# help me antilog problem

Thread Starter

#### Xufyan

Joined Aug 3, 2010
114
look at the attached multiplier circuit,

the circuit works like this,

First i took log of two inputs than take the anti log of the sum of both inputs in order to get two input multiplier but the circuit is not working correctly , please spot the problem

#### wayneh

Joined Sep 9, 2010
17,152
Both of the input op-amps are configured to have a positive voltage always applied to their inverting inputs, so both will have an output near the negative rail.

Are you accounting for voltage drop across the diodes?

And you are aware that taking the log or antilog is not the same as dividing or multiplying by ten, right?

Thread Starter

#### Xufyan

Joined Aug 3, 2010
114
Both of the input op-amps are configured to have a positive voltage always applied to their inverting inputs, so both will have an output near the negative rail.

Are you accounting for voltage drop across the diodes?

And you are aware that taking the log or antilog is not the same as dividing or multiplying by ten, right?
What do you mean i didn't get you ?
when we takes log of two input and sum the two and then antilog the sum we'll get the two input multiplied, RIGHT ?
this is what mentioned in the book with example,

Antilog (First Input + 2nd Input ) = First Input x 2nd Input

???

#### Georacer

Joined Nov 25, 2009
5,182
I 'll second Xufan's statement.
$$e^{loga+logb)=ab$$

I can't tell about the circuit though...

#### wayneh

Joined Sep 9, 2010
17,152
I'm not questioning the math idea, but I don't think the circuit is accomplishing the desired math. The circuit appears to divide by ten, add, and then multiply by ten. That's not the same as taking the log and antilog. I'm no expert in op-amp circuits and could easily be wrong. That's why I asked if the OP was aware of the difference.

#### steveb

Joined Jul 3, 2008
2,436
look at the attached multiplier circuit,

the circuit works like this,

First i took log of two inputs than take the anti log of the sum of both inputs in order to get two input multiplier but the circuit is not working correctly , please spot the problem
Did you look carefully at the specified reverse saturation current on your 1N914 diode? What value is your simulation software assuming in the model of this diode?

It looks to me that you assumed a value of 1 pA for reverse saturation current when you chose the 10000 K and 100 ohm resistors, but if the diode model assumed something much larger then it will affect the scaling of the output.

Thread Starter

#### Xufyan

Joined Aug 3, 2010
114
Did you look carefully at the specified reverse saturation current on your 1N914 diode? What value is your simulation software assuming in the model of this diode?

It looks to me that you assumed a value of 1 pA for reverse saturation current when you chose the 10000 K and 100 ohm resistors, but if the diode model assumed something much larger then it will affect the scaling of the output.
the diode saturation current is not mentioned in Multisim can you please tell me what is the right diode to choose and what are the right values for the resistors? ??

#### steveb

Joined Jul 3, 2008
2,436
the diode saturation current is not mentioned in Multisim can you please tell me what is the right diode to choose and what are the right values for the resistors? ??
I'm surprised that you can't look at the diode model in Multisim, but anyway it's not critical to get the circuit working.

Simply increase R6 until you get the correct answer on the output.

I did a quick derivation with the assumption that the input voltages are the same and each input channel is identical. Also, I assume all diodes are identical physically and at the same temperature. This then acts like a square function, rather than a multiply function. I got the following formula.

$$v_o={{{v_i^2\; R_6}\over{R^2\; I_0}}$$

where R=R1=R2 and I0 is the saturation current of the diode.

I'll give the same warning on this circuit as I did when you were working with the log circuits previously. That is, this circuit is completely impractical, and you are seeing the reason now. The circuit is sensitive to changes in diode saturation current and saturation current is highly sensitive to temperature. You can get this working in a simulation no problem, but it will never work in the real world, unless you control the temperature of the diodes, and even then you will be hunting around for the correct resistor value, or tuning the circuit by changing the diode temperatures. Further, you would have to implement all diodes on a single chip. You can actually buy matched transistors on a single IC and wire them up as diodes, but you still would need to control the chip temperature.

#### MrChips

Joined Oct 2, 2009
23,545
If you are attempting to build a multiply circuit, this is not the way to do it. I would be looking at a variable gain amplifier.

While you can build log and anti-log circuits as shown, do not expect to get the output voltages to match the expected numerical results. There are too many constants and parameters that have to be taken into account in the diode formula. These have been ignored in the circuit.

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Thread Starter

#### Xufyan

Joined Aug 3, 2010
114
Yes , i really don't know how to check Is in multisim as i am newbie would you like to guide me to check it,

Edit: Got it , its 64pA And,

I derived the equation you mentioned in our last post thankyou,

but the last question is,
the final equation doesn't include kt/q then how it effects the output

I'm surprised that you can't look at the diode model in Multisim, but anyway it's not critical to get the circuit working.

Simply increase R6 until you get the correct answer on the output.

I did a quick derivation with the assumption that the input voltages are the same and each input channel is identical. Also, I assume all diodes are identical physically and at the same temperature. This then acts like a square function, rather than a multiply function. I got the following formula.

$$v_o={{{v_i^2\; R_6}\over{R^2\; I_0}}$$

where R=R1=R2 and I0 is the saturation current of the diode.

I'll give the same warning on this circuit as I did when you were working with the log circuits previously. That is, this circuit is completely impractical, and you are seeing the reason now. The circuit is sensitive to changes in diode saturation current and saturation current is highly sensitive to temperature. You can get this working in a simulation no problem, but it will never work in the real world, unless you control the temperature of the diodes, and even then you will be hunting around for the correct resistor value, or tuning the circuit by changing the diode temperatures. Further, you would have to implement all diodes on a single chip. You can actually buy matched transistors on a single IC and wire them up as diodes, but you still would need to control the chip temperature.

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Thread Starter

#### Xufyan

Joined Aug 3, 2010
114
Yes i know that there are too many constants in Diode equation but my teacher asks me to cancel their effects, i read somewhere that it is possible if i connect thermister (Temperature variable resistor , I think) on positive input and set its value equal to the inverse of kt/q value , i don't know this is correct way to cancel the effect.

If you are attempting to build a multiply circuit, this is not the way to do it. I would be looking at a variable gain amplifier.

While you can build log and anti-log circuits as shown, do not expect to get the output voltages to match the expected numeral results. There are too many constants and parameters that have to be taken into account in the diode formula. These have been ignored in the circuit.

#### steveb

Joined Jul 3, 2008
2,436
Yes , i really don't know how to check Is in multisim as i am newbie would you like to guide me to check it,

Edit: Got it , its 64pA And,

I derived the equation you mentioned in our last post thankyou,

but the last question is,
the final equation doesn't include kt/q then how it effects the output
I never said that kT/q was what was affecting the answer. I suggested that you check Io which does effect the answer.

Look at the formula you derived. In order for your output to be the square of the input, then $${{R6}\over{R^2 Io}}$$ must equal one.

Using the equation, it would seem that the Io must equal 1 pA for your circuit to work. There is a descrpency however. You said that the value of Io is 64 pA, but this does not seem to agree with the formula. Remember that Io is also temperature dependent, so maybe the value in simulation is different than you think.

Anyway, dont ignore MrChips' comment. These cancellations of kT/q and Io involve a very big set of assumptions. We are assuming that the inputs are equal and that all diodes are exactly the same and are at the exact same temperature. This isn't going to happen in the real world. I hope your teacher is stressing these practical concerns. It's fine to work on these things to learn, but the lesson should include the practical information as well. If the teacher is going to make you simulate this circuit, he should then make you build it so you see the problems. Try building this with descrete separate diodes and no temperature control and see if it works. It never will work. Then try it using diodes all built into one IC chip and mount the chip on a thermoelectric cooler with temperature control. You might be able to get something functional after tweaking the value of R6, but it will still be very very sensitive, and probably not all that precise.

• Xufyan
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