Logarithmic amplifier

Thread Starter

tassat

Joined Oct 11, 2014
7
I was given basically this image:
as a guide and told to build a logarithmic amplifier and verify that its output is proportional to the logarithm of its input. I have a 411 for the op amp, a 10k for the resistor, and a 1N914 for the diode. I'm driving the circuit with a 10khz sine wave, but my output just looks like a square wave. Is this what I'm supposed to get, and if so, how is that proportional to the logarithm of the input? If not, what should I do differently?
 

ericgibbs

Joined Jan 29, 2010
18,766
hi,
I would use a variable DC source as Vin, start at 0Vin and increment Vin in say 0.1V steps, note the Vin and Vout for each step, then paper plot Vin versus Vout.

Post your results.;)
E
 

Thread Starter

tassat

Joined Oct 11, 2014
7
Ok, I used a DC power supply and got the following (all in volts):

In Out
1.0 -.497
5.0 -.575
7.0 -.592
10 -.609
12 -.617
15 -.629

So that's, according to my calculator, out=-.497+-.049*ln(in).

So my results are proportional, thanks, that was honestly super helpful. If I can ask, what is the -.497?
 

ericgibbs

Joined Jan 29, 2010
18,766
hi,
I did say 0V and the in 0.1V steps, 1V is too big a step.!

To get more detail, you should start at 0V and increment in 0.01V steps.

Its an Inverting amplifier so the polarity of the Vout will be opposite of Vin.

Try 0V thru -1V , in -0.01V steps upto -1V, lets see what you get.
E
 

ericgibbs

Joined Jan 29, 2010
18,766
So you want the voltage I apply to be negative?
Yes, if you have a negative variable voltage source to use use for the test.

You may get a result you don't expect, can you tell us why that is,?:rolleyes:
 

Thread Starter

tassat

Joined Oct 11, 2014
7
I don't know if I have a negative variable voltage source. I can connect the positive from my power supply to ground and the negative to the input and the voltage will be negative right?
 

ericgibbs

Joined Jan 29, 2010
18,766
hi,
If you cannot do that test, think what the Vout may be for a negative Vin input.?
Clue: which way will the diode be biased.?
E
 

ericgibbs

Joined Jan 29, 2010
18,766
If your Vin power supply source is floating, ie: the neg is not connected to ground, then you can connect the positive to 0V and the negative to the Vin.

I guess you know the gain of an Inverting OPA is: Vout= Vin * Rfeedback/Rin.?
As I asked you earlier if the Vout goes positive with respect to the Input, will the diode be biased ON or OFF.?
 

Thread Starter

tassat

Joined Oct 11, 2014
7
The diode would be biased off I think.

How would it make sense for the output to be proportional to the logarithm of the input if the input is negative?
 

ericgibbs

Joined Jan 29, 2010
18,766
When the diode is reverse biased its resistance is very high, many megohms, so the gain Vout= Vin* Rmegohm/Rin is the 'open loop' gain of the OPA, which can be 10,000's!
So the output voltage would tend towards the OPA's positive supply rail, with only a few -mV on Vin.

The Vout is only the Log of Vin when when is positive, that is when Vout is going negative.

Do you follow OK.

To get the best plot you should increase +Vin in very small increments

E
 

Thread Starter

tassat

Joined Oct 11, 2014
7
Ok, I think I understand, when using a small increment I got (all in volts)

in out
0 -.370
.1 -.387
.2 -.425
.3 -.452
.4 -.462
.5
.6
.7
.8
.9
1
 

ericgibbs

Joined Jan 29, 2010
18,766
If you use the Table of values you posted to produce a graph plot, does it look logarithmic.?
[you could ignore the minus sign and create a positive going plot]

You will get a better resolution if you use 0.01Vin increments and not 0.1Vin
 
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