steps please,about design considerations)?

ReplyDeletesometimes I found many designers limit the primary inputs fan out to be 1

ReplyDeleteset_max_fanout 1 [all_inputs]

but i can not understand why. Kindly can you explain it.

Thanks

Hi Sir,

ReplyDeleteThank's a-lot for sharing the information & the knowledge that you have which is best of its quality.I would like to thank you once again for taking time in solving queries by giving reply..which rarely people do....I only wish that your knowledge will helps other's in achieving their goals.... and you may live prosperously... by adding more & more posts...

Thanks & Regards

what is the diffence between max_fanout and max_cap

ReplyDeletefanout relates to current whereas cap relates to strength

DeleteWhat concept using in this method

ReplyDelete