Post by thread:[EE]: Differential amplifier circuit?

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 -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

> 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_OUTolinTakeThisOuTembedinc.com, http://www.embedinc.com -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

> 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. ******************************************************************** Olin Lathrop, embedded systems consultant in Littleton Massachusetts (978) 742-9014, .....olinKILLspam@spam@embedinc.com, http://www.embedinc.com -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

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. -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

> 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 -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

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 <jayKILLspamILHAWAII.NET> writes: {Quote hidden}> 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 > > -- > http://www.piclist.com hint: The list server can filter out > subtopics > (like ads or off topics) for you. See http://www.piclist.com/#topics > > FCC Rules Online at http://hallikainen.com/FccRules Lighting control for theatre and television at http://www.dovesystems.com ________________________________________________________________ GET INTERNET ACCESS FROM JUNO! Juno offers FREE or PREMIUM Internet access for less! Join Juno today! For your FREE software, visit: dl.http://www.juno.com/get/web/. -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

----- 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 -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email .....listservKILLspam.....mitvma.mit.edu with SET PICList DIGEST in the body

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. -- Ned Konz currently: Stanwood, WA email: EraseMEnedspam_OUTTakeThisOuTbike-nomad.com homepage: http://bike-nomad.com -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listservspam_OUTmitvma.mit.edu with SET PICList DIGEST in the body

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 -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email @spam@listservKILLspammitvma.mit.edu with SET PICList DIGEST in the body

>> > 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 -- http://www.piclist.com hint: To leave the PICList KILLspampiclist-unsubscribe-requestKILLspammitvma.mit.edu

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}> > 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. > > > ******************************************************************** > Olin Lathrop, embedded systems consultant in Littleton Massachusetts > (978) 742-9014, RemoveMEolinTakeThisOuTembedinc.com, http://www.embedinc.com > > -- > http://www.piclist.com hint: The PICList is archived three different > ways. See http://www.piclist.com/#archives for details. > > -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

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 -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

> 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. ******************************************************************** Olin Lathrop, embedded systems consultant in Littleton Massachusetts (978) 742-9014, spamBeGoneolinspamBeGoneembedinc.com, http://www.embedinc.com -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

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 <TakeThisOuTvasileEraseMEspam_OUTL30.ITIM-CJ.RO> writes: {Quote hidden}> 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 FCC Rules Online at http://hallikainen.com/FccRules Lighting control for theatre and television at http://www.dovesystems.com ________________________________________________________________ GET INTERNET ACCESS FROM JUNO! Juno offers FREE or PREMIUM Internet access for less! Join Juno today! For your FREE software, visit: dl.http://www.juno.com/get/web/. -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

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 -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

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 -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

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}> 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 > <RemoveMEvasileTakeThisOuTL30.ITIM-CJ.RO> writes: > > 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 > > FCC Rules Online at http://hallikainen.com/FccRules > Lighting control for theatre and television at http://www.dovesystems.com > > ________________________________________________________________ > GET INTERNET ACCESS FROM JUNO! > Juno offers FREE or PREMIUM Internet access for less! > Join Juno today! For your FREE software, visit: > dl.http://www.juno.com/get/web/. > > -- > http://www.piclist.com hint: PICList Posts must start with ONE topic: > [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads > > -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics