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

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  • Start date
141
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Summary:

log-antilog opamp question

Main Question or Discussion Point

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

tech99
Gold Member
1,720
575
It looks as if Is is just a constant, a characteristic of the transistor, which we must have to define the characteristic.
 
eq1
Gold Member
120
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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?
 
141
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
 
eq1
Gold Member
120
49
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.
 
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eq1
Gold Member
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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]
 
141
5
Hi thanks again for the reply. It was Question 8 I attempted.
 
eq1
Gold Member
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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]
 
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.
 
141
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.
 
eq1
Gold Member
120
49
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
 

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