Analog Multiplier

[Chandu06450]

Hi friends!!! This is one of the projects handling by my class as mini project. But unfortunately working of this circuit(analog multiplier) is not getting for us. Please help us in solving it.

 

[dknguyen]

I'm gonna assume you didn't design the circuit and that the circuit's designers are correct. So with all the information you gave us, the only thing we can say is "well, if you built it like the schematic then your circuit should work."

We need more information about YOUR circuit and how it's not working as the problems are not with the design of the circuit but the way that it was built.

 

[Chandu06450]

If We design the circuit, why I ask its working to you?? . It multiplies the voltage E1 and E2 and divides the voltage E3. so, result is E1.E2/E3. It uses the log properties of transistors(most probably gummel plot). we try to simulate it before making the hardware, but again a problem. We use pspice student version and it didn't support for this diagram(no. of nodes are more than its limit).

 

[dknguyen]

If you design the circuit and it is not working, then the schematic you give us might be wrong and we might be able to find something. But if the schematic is made by someone else (probably someone professional) then it is probably correct and giving us the schematic won't tell us anything about what might be wrong because the schematic is right, but the circuit was built wrong.

WHether or not the circuit was designed by you, or you're building it from someone else's schematic we need more information.

 

[Chandu06450]

Actually my question is not to find wrong in schematic. Question is please explain the working of analog multiplier provided schematic there.

 

[Roff]

Actually my question is not to find wrong in schematic. Question is please explain the working of analog multiplier provided schematic there.

I recognize that as a National Semiconductor schematic.

In a BJT, Ic≈Ie=Is*(e^(Vbe/Vt)-1)), the well-known diode equation.
For Ic>>Is, Ic≈Is*e^(Vbe/Vt)
Ic/Is=e^(Vbe/Vt)
Taking the log of both sides,
Vbe/Vt=ln(Ic/Is)

In the attached schematic,
V(A)=Vbias-Vt*ln(I1/Is)
V(B)=Vbias-Vt*ln(I1/Is)-Vt*ln(I2/Is)
V(C)=Vbias-Vt*ln(I3/Is)
Vbe4=V(C)-V(B)
substituting,
Vbe4=Vbias-Vt*ln(I3/Is)-(Vbias-Vt*ln(I1/Is)-Vt*ln(I2/Is))
simplifying,
Vbe4=Vt*(ln(I1/Is)+ln(I2/Is)-ln(I3/Is))
Vbe4=Vt*ln(I1*I2/(I3*Is))

From the diode equation,
I4≈Is*e^(Vt*ln(I1*I2/(I3*Is))/Vt)
I4/Is=e^ln(I1*I2/(I3*Is)
Taking the log of both sides,
ln(I4/Is)=ln(I1*I2/(I3*Is))
I4=I1*I2/I3

Since R1=R4=R9=R11=10k,

Vout=V1*V2/V3
Q.E.D.

I hope I didn't make any typos.

EDIT: In order for Is to drop out of the equations, all transistors must have equal Is. Therefore, they should all be on the same chip, such as a CA3046. I suspect the asterisks on the schematic refer to a comment in the original app note to this effect.

 

[Chandu06450]

Thank u dude......

 

[eblc1388]

Roff,

You're correct about the circuit is from NS application note. It is AN4-6 dated April 1968.

Other points to note are:

(1) the circuit gives 1% accuracy for 500mV to 50V input, so input lower than 500mV has bigger inaccuracy

(2) the "*" on the transistors are described in the AppNote as "should be of the same temperature"

(3) the transistors should be installed with short leads and no socket as lead resistance even as low as 0.25 ohm can affect accuracy

 

[Hero999]

The op-amp used is ancient, use a modern instrumentation quad op-amp.

Why not just use a conventional analogue multiplier IC?

Chandu06450]We use pspice student version and it didn't support for this diagram(no. of nodes are more than its limit).

Then use some proper software which doesn't have such restrictions, try LTSpice which can be found using Google.

 

[Roff]

The op-amp used is ancient, use a modern instrumentation quad op-amp.

The word "instrumentation" is misleading. An instrumentation amp won't work here. A modern quad op amp would.
I know what you meant, but noobs might not.

 

[Freq Band]

I recognize that as a National Semiconductor schematic.

In a BJT, Ic≈Ie=Is*(e^(Vbe/Vt)-1)), the well-known diode equation.
For Ic>>Is, Ic≈Is*e^(Vbe/Vt)
Ic/Is=e^(Vbe/Vt)
Taking the log of both sides,
Vbe/Vt=ln(Ic/Is)

In the attached schematic,
V(A)=Vbias-Vt*ln(I1/Is)
V(B)=Vbias-Vt*ln(I1/Is)-Vt*ln(I2/Is)
V(C)=Vbias-Vt*ln(I3/Is)
Vbe4=V(C)-V(B)
substituting,
Vbe4=Vbias-Vt*ln(I3/Is)-(Vbias-Vt*ln(I1/Is)-Vt*ln(I2/Is))
simplifying,
Vbe4=Vt*(ln(I1/Is)+ln(I2/Is)-ln(I3/Is))
Vbe4=Vt*ln(I1*I2/(I3*Is))

From the diode equation,
I4≈Is*e^(Vt*ln(I1*I2/(I3*Is))/Vt)
I4/Is=e^ln(I1*I2/(I3*Is)
Taking the log of both sides,
ln(I4/Is)=ln(I1*I2/(I3*Is))
I4=I1*I2/I3

Since R1=R4=R9=R11=10k,

Vout=V1*V2/V3
Q.E.D.

Could you restate that without using all those annoying numbers ??


=FB=

 

[Roff]

Could you restate that without using all those annoying numbers ??


=FB=

LOL!
OK, you can do multiplication and division by adding and subtracting logarithms, the taking the antilog of the result.