How to derive a log-antilog opamp square law transfer function?

  • Thread starter fonz
  • Start date
  • #1
146
5
Summary:
log-antilog opamp question
Firstly, this is not a homework question. I found a worksheet online with an example of a square law circuit built using log-antilog operational amplifiers. I tried to derive the transfer function but I can't seem to eliminate the reverse saturation current term ##I_S##. I would really appreciate some help with this problem.
I have included the schematic below:

01019x01.png


Here is my attempt:

Assuming all resistors are equal value, all transistors are perfectly matched and at equal temperature and all op-amps are ideal.

Output of first stage:

##V_{O1} = - V_T ln(\frac{V_{IN}}{R I_S})##

Output of second stage:

##V_{O2}= 2V_{O1}##

Output of third stage:

##V_{O3} = - RI_Se^{\frac{V_{O2}}{V_T}}##

Sub 1 into 2:

##V_{O2} = -2V_Tln(\frac{V_{IN}}{RI_S}) = - V_Tln(\frac{V_{IN}^2}{R^2I_S^2})##

Sub into 3:

##V_{O3} = - RI_Se^{-ln(\frac{V_{IN}^2}{R^2I_S^2})} = -RI_S\frac{R^2I_S^2}{V_{IN}^2} = - \frac{R^3I_S^3}{V_{IN}^2}##
 

Answers and Replies

  • #2
tech99
Gold Member
2,157
803
It looks as if Is is just a constant, a characteristic of the transistor, which we must have to define the characteristic.
 
  • #3
eq1
196
71
Is is the saturation current of the diode and it is a physical property of the device just like the resistance of the resistor. https://en.wikipedia.org/wiki/Saturation_current Usually one DC biases the diode (or the BE's PN junction in this case) at a couple of operating points and then curve fits to the schottky diode equation to get it.

BTW, Is it usual to make a log and anti-log amplifier by using the virtual short of the op amp to keep the NPN's BC off? (i.e. base is ground, collector is virtual ground) I guess the opamp is usually compensated to be very slow (10s of KHz) so it's probably 6-to-one half dozen to the other but it seems like just shorting BC with a wire is a safer way to make a diode from a BJT. I am wondering if anyone knows the pro/con of grounding the base in a log amplifier. Maybe the turn on properties are better?
 
  • #4
146
5
Hi thanks for the replies. I probably should have been clearer in my original post. I am asking whether I have derived the correct transfer function for this circuit not what the reverse saturation current is in relation to the operation of a BJT.

Thanks
 
  • #5
eq1
196
71
Looks right to me and it passes the sniff test because -Exp[-2Ln(x)]=-1/(x^2).

Why do you think you need to eliminate Is?

In this circuit the BJT is masquerading as a diode because V(BC)=0. Is is just an intrinsic property of that diode just like R is an intrinsic property of the resistor.
 
  • #7
eq1
196
71
You mean question 7 right? As an analog engineer, I read "This circuit takes the square root of the input signal (y = √x)." as the output of the circuit is proportional to Sqrt[x] but we could also just play a tricky game. Choosing R=1/Is -> y=Sqrt[x]
 
  • #8
146
5
Hi thanks again for the reply. It was Question 8 I attempted.
 
  • #9
eq1
196
71
Actually, now that I think about it the transfer function given for question 8 is wrong. The middle stage should be inverting.

If the middle stage had G=-2 then y=-x^2=-Exp[-2(-Ln[x])] when R=1/Is. (One can also measure x in units of Vt and ignore it.)

With the non-inverting middle stage the transfer function is the one you found.

1/x^2 and x^2 very different functions and there is no proportionality constant that will make them the same.

P.S. Question 7 should also have G=-1/2, not 1/2, to give y=Sqrt[x]
 
  • #10
625
138
If for simplicity we assumed that the first stage output is -In(x) and the second stage gain is 2 and the last stage gain is e(x) we have y = e^( -In(x)*2 ) = 1/(x^2). Of course, the first and the last stage are not ideal they add offset (Is*R) but this is normal.
 
  • #11
146
5
Thanks forthe replies. I thought that ##I_S## was an incredibly small number, something like ##10^{-9}## which would mean that unless the resistance was huge the output would be really small.
 
  • #12
eq1
196
71
unless the resistance was huge the output would be really small.

That's just because the model we're using for the diode is overly simplistic. Here is a simple example with a better model. It compares the output of gain curve of the log amp to the ideal function when the resistor is 1Meg using a standard diode.

% cat log.sp
*
E1 y 0 0 n 10K
R1 n x 1Meg
X1 n y 1N4148
V1 x 0 PWL(0 0 1 1)
******************************************
*NXP Semiconductors
.SUBCKT 1N4148 1 2
*
* The resistor R1 does not reflect
* a physical device. Instead it
* improves modeling in the reverse
* mode of operation.
*
R1 1 2 5.827E+9
D1 1 2 1N4148
*
.MODEL 1N4148 D
+ IS = 4.352E-9
+ N = 1.906
+ BV = 110
+ IBV = 0.0001
+ RS = 0.6458
+ CJO = 7.048E-13
+ VJ = 0.869
+ M = 0.03
+ FC = 0.5
+ TT = 3.48E-9
.ENDS
.tran 100m 1
.print tran v(y) exp(-v(x))
.end
% ngspice -b log.sp # note: for this simulation v(x)=t so the tran is also drawing the gain curve.
--------------------------------------------------------------------------------
Index time v(y) exp(-v(x))
--------------------------------------------------------------------------------
0 0.000000e+00 -9.64830e-13 1.000000e+00
1 1.000000e-04 -1.03575e-03 9.999000e-01
2 2.000000e-04 -2.12523e-03 9.998000e-01
3 4.000000e-04 -4.24671e-03 9.996001e-01
...
13 1.056000e-01 -1.59227e-01 8.997845e-01
14 1.256000e-01 -1.67437e-01 8.819676e-01
15 1.456000e-01 -1.74477e-01 8.645034e-01
...
39 6.256000e-01 -2.45224e-01 5.349404e-01
40 6.456000e-01 -2.46765e-01 5.243478e-01
41 6.656000e-01 -2.48259e-01 5.139651e-01
...
57 9.856000e-01 -2.67507e-01 3.732152e-01
58 1.000000e+00 -2.68219e-01 3.678794e-01
 

Related Threads on How to derive a log-antilog opamp square law transfer function?

Replies
1
Views
849
Replies
3
Views
2K
Replies
11
Views
6K
Replies
1
Views
1K
Replies
7
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
1
Views
847
Replies
2
Views
909
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
1K
Top