Analog multiplier

Thread Starter

atferrari

Joined Jan 6, 2004
4,275
I run across the schematic shown in page 13 of the RC4136's datasheet (attached). After 3 days trying to have it "working" in LTSpice (circuit attached) I realized I need help.

I tend to think I know the basics of log / antilog amps and I understand this circuit should give (hopefully) a Vout = Vy * Vx / Vz. Instead, the prevailing result I get is Vout slightly above Vx, no matter what Vy or Vz are.

With different op amps, or no matter how many diodes I pile up at the top (to get up to around 3.5V) to polarize the first stages, I always get a result not even close to the expected (and most of the times with >1 MHz noise riding the output).

I am puzzled by what is wrong: my simulation or the circuit?

If anyone is tempted to tell about the so many pitfalls of a circuit like that implemented with discretes or how old / lousy the RC4136 is, PLEASE don't. I am asking about a simulation and / or eventual flaw of a design.

Helps is appreciated.
 

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Brownout

Joined Jan 10, 2012
2,390
Sorry, I can see your circuit, as I don't have LTSpice on my laptop. I can only warn that some component models don't work very well, especially diodes. I would maybe try different diodes.

Sorry for the shot in the dark :)
 

Thread Starter

atferrari

Joined Jan 6, 2004
4,275
Sorry, I can see your circuit, as I don't have LTSpice on my laptop.
Here you have it, Señor Brown!

Leaving aside the diodes could be not properly modeled, I have no idea what are they for at the output of each amp.

Sorry for the shot in the dark :)
As long as you do not shoot me...... :D :D :D
 

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Brownout

Joined Jan 10, 2012
2,390
Leaving aside the diodes could be not properly modeled, I have no idea what are they for at the output of each amp
I'm not exactly sure either, but here's a guess: Each log amp needs to be stabilized for temperature. The diodes might shunt current on a temp dependent bases, which denies current to the log transistors, giving them some temp stability.
 

t06afre

Joined May 11, 2009
5,934
Analog multipliers built by discrete components have many issues at least far more than the integrated ones. Analog Devices have a line of analog multipliers still i think. You better check out one of those
 

Thread Starter

atferrari

Joined Jan 6, 2004
4,275
Analog multipliers built by discrete components have many issues at least far more than the integrated ones. Analog Devices have a line of analog multipliers still i think. You better check out one of those
Thanks for replying.

Quoting the OP:

If anyone is tempted to tell about the so many pitfalls of a circuit like that implemented with discretes or how old / lousy the RC4136 is, PLEASE don't. I am asking about a simulation and / or eventual flaw of a design.
 

#12

Joined Nov 30, 2010
18,217
The LT1022 can not work with inputs within 2 or 3 volts of the rails and the outputs miss the rails by 2 to 3 volts. This might be your entire problem. The chip you chose is just not right for the job.
 

Brownout

Joined Jan 10, 2012
2,390
I don't know if this helps but here is a circuit that does simulate. The 1st and 3rd stages are log and anit-log stages, which are the baises of multipler/divider circuits. This circuit would be impractical to build.
 

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Thread Starter

atferrari

Joined Jan 6, 2004
4,275
The LT1022 can not work with inputs within 2 or 3 volts of the rails and the outputs miss the rails by 2 to 3 volts. This might be your entire problem. The chip you chose is just not right for the job.
Hola Nro. 12,

Could show one that is right for the job?
 

#12

Joined Nov 30, 2010
18,217
Now I see that you gave it enough voltage to work with. Sorry. My bad.
I thought you were trying to use zero and +5 volts for the chips.
 

t06afre

Joined May 11, 2009
5,934
@Agustín
It looks like you have forgotten the KISS rule. Just for simulation purpose "a" multiplier or divider may I suggest these simple model files:D
Rich (BB code):
*Multiply Voltages
.SUBCKT MULTV 1 2 3
BX 3 0 V=V(1)*V(2)
.ENDS MULTV
Rich (BB code):
*Divide Voltages
.SUBCKT DIVV 1 2 3
BX 3 0 V=V(1)/V(2)
.ENDS DIVV
 
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Thread Starter

atferrari

Joined Jan 6, 2004
4,275
Thanks to suggestions in another forum, grounding the base of Q2 and Q6, the attached circuit works with positive values at the inputs since the action inside occurs in the negative values realm.

As you could see, it simulates properly 6*20/10 = 12. Not bad, eh?

Pending:to understand why the designer polarized those bases with positive values.
 

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Thread Starter

atferrari

Joined Jan 6, 2004
4,275
At least I am learning something new today.;)
Allen
Same for me, good for both!!

It took me some time to realize what the op amp does: output the necessary voltage (a Vbe that corresponds to the input current). Matched transistors, a must.

In this particular circuit the op amps' outputs are all negative.

What simulator are you using? Free by any chance?
 
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