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a doubt while designing a cmos inverter in cadence 614

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sumansamui

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hello friends i have designed a cmos symmetric inverter using gdpk180 library .....in this technology the nominal value of vdd is 1.8volt......but if i changes the vdd=0.5 or 1v cmos is still performing the same inverting action.how this is happening?
 

FvM

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Depends on the transistor threshold voltage. Most likely, you get lower transistor currents and thus lower speed at lower supply voltage.
 

keith1200rs

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The working voltage of a process is usually the maximum. You can design for use at lower voltages provided you have enough voltage to turn the transistors on.

Keith
 

sumansamui

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but as far as i know mininmum vdd should be=Vtn +|Vtp|
 

keith1200rs

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Not really. You only need one transistor turned on at a time for an inverter, not both.

Keith
 

sumansamui

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but in that case we will get only one cycle over n over causing hysteresis ........that is expectable but i m getting normal inverting action having full voltage swing
 

keith1200rs

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I am not sure what you are expecting? Yes, the positive and negative going thresholds could be different but it will still be an inverter and have full Vss to Vdd output swing.

Keith
 

FvM

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You'll need to look for dynamic behaviour (delay, rise and fall time) or output current strength to see an effect of the low supply voltage.
 

sumansamui

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some times i am getting -ve delay if vdd=3V or above .whay -ve delay actually means?

---------- Post added at 10:05 ---------- Previous post was at 10:03 ----------

i am expecting a hysteresis curve......if vdd <vdd(min)
 

keith1200rs

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I don't know how you are measuring delay but a negative delay to me means the output is changing before the input does. That doesn't make sense. You need to show your circuit and transient simulation waveforms.

Keith
 

sumansamui

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thanks for reply .i have solved the problem .
 

muffassir

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Hi frnd,

u need to see book by Rabaey for it see chapter5 on inverters.

Even below the threshold the inverter will work even at values of o.02 V...the graphs are self explainatory...read the book.





[REMOVED - MODERATOR. Every post says "Did you find this post helpful? Click: Yes" - you don't need to repeat it]
 

vlsi_whiz

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Hi,

First of all P and N transistors in CMOS designs for digital ckts cannot have the same W/L ratios, ie, the transistor cannot or should not be symmeterical. This is because of the mobility difference in P and N transistors. N transistors have higher mobility than P and so to offset this difference, the P transistor is always designed at least 3 times larger than the N transistor. This will give you equal rise and fall times or equal voltage swings from HIGH to LOW.

Secondly, if you reduce the VDD of a digital circuit, it will still work as designed, but the output voltage swing will be lower, ie, you will not get full +3V (VDD) swing for logic 1, etc. You cannot keep on decreasing the VDD of a transistor below a certain level as this voltage is what sets up the electric field in the MOS that is responsible for the proper formation of the channel.
If Vdd is reduced too low, the channel will not get formed properly leading to poor functioning of the device.
 

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