# Homework Help: Logartihmic amplifier

1. Jul 10, 2009

### smuscat

I am trying to understand the derivation of a logarithmic amplifier as shown in the figure attached.

The diode has a non linear characteristics represented by V=ClnI,where C is a constant.
Then since the current through the feed back loop is the same as the current through the input resistance and the potential difference across the diode is -Vout,we have

Vout=-Cln(Vin/R)=Kln(Vin) where R is input resistance.

What I cant understand is the step in bold,what mathematical rule is used between the two equations?

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2. Jul 10, 2009

### Redbelly98

Staff Emeritus
Vout = -Vdiode = _____

Can you take it from there?

EDIT:

Just realized, there are three equations highlighted in bold. Which 2 equations do you mean?

3. Jul 10, 2009

### Bob S

Write down the current across a diode as a function of the voltage (hint-it is an exponential). It turns out that an "ideal" diode is better represented by a NPN transistor, with the collector and base tied together for the anode, and the emitter for the cathode. Unfortunately, the logarithmic amplifier made from diodes has a temperature coefficient due to (kT/q) in your equation (another hint).

4. Jul 10, 2009

### Phrak

$$V_{out} = -C \ ln(V_{in}/R)$$

$$V_{out} = -C \ ( ln \ V_{in} - ln \ R )$$

$$V_{out} = -C \ ln \ V_{in} + C \ ln \ R \neq K\ ln \ V_{in}$$

In addition, why are there dimensioned units in the log argument?

You need better source material--the equation is beyond wrong.

Last edited: Jul 11, 2009
5. Jul 11, 2009

### Redbelly98

Staff Emeritus
Good catch Phrak, I completely missed that

6. Jul 11, 2009

### smuscat

I tried to think about it given V=clnI make I subject of the formula gives I=e^v/c but I cant go any further unfortunately.

7. Jul 11, 2009

### Phrak

I seem to recall someone by the name of Redbelly guiding me out of a big whopper of a mistake not too long ago.

Last edited: Jul 11, 2009
8. Jul 11, 2009

### fleem

I believe the third expression has presumed R is a constant. Specifically, K is chosen for a certain R. It shouldn't be written that way (all three expression equal) because the second equals sign has that qualification (that R is a constant and its effect is included in K) but the first equals sign does not have that qualification.

Good catch. Its really a ratio with the denominator constant (at a given temperature). See http://en.wikipedia.org/wiki/Diode_modelling

9. Jul 12, 2009

### smuscat

I understood your first point that R (input resistance) is included in K but I cant understand is how R is included in K.In the previous post by Phark showed that
Vout=-ClnVin+ClnR is not equal to KlnVin.Can you state the steps how you concluded that?

Sorry for my luck of understanding.
Thanks

10. Jul 12, 2009

### Redbelly98

Staff Emeritus
That was what I thought at first too, but Phrak showed us that it is wrong. R could be included in an additive constant, but not in a multiplier because of the properties of logarithms.

Have you been told the diode equation, which is the basis for this discussion?

For a diode,

i = iS (eV / V0 - 1)​

where iS and V0 are constants for a particular diode.

But, iS is often in the nA or pA range for many diodes, so the "-1" can be ignored when the current is much higher than that. With that approximation we have

i = iS eV / V0

To get the correct equation for gain in the op-amp circuit, you need to do 2 more things:

• Solve the above equation for V
• Use Ohm's law for the resisitor R to get i in terms of Vin and R

As a final step, you could also combine all the constants in the final equation into just 1 or 2 constants such as C1 and C2. Remember that R is also considered as a constant in this particular circuit.

11. Jul 12, 2009

### smuscat

I tried follow the above steps by Redbelly,please tell if I did wrong or not

given I=Ise^Vout/Vo

I made Vout subject of the formula gives Vo(lnI-lnIs)=Vout

using Ohms Law did I in terms of V and R

Vo(lnVin/R-lnIs)=Vout

Vo(lnVin-lnRIs)=Vout

VolnVin-VolnRIs=V (I think that VolnRIs ~ 0 ,because Is is approx 0)

Therefore VolnVin=Vout or KlnVin=Vout

Thanks for all

12. Jul 12, 2009

### Redbelly98

Staff Emeritus
Looks good up to here.

Not quite. If Is is "≈0", then ln(R Is) would tend to -∞, not zero.

As I see it, the ln(R Is) term should be kept, and perhaps incorporated into a single constant if you wish. This does disagree with the equation given in Post #1.

13. Jul 13, 2009

### smuscat

Thank you for your replay the final answer is I think VolnVin-C=Vout

Might the equation be KlnVin=Vout or VolnVin=Vout for a particular R I think?
In other words Voln(RIs)~0

14. Jul 13, 2009

### Redbelly98

Staff Emeritus
Technically yes, for a particular R in suitable units ... but this would not be a practical value for R.

We would need R Is = 1, in some system of units, for the ln to be zero. But for realistic values of Is of 10-12 to 10-8 Amps, R would have to be between 108 to 1012 ohms! It's simply not practical to use such large resistor values in most circuits.

So there really needs to be a constant term added (or subtracted) in the expression.

15. Jul 13, 2009

### smuscat

Thanks to everybody, many thanks in particular to Redbelly98.
Gooday!