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- Path: sparky!uunet!olivea!decwrl!csus.edu!netcom.com!netcom!mjohnson
- From: mjohnson@netcom.Netcom.COM (Mark Johnson)
- Newsgroups: comp.lsi
- Subject: A problem for CMOS designers
- Message-ID: <MJOHNSON.92Aug18191600@netcom.Netcom.COM>
- Date: 19 Aug 92 03:16:00 GMT
- Organization: Netcom Online Communications Service
- Lines: 74
- In-Reply-To: mjohnson@netcom.Netcom.COM's message of Mon, 17 Aug 92 20:23:26 GMT
-
- A fun design problem just crossed my desk & I thought the
- readers here might enjoy it. It's about the right
- complexity to be one of three or four problems on a one
- week homework assignment in an IC design course for grad
- students. At least, in my opinion :-).
-
- The challenge (and thus the fun) in the problem is figuring
- out how to deal with signals that are less than one threshold
- voltage away from the power rail. My solution, for example,
- contains a weirdness that would have to be called "seldom
- employed" :-).
-
- ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
- PROBLEM: You have a plain ordinary CMOS 2 micron technology
- to work with; it has an enhancement-mode NMOS
- transistor (VT = +0.75 volts) and an enhancement-mode
- PMOS transistor (VT = -0.75 volts). By design rule
- law, all P-wells must be tied to 0.00 volts and
- all N-wells must be tied to +5.00 volts.
-
- Also by design rule law, you are not allowed to
- construct explicit resistors or capacitors or BJTs;
- all you have to work with are N-FETs and P-FETs.
-
- ---> Design an "inverting circuit" which switches
- ---> at Vin=0.4 volts. That is, when you plot
- ---> Vout versus Vin (the so-called DC transfer function),
- ---> a sharp transition from high to low occurs at
- ---> Vin=0.4 volts.
-
-
- SPECIFICATIONS
-
- Speed is unimportant; if your circuit takes eight
- hundred nanoseconds to sense the input voltage
- and then switch its output, that's perfectly fine.
-
- It's OK for your circuit to be multi-stage;
- so if you happen to get a noninverting topology,
- just go ahead and slap a CMOS inverter on the
- end. Hell, slap on three inverters. :-)
-
- Your circuit must not source or sink any current
- to/from the input.
-
- The value "0.4 volts" need not be too precisely
- achieved. To be concrete, here are the
- Vout versus Vin specs (for nitpickers only).
- To make life easy, these specs need to be
- met only when the fab process parameters are
- at their typical (nominal) values, and the
- VDD voltage is exactly 5.000 volts, and the
- die is at room temperature (298 Kelvin).
-
- (i) When Vin<0.2 , you must have Vout>4.9
- (ii) When Vin>0.6 , you must have Vout<0.1
-
- If your circuit has hysteresis (which I don't
- recommend), both branches of the hysteresis
- loop must fit in the region defined in (i)
- and (ii) above.
- ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
-
- Practically every set of real life CMOS process parameters
- is just fine for doing the problem; what we seek is an "Eureka"
- insight in circuit _topology_, moreso than a precise set of
- device sizes ready for mask layout. So the Correct Answer
- to the problem is a topology and a prose explanation of why
- that topology will give an "inverter" that switches at
- 0.4 volts (rather than at 0.1 volts or 0.7 volts or at VDD/2).
-
-
- Enjoy!
- -- M. Johnson
-