Xref: utzoo comp.ai.neural-nets:199 sci.electronics:3679
Path: utzoo!utgpu!water!watmath!clyde!att!osu-cis!tut.cis.ohio-state.edu!husc6!bu-cs!buengc!bph
From: bph@buengc.BU.EDU (Blair P. Houghton)
Newsgroups: comp.ai.neural-nets,sci.electronics
Subject: Re: Sigmoid transfe function
Message-ID: <917@buengc.BU.EDU>
Date: 23 Aug 88 21:10:40 GMT
References: <1982@aecom.YU.EDU>
Reply-To: bph@buengc.bu.edu (Blair P. Houghton)
Followup-To: comp.ai.neural-nets
Organization: Boston Univ. Col. of Eng.
Lines: 140

In article <1982@aecom.YU.EDU> krishna@aecom.YU.EDU (Krishna Ambati) writes:
>
>Concerning circuits that yield a tanh transfer function, I wonder
>whether the following will work?
>
>			     D1
>
>			____|/|___
>		       |    |\|   |
>		    ---|          |---
>		   |   |____|\|___|   |
>		   |        |/|       |
>		   |         D2       |
>		   |                  |
>		   |                  |
>		   |         C        |
>		   |--------| |-------|
>		   |                  |
>		   |         R        |
>		   |-------/\/\-------|
>		   |                  |
>		   |                  |
>		   |       |\         |
>    V(in) .-------|-------|-\        |
>			   |  \_______|_____. V(out)
>			   |  /
>		   |-------|+/
>		   |       |/
>		   |
>		  GND
>
>What is the transfer function of this circuit?

It's not tanh(Vin); that's for sure.  I do want to thank you
for unrusting my op-amp design talents.  This was some fun!

First, put a significant resistance (call it Ri) in series with Vin, or
you'll get only a step at Vin crossing 0 where Vout goes from Vd to -Vd.
Vd is the threshold voltage of the diodes (assume ideal diodes
with forward resistance zero, reverse infinite; this assumption
doesn't change the result too much.)

Call the potential at the inverting input Vinv.  It is the op-amp's
duty to keep this voltage equal to the noninverting-input voltage,
which you have tied to ground, or suffer nonlinearity.

Now we have three cases:

1.  |Vout - Vinv| < Vd, which is to say |Vout| < Vd, since this is the
	linear region, and the op-amp may do its job proudly.

	We find

		Vout            R
		----  = - ------------- ;
		 Vin      Ri + s C R Ri

	which, for low frequency (s) is approximately
	
		Vout       R
		----  = - --- ; 
		 Vin       Ri
	
	and, it is a lowpass filter, so any frequency above 1/(RC) is
	significantly attenuated.  For the purpose of "sigmoidality"
	we worry about all frequencies, especially the lower
	ones.  But, neurons drive themselves and their peers.
	I won't worry here about what the lowpass characteristic
	does to neuromimetic activity.

2.  Vout - Vinv >= Vd;

	Now the resistance of D1 drops to zero.  R and C are well and
	truly shorted, and any current from the op-amp output can be
	fed back to the inverting input.  Most importantly, that
	amount of current will be fed back which is equal to the
	amount taken by Vin; i.e.,

		      Vinv - Vin    Vout - Vd - Vin
		I =   ---------- =  --------------- ;
			  Ri              Ri

	and the only truth about Vout is 

		Vout = Vd + I Ri + Vin = Vout

	This means that the op-amp sets Vout at its convenience.
	Its convenience is to set Vout to keep Vinv = Vnoninv = 0;
	hence

		Vout = Vd .

	Here is where things can get ugly if you include the series
	resistance of the diodes.  Practically, it only adds a little
	bit to the Vout vs Vin line, causing it to slope upward a bit,
	rather than being flat.  One other effect of reality is a
	slight rounding at the point where regions 1 and 2 meet.

3.  Vout - Vin <= Vd;

	See region 2 for the currents; since Vin and Vinv have both
	switched sign, I will have also.  (Betcha didn't know that
	about me :-)  The output volts are just

		Vout = -Vd .

The full effect is


			       | Vout
			       |
	xxxxxxxxxxxxxxxxxxxx   - Vd
			    x  |
			     x |
			      x|
        -----------------------x--------------------
			       |x		  Vin
			       | x
			       |  x
			   -Vd -   xxxxxxxxxxxxxxxxx
			       |

So, what you have just invented (not really :-( ) is a double-sided
clipping circuit that operates as a low-pass filter for small-signal
input.

With the slight rounding at the corners, this can be considered
a "sigmoid", but you have to emphasize the "-oid" heavily.

I think for the purposes of the Hopfield model it will work; just
feed this one's output through another inverting amp with gain
set so that Vin = Vd gives Vout = 1.

It's easier to leave out the diodes and capacitor and run the thing with
supply voltages of +/-1v. the op-amp will clip on its own.

				--Blair
				  "Even an Electrical Engineer
				   could do such a thing; I'm
				   sure of it..."