OT: milestone, the x-Chapters

OK that is dumb. Can you chain together TIA stages? Lowest gain first.

+---R2---+ | |\ | +---R1---+-|-\ | | |\ | >---+ I_in---+-|-\ .-|+/ | >---+ |/ Gnd-|+/ |/

I'm not sure that works.. and I have to do something when the 2nd TIA rails... short it's inputs or something.

George H.

Reply to
George Herold
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Here's a stupider question, since I'm no engineer. Wouldn't the second stage be a regular opamp? I mean the TIA output is voltage, so....

Reply to
Tom Del Rosso

Is this an electronics discussion or a book promotion?

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
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Reply to
John Larkin

Your circuit is probably way better than is justified by the photodiode. The response of a pd varies wildly with wavelength, temperature, angle of illumination, and light distribution over the surface. Big ones can be leaky too.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

A1: yes it is. A2: does it matter? A3: it is fun, who cares.

Reply to
Tom Gardner

We try to drive the SJ conductance to infinity. One can imagine a scheme where an extra factor is picked up, but it's hard to imagine getting 10^8 or 10^10 operating range, you pick a number.

This fellow's creation can do that. OK, time for another hint, if it isn't obvious by now, there's one input amp, and one feedback resistor chain.

The entire circuit has output ranges, yes, but multiple ranges are simultaneously active. You can have as many ranges, with whatever scaling factors, 10, 30, 100, 300, etc., as you like. We picked 300 for star-to-sunlight, save parts.

Your ranges could go from 100fA to 10 amps.

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 Thanks, 
    - Win
Reply to
Winfield Hill

Well the perhaps silly idea is to have to current go through R1 and R2. Say R1 = 10 k and R2 = 1 Meg, and there is 1 uA of current, then ignoring offset voltages the output of the second opamp should be 1.010 V, and the first 10 mV... I guess I'll have to spice to check it.

I seem to recall some chained TIA circuits, but I couldn't find them online, and forgot what they looked like.

George H.

Reply to
George Herold

When the current is high enough in R2 to rail the 2nd opamp, how can the first one go to its proper value? But, yes, the current has to go through both R1 and R2, at least for low currents (oops, another hint).

--
 Thanks, 
    - Win
Reply to
Winfield Hill

Yes, certainly, I'd say. But he'd have applied for the patent in 2011, and it would have been granted by now. He left the idea to all of us.

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 Thanks, 
    - Win
Reply to
Winfield Hill

Yes, that would give you dynamic range, just like range-switching the feedbackresistor. But the cool thing about this design is, there's no range switching. Multipule ranges are active at the same time, so there's never a momentary loss of data. BTW, it also works going from positive to negative and back.

--
 Thanks, 
    - Win
Reply to
Winfield Hill

Right, you've got to short the inputs of the 2nd opamp when it rails.. I guess you can do that with a fet and signal from the uC.

George H. But, yes, the

Reply to
George Herold

And since there's only one input stage, that amp somehow never rails.

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Reply to
Tom Del Rosso

In x-ray systems the get those kind of dynamic ranges using CCO chips, Current Controlled Oscillators.

Steve

Reply to
sroberts6328

Correct. Tom, if you scratch the word, somehow, then I'd think you know how it was done. You're very close now to the answer.

--
 Thanks, 
    - Win
Reply to
Winfield Hill

So the 2 opamps in the second stage produce 2 oututs that both go to an ADC. A uP chooses an input but there is no other switching. There are

2 JFETS but they aren't used to change the gain of any stage.

Since there is no gain switching, is there some kind of feedback that causes a change in "bias" on the input stage? But no, that would oscillate.

Reply to
Tom Del Rosso

On a sunny day (Wed, 26 Jun 2019 04:48:16 -0400) it happened "Tom Del Rosso" wrote in :

It would oscillate, but you could only use the value that makes sense, start: nothing railed: use the higest gain if 1 railed, use the next one up else if 2 railed activate JFET1, test again if still 2 railed also activate JFET2 and test again. if still 2 railed sound the super-nova alarm else use the one that is not railed .. .math.. output go to start

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Reply to
Jan Panteltje

The first stage will have to lowest gain.. Still one could probably make it rail with a laser. I want to see what the various step responses look like.

Win, If we all promise to buy the X-chapters, will you post 'scope shots? (or do I have to build it myself. :^)

George H.

Reply to
George Herold

Right something like that.. but you could use the signal from the lower gain stage to see if the upper one has 'unrailed'. GH

Reply to
George Herold

John, you are usually not so unkind.

Reply to
John S

+1
Reply to
John S

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