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'[EE]: Differential amplifier circuit?'
2001\10\25@205712 by Jay Hanson

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Can anyone point me at a simple differential amplifier circuit using a
common operational amplifier chip and a single +12 volt supply?  I want to
amplify the difference between two photodiodes.  Frequency in the kilohertz
range.

Tnx,
Jay

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2001\10\26@074230 by Gerhard Fiedler

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At 14:55 10/25/2001 -1000, Jay Hanson wrote:
>Can anyone point me at a simple differential amplifier circuit using a
>common operational amplifier chip and a single +12 volt supply?  I want to
>amplify the difference between two photodiodes.  Frequency in the kilohertz
>range.

you should use a buffer amp for each of the photodiodes. there are a few
possibilities, they all come down to that you usually want 0 V across the
diode (i.e. measure the shortcut current). a simple circuit is the anode to
ground, the cathode to the inv. input, the non-inv. input to ground and a
feedback resistor from output to the inv. input. (depending on the range,
you may need an opamp that drives the output down to the negative supply.
but this feature is quite common.)

then you have two low impedance outputs and can use a standard diff amp.

ge

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2001\10\26@083246 by Olin Lathrop

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> Can anyone point me at a simple differential amplifier circuit using a
> common operational amplifier chip and a single +12 volt supply?  I want to
> amplify the difference between two photodiodes.  Frequency in the
kilohertz
> range.

Most opamps will do.  You don't need high imput impedance, low supply, or
high speed.  Photodiodes only produce a few 100 mV, so make sure you get an
opamp whose inputs can go close to the negative rail.  The biggest source of
error will probably be the matching between the resistors to get the common
mode rejection as low as possible.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, spam_OUTolinTakeThisOuTspamembedinc.com, http://www.embedinc.com

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2001\10\26@090616 by Olin Lathrop

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> you should use a buffer amp for each of the photodiodes.

I don't think this is needed.  Photodiodes aren't *that* high impedance.
You can easily find resistor values for your diff amp without loading the
photodiodes.  For example, 100K resistors should be plenty big enough to not
load the photodiodes, but well below the input impedance of the opamp.


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Olin Lathrop, embedded systems consultant in Littleton Massachusetts
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2001\10\26@094352 by Gerhard Fiedler

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At 08:52 10/26/2001 -0400, Olin Lathrop wrote:
> > you should use a buffer amp for each of the photodiodes.
>
>I don't think this is needed.  Photodiodes aren't *that* high impedance.

I guess this depends on what you want to do, mainly on the bandwidth and
precision required. For the bandwidth, it's not the (resistive) impedance,
it's the capacitance that is the problem, which forms a low pass with the
load resistance.

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2001\10\26@113812 by Bob Blick

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> At 14:55 10/25/2001 -1000, Jay Hanson wrote:
> >Can anyone point me at a simple differential amplifier circuit using a
> >common operational amplifier chip and a single +12 volt supply?  I want to
> >amplify the difference between two photodiodes.  Frequency in the kilohertz
> >range.

When amplifying photodiodes, do not treat them as voltage output devices!

Usually you build a current amp. You could build two as buffers before
your differencing amp, but you could also do it with one amp which might
be good enough for your application.

The LM3900 is a Norton opamp, you can easily put your two photodiodes from
positive then into each input of one of the amps. Then bias the amp with a
resistor(say 10 meg) from positive to the noninverting input, and a 5 meg
resistor from the output to the inverting input.

Cheerful regards,

Bob Blick

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2001\10\26@122913 by Harold M Hallikainen

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       I'd use an instrumentation amp chip, but have a look at
http://www.hallikainen.org/cuesta/et113/InstrumentationAmpAnalysis.pdf .
The basic diff amp is from points F and J to the right. Adding the stuff
to the left makes it an instrumentation amp. Looking only at the
circuitry to the right of F and J, the 1K resistors can be called R1 and
the 2K called Rf.  The gain is then Rf/R1.
       Note that the differential amp is a single amp being used as both an
inverting and non-inverting amp. The gain of the inverting side (from J
to the output) is -Rf/R1. The gain from the non-inverting input of the op
amp to the output is 1+(Rf/R1). The voltage divider on the non-inverting
input drops this gain down to Rf/R1 so it matches the inverting side.
       Looking to the left side of the circuit, if we call the 2K resistors Rf
and the 1K R1, the gain is 1+(Rf/(R1/2)) or 1+(2Rf/R1). This gain gets
multiplied by the Rf/R1 of the right side to get the overall gain.
       If you use JUST the right side of the circuit (the diff amp instead of
the instrumentation amp), the input resistance is low and unbalanced. The
instrumentation amp has a high input resistance, the input is balanced,
and the CMRR is higher.
       Due to the precision of resistor matching required to get a reasonable
CMRR (CMRR=Ad/Acm), again, I'd go with an instrumentation amp on a chip.

Harold

On Thu, 25 Oct 2001 14:55:52 -1000 Jay Hanson <jayspamKILLspamILHAWAII.NET> writes:
{Quote hidden}

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2001\10\26@124355 by Jay Hanson

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Thank you all for your great help!!!  I may be back for more if I get stuck
trying to make it work.

Thanks again,
Jay

{Original Message removed}

2001\10\27@043421 by Chris Carr

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----- Original Message -----
From: "Bob Blick"
>
> The LM3900 is a Norton opamp, you can easily put your two photodiodes from
> positive then into each input of one of the amps. Then bias the amp with a
> resistor(say 10 meg) from positive to the noninverting input, and a 5 meg
> resistor from the output to the inverting input.
>
I believe that the LM3900 has been obsolete for some time now
(but I bought so many when they were current that I still have some in
my junk box) LM259 is still available and if I remember correctly
Motorola (On Semiconductor) used to (still do ?) produce a Norton Amplifier
but it was never a popular as the LM3900.

Are they not also known as Transconductance Amplifiers? Just a random
thought, too early in the morning, insufficient caffeine in the system.
Probably wrong.

Regards

Chris Carr

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2001\10\27@121437 by Ned Konz

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On Friday 26 October 2001 11:42 pm, Chris Carr wrote:
> ----- Original Message -----
> From: "Bob Blick"
>
> > The LM3900 is a Norton opamp, you can easily put your two photodiodes
> > from positive then into each input of one of the amps. Then bias the amp
> > with a resistor(say 10 meg) from positive to the noninverting input, and
> > a 5 meg resistor from the output to the inverting input.
>
> I believe that the LM3900 has been obsolete for some time now
> (but I bought so many when they were current that I still have some in
> my junk box) LM259 is still available and if I remember correctly
> Motorola (On Semiconductor) used to (still do ?) produce a Norton Amplifier
> but it was never a popular as the LM3900.

This is what, a 30 year old chip?

The Motorola MC3401 was close or equivalent. But I think it's discontinued
now too.

The LM259/LM359 is a dual; I don't know if there's any quads any more.

> Are they not also known as Transconductance Amplifiers? Just a random
> thought, too early in the morning, insufficient caffeine in the system.
> Probably wrong.

No, a transconductance amplifier has a current-mode output, not input.

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2001\10\27@151456 by Chris Carr

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Ned Konz wrote :
> On Friday 26 October 2001 11:42 pm, Chris Carr wrote:
> > ----- Original Message -----
> > From: "Bob Blick"
> >
> > > The LM3900 is a Norton opamp, you can easily put your two photodiodes
> > > from positive then into each input of one of the amps. Then bias the
amp
> > > with a resistor(say 10 meg) from positive to the noninverting input,
and
> > > a 5 meg resistor from the output to the inverting input.
> >
> > I believe that the LM3900 has been obsolete for some time now
> > (but I bought so many when they were current that I still have some in
> > my junk box) LM259 is still available and if I remember correctly
> > Motorola (On Semiconductor) used to (still do ?) produce a Norton
Amplifier
> > but it was never a popular as the LM3900.
>
> This is what, a 30 year old chip?

Thanks, I like been reminded that I'm considerably closer to the Grim Reaper
than
the time I popped into this world  8-)

It does make you realise the problems that they must have keeping ancient
planes like the Concorde in the air, that also is 30 years old. It must rely
on 555 timers for everything electronic.

>
> The Motorola MC3401 was close or equivalent. But I think it's discontinued
> now too.

That number looks sort of familiar, you are probably right about it's lack
of availablity.
Can we construct a replacement with a PIC ?  8-)
>
> The LM259/LM359 is a dual; I don't know if there's any quads any more.

Probably correct but in the old days when we used Maths, 2 Duals made a
Quad.
Now we have had, probably in the interests of Globalisation, Math imposed
things are probably different......Sorry I couldn't resist the temptation.
>
> > Are they not also known as Transconductance Amplifiers? Just a random
> > thought, too early in the morning, insufficient caffeine in the system.
> > Probably wrong.
>
> No, a transconductance amplifier has a current-mode output, not input.
>
Yes, I've just had a look at the datasheets for a LM13700 and you are
correct
I must keep my caffeine levels up.

This question of obsolescence may be worth exploring as an OT

Regards

Chris Carr

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2001\10\28@002552 by Bob Blick

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>> > but it was never a popular as the LM3900.
>>
>> This is what, a 30 year old chip?

I'm sure they are still available for about 0.20 each, but it's easy as pie
making a current differencing amp with three transistors(if anyone is old
enough to have heard of them, if not, they have three terminals and make a
poor substitute for vacuum tubes :-)

Cheers,

Bob

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2001\10\29@034031 by Vasile Surducan

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Olin, If you need to have a linear output response ( voltage/light ) than
YOU MUST use a current to voltage operational amplifier converter and
measure the photodiodes output current rather than output voltage.
If not, than you'll have a non linear response and a lot of trouble by
linerising.
See the Burr Brown optical chips. ( OPT 301 etc...)
Of course the expensive BB chips could be replaced with cheapers structure
at the same electrical performance.
Regards, Vasile

On Fri, 26 Oct 2001, Olin Lathrop wrote:

{Quote hidden}

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2001\10\29@035319 by Vasile Surducan

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On Sat, 27 Oct 2001, Chris Carr wrote:

> ----- Original Message -----
> From: "Bob Blick"
> >
> > The LM3900 is a Norton opamp, you can easily put your two photodiodes from
> > positive then into each input of one of the amps. Then bias the amp with a
> > resistor(say 10 meg) from positive to the noninverting input, and a 5 meg
> > resistor from the output to the inverting input.
> >
> I believe that the LM3900 has been obsolete for some time now
> (but I bought so many when they were current that I still have some in
> my junk box) LM259 is still available and if I remember correctly
> Motorola (On Semiconductor) used to (still do ?) produce a Norton Amplifier
> but it was never a popular as the LM3900.
>
> Are they not also known as Transconductance Amplifiers? Just a random
> thought, too early in the morning, insufficient caffeine in the system.
> Probably wrong.
>
 Yes, they are transconductace ampolifier. The principle mentioned by BOB
is exactly. If don't like LM3900 ( which is an ugly chip really ! )
any operational amplifier coud be a current to voltage converter from a
photodiode. Of course the operational input current must be at least 10
times smaller than the minimum photovoltaic cell current. ( any photodiode
which is not biased acts like a photovoltaic cell )
Vasile

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2001\10\29@160636 by Olin Lathrop

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> Olin, If you need to have a linear output response ( voltage/light ) than
> YOU MUST use a current to voltage operational amplifier converter and
> measure the photodiodes output current rather than output voltage.
> If not, than you'll have a non linear response and a lot of trouble by
> linerising.
> See the Burr Brown optical chips. ( OPT 301 etc...)
> Of course the expensive BB chips could be replaced with cheapers structure
> at the same electrical performance.

Yes, I agree that the voltage of a photodiode is very non linear (closer to
logarithmic) with light intensity.  However, I don't remember the original
poster saying he cared about that.  If you need linearity and high
resolution, then I agree you should amplify the current.  If you don't need
either, you can hook the photodiode directly to the A/D and do the rest in
software.


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(978) 742-9014, spamBeGoneolinspamBeGonespamembedinc.com, http://www.embedinc.com

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2001\10\29@161005 by Harold M Hallikainen

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       Since it's light being measured, you can reduce parts count by using a
light to frequency converter, putting all the analog stuff on the light
detector. This worked out very well in a product for me.  See
http://www.taosinc.com/pdf/tsl235.pdf

Harold

On Mon, 29 Oct 2001 09:35:46 +0200 Vasile Surducan
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2001\10\29@164740 by Gerhard Fiedler
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At 15:46 10/29/2001 -0500, Olin Lathrop wrote:
>Yes, I agree that the voltage of a photodiode is very non linear (closer to
>logarithmic) with light intensity.  However, I don't remember the original
>poster saying he cared about that.  If you need linearity and high
>resolution, then I agree you should amplify the current.  If you don't need
>either, you can hook the photodiode directly to the A/D and do the rest in
>software.

Maybe make this "if you need linearity and high resolution or high speed"...

ge

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2001\10\30@054646 by Vasile Surducan

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On Mon, 29 Oct 2001, Olin Lathrop wrote:

> > Olin, If you need to have a linear output response ( voltage/light ) than
> > YOU MUST use a current to voltage operational amplifier converter and
> > measure the photodiodes output current rather than output voltage.
> > If not, than you'll have a non linear response and a lot of trouble by
> > linerising.
> > See the Burr Brown optical chips. ( OPT 301 etc...)
> > Of course the expensive BB chips could be replaced with cheapers structure
> > at the same electrical performance.
>
> Yes, I agree that the voltage of a photodiode is very non linear (closer to
> logarithmic) with light intensity.  However, I don't remember the original
> poster saying he cared about that.  If you need linearity and high
> resolution, then I agree you should amplify the current.  If you don't need
> either, you can hook the photodiode directly to the A/D and do the rest in
> software.
>
>
 Agree also. Unfortunately, for me the easyest way is still hardware...
 Best, Vasile

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2001\10\30@055108 by Vasile Surducan

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Yes, excellent ideea for simple isolation. A cheaper i/f or v/f converter
can be done also with 555 and a current mirror or with operational
amplifiers.
I've guess TLC235 is very cheap isn't it ?
Vasile


On Mon, 29 Oct 2001, Harold M Hallikainen wrote:

{Quote hidden}

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