Op amp production of y=ln(x)

Warren · 2011-02-11T19:43:27+00:00

I am looking to implement a true ln(x) function inan op amp circuit, operating at audio frequencies.The goal is to feed a signal x into a quadraticequation (positive values only), and spit out avalue y using analog computing techniques. I amcurrently ignoring temperature stability forsimplicity.To perform the exponentiation, I planned onconverting the instantaneous signal x with alogarithm ln(x). Once I have ln(x), it is possibleto square and do x**4 using multiplication (priorto the antilog). A summing stage at the end willadd/subtract the terms involved.For example: y = a * x**4 - b * x**2should be easily done, if I can get a proper ln(x)and it's antilog.The following page under the heading "Logarithmicoutput" shows a typical circuit: +------|>|---+ | D1 | | | | | | |Vin ----R----+--| - | | >-----+------ Vout +--| +/ | | / | |/ | ----- --- -The wiki sez: Vout = -VT * ln ( Vin / ( Is * R ) )where: VT ~= 25 mV at room temp. Is = saturation current for the diode D1The problem that I have is that I want to eliminatethe denominator underneath Vin. Ideally, I want toget: Vout = ln ( Vin ).VT is simple enough to work around with acompensating gain stage.For the pesky Is * R denominator, I can make Rlarge, perhaps 1M. At some point though, if R istoo large, noise becomes a factor.For Is, I can use a larger diode, such as a1N4004 rectifier diode, with larger Is value(presumably somewhere in the 1E-09 range).But even with large R and a larger Is, I still have a denominator that is < 1, leading to a larger x in ln(x). I would like to get x = Vin.Is there an alternate circuit design that can achieve this? Or are there ready-made chips that would do this more easily? I'll obviously need the antilog of the same.Warren

Feb 11, 2011 43 Replies

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Warren 15 years ago

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Phil Hobbs 15 years ago

No, it takes care of the kT/e*ln(I_s) term. I_s is a device parameter, but it has the dimensions of a really really small current. It's the scale factor in the Ebers-Moll equation, so it's vitally important.

Feedback amplifiers have asymmetrical gain--the noninverting gain A_plus = 1 - A_minus. Thus if you make R_in = R_f, you get an inverting gain of -1 and a noninverting tain of +2. If you don't understand this, take the time to think carefully about it, and be ready to take a few partial derivatives in the process. You won't get far in analogue electronics without understanding this clearly.

It's more useful to look at the two halves of the pot as conductances--current divides itself according to the ratio of the conductances. Conductances in parallel driven by a current are the dual of resistances in series driven by a voltage.

Right.

I might be a little worried about the offset voltage drift. A constant offset voltage makes a fixed gain error that you can take out with the pots, but drift makes the gains change with temperature.

Should work fine.

I showed that in the schematic--the series RC to the signal plus a series R to V_ref effectively forces the input voltage to centre around V_ref. If you want 3V p-p on the input, use a 2.5V reference and increase the resistances to keep the collector currents around 100 uA.

Microamps. The 2.5V reference produces a 125 uA DC current, so going +-

1.5V from there makes the current vary between 50 uA and 200 uA. That's a ratio that the transistors can cope with pretty readily.

Correction tape is a wonderful thing.

Cheers

Phil Hobbs

Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 email: hobbs (atsign) electrooptical (period) net http://electrooptical.net

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Phil Hobbs 15 years ago

Cool. Good luck with it!

Cheers

Phil Hobbs

Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 email: hobbs (atsign) electrooptical (period) net http://electrooptical.net

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Warren 15 years ago

Phil Hobbs expounded in news:RIqdnZLkdrwOOsHQnZ2dnUVZ snipped-for-privacy@supernews.com:

Sorry, my bad.

I usually draw the (-) on top and I was glibly thinking of them in terms of inverting amps. I also failed to see "quadratic" scribbled on the other branch (yes its time for new lenses again). So I thought you were summing for the cubic. But now it is very clear and my world is OK again.

One idea I'm toying with is to set Vref to 1v, so that the ln(x) units are easier for testing. For example, Vin=1v, should produce a ln(x) value of zero. But I'll have to check the currents. More LTspice testing.

Then I want to test certain DC Vin values, to see if the Vout agrees with the f(x) involved.

Warren

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