Current pump to replace photodiode in transimpedance amplifier circuit - Page 1
Forum section: electronics / beginners

author: Bigman
author_rank: Regular Contributor
timestamp: May 14, 2015, 02:28:56 pm
content: Hello,I have successfully built my first transimpedance amplifier with a photodiode as the current source. Next, I want to analyse the TIA's behaviour (such as bandwidth) by real measurements. Instead of modulating light, I think it is much easier to replace the photodiode with a voltage-controlled current source which I can modulate instead (approx. 0.1uA).For the design of a voltage-controlled current source I have found the Howland Current Pump in TI's Appplication Note AN-1515 (http://www.ti.com/litv/pdf/snoa474a).My idea is to use my Waveform Generator to control this Howland Current Pump. So my full "system" compromises three elements:Waveform GeneratorHowland Current PumpTransimpedance AmplifierBut I am uncertain how to connect those three elements together. The attached picture shows with the dashed lines my assumptions:it my look wired, but I really need that 0.7V DC-offset at the inverting input at the TIA-OpAmp. (I achieve this by using a rail-splitter at the 3.6V battery, so I have a dual supply for the OpAmp). My assumption is, that I have to connect the "-Input" of the Howland Current Pump with this DC-Offset;I am wondering about the BNC-connector housing of my Waveform Generator. Is this housing connected inside the waveform generator with GND of my house's power supply? Should I connect this with the 0V from the battery or with the 0.7V of my offset?Thanks in advance for all your comments!

author: ConKbot
author_rank: Super Contributor
timestamp: May 14, 2015, 03:33:15 pm
content: Resistor? If its the same value as your transimpedance amp's transfer function then Vin = VoutAs far as your waveform generator, depends. Usually the front panel specifies how much or if you can float the shield of the bnc above earth ground.

author: Bigman
author_rank: Regular Contributor
timestamp: May 14, 2015, 03:44:03 pm
content: Quote from: ConKbot on May 14, 2015, 03:33:15 pmhonestly speaking, I am totally lost ... pls. be patient with me. I am a newby and have no glue about which resistor your are speaking (again, as I newbie I may be a bit slow).However: on the front of the waveform-generator the outputs are labled with 50 Ohm. Well ... but isn't this "just" telling me: the output of the waveform-generator has a resistance of 50 Ohm ...

author: ConKbot
author_rank: Super Contributor
timestamp: May 14, 2015, 04:34:06 pm
content: On a keysight/agilent 33210A for example,https://mse.gist.ac.kr/~flexible/data/file/equipment/873648049_T7A1xeQO_2.Agilent_33210A_2810_MHz_FunctionArbitrary_Waveform_Generator29.jpgon the right of the output there is a marking "42Vpk" indicating you can float the shield of the output BNC upto 42V above earth ground. If the symbol is just a earth ground symbol, it is directly grounded, and you have to take care where you hook it, much like a scope ground lead. If your scope and sig gen are both plugged into the wall, measure resistance between the BNCs on both and you can see if the sig-gen is floating or earth referenced. The equipment manual should specify too.EDIT: looked up the SDG1010, the symbol indicates its earthed directly, so no floating the SG output.As far as the first comment, a resistor makes a good voltage to current converterAlso, in your schematic, you have a bias, but both ends of your photodiode are hooked up to op-amp inputs, meaning they will be at the same potential, no bias applied ( By the rules of ideal op-amps) If youre trying to reverse bias your photodiode, then the anode of the photodiode should be connected to your 0v, so 0.7v of reverse bias would be applied. Changing this would mean your sig gen output - current pump in and out -, and scope reference lead, will all be attached to 0V, and everything should work ok.

author: Bigman
author_rank: Regular Contributor
timestamp: May 14, 2015, 05:13:54 pm
content: First: thank a lot for the kind reply. Now I understand.Quote from: ConKbot on May 14, 2015, 04:34:06 pmat the waveform-generator's output, besides 50 Ohm, there is also the GND-symbol. So it's directly grounded (as you have also found it in the manual) ... hmmm ... this bring me to the following assumption: for sure, I should not connect it to my 0.7V. Because the waveform-generators GND would pull those 0.7V down to GND. So, I assume, I have to connect that waveform-generator-GND with NOTHING (perhaps only with the housing of my circuit for shielding???)Quote from: ConKbot on May 14, 2015, 04:34:06 pmahhh ... now I think I got it. I know the waveform-generator's voltage at the output, I also know the voltage at the inverting-input of the TIA-OpAmp (=0.07V). I also have the waveform-generator's output resistance. Hence, based on Ugenertor- 0.07V = Rsum* ItargetI should be able to controll my Itargetby tuning Ugeneratorand Rsum. But wouldn't the voltage from the waveform-generator bias my photodiode? Respectively, how could I avoid biasing my photodiode .... hmmm ... I need to draw some schematic to get a better understanding of this. As soon as I have something reasonable, will post it.Quote from: ConKbot on May 14, 2015, 04:34:06 pmNo, I do not want to reverse-bias the photodiode. To avoid a bias, the TIA's non-inverting input and the photodiode's anode have the same potential. That this common potential is 0.7V is on purpose, since I want to have a DC-offset. But I see your point, that my Current Pump will generator a voltage different than 0.07V at the photodiode's cathode which will bias it. But I can't follow you, how to avoid this bias ("but both ends of your photodiode are hooked [...]"However, ConKbat, thanks a million time for your kind effort so far!

author: Bigman
author_rank: Regular Contributor
timestamp: May 14, 2015, 06:07:45 pm
content: Quote from: Bigman on May 14, 2015, 05:13:54 pmthis doesn' feel right. I mean, then I would never connect the waveform-generator's GND to any circuit under test ... ahhh ... I am now so confused. I still have no glue, how I can exchange the photodiode with a voltage controlled current source ... perhaps I have to sleep over it first.

author: Bigman
author_rank: Regular Contributor
timestamp: May 14, 2015, 09:02:55 pm
content: Okay, I think this Howland Current Pump was "overdone" and it seems I "missed the forest for the trees". Such a "Howland Current Pump" solution would have been only necessary, if the load-resistor could be unstable - which shouldn't be the case in my situation.So, pls. allow me a new attempt, based on the suggested "resistor". Since the waveform-generator already has a 50 Ohm internal resistance at it's output, I didn't see the need for adding an additional resistor. Since I want the current to swing with a 0.1uA peak-to-peak, the voltage had to swing Vpp=0.005mV. Well, the smallest Vpp my waveform-generator can provide is 2mV. But I think, by adding an additional resistor Rvd, I will have a voltage divider, that will divide my Vpp=2mV from the generator down to the desired 0.005mV.

author: ConKbot
author_rank: Super Contributor
timestamp: May 14, 2015, 09:25:15 pm
content: I'd just put a series resistor in series with the function generators output that is equal to your Rf in the TIA, hook it to the Cathode of the photodiode, and the negative of your signal generator to the 0V rail of your TIA. Whatever voltage you put in, the inverse should come out of the TIA after you account for the 0.7V offsetIf your Rf was 10k for example, and the resistor you use to inject current is 10k also, if you applied 0.6V(relative to 0V) to the injection resistor, there would be 0.1V difference from the inverting input to the sig-gen, so 10uA would flow towards the sig-gen, which means 10uA would have to flow across Rf, and the output of the amplifier would have to rise to 0.8V Since youre looking at 0.1uA, im assuming your resistors are higher in value, but the principal is the same.

author: Bigman
author_rank: Regular Contributor
timestamp: May 15, 2015, 08:03:19 am
content: Quote from: ConKbot on May 14, 2015, 09:25:15 pmThis is very kind of you. After I realized, how stupid my first approaches have been, I feard you may think "this guy is worse than useless". Attached my latest update. Pls. allow me following questions to make sure, I don't make it wrong at the "last mile":the "injection resistor" Rinjshould have the same value as the "feedback resistor" Rfto avoid biasing the photodiode. (Indeed, first I got you wrong and thought you mean I should take Rfout since I have now Rinj);I do not have to unsolder the photodioe. Just do darken it totally is okay. This would have the beauty, that I still would see the effects coming from the photodiode's capacity (important to see effects on the bandwidth);I have to change my sig-gen "Output load setting" to Hi-Zif that is really it, than you are my hero(oh man, now it looks so simple. Thanks again a million time!)

author: tggzzz
author_rank: Super Contributor
timestamp: May 15, 2015, 09:24:16 am
content: If the TIA is operating in the linear region, which it will be provided the output isn't saturated nor the input changing too fast, then the inverting input will be at the same potential as the non-inverting input - give or take the input offset voltage.So, if you want to inject a current I, then simply connect a resistor R between the non-inverting input and a signal generator, and set the signal generator's output to V = IR.Complications... The PD will have a capacitance; if you remove the capacitance then the high frequency operation will change. The resistor R will lower the impedance at the non-inverting input, which may cause frequency and offset changes. Those effects are likely to be less than if you use a "complex" current generator.As a separate issue, be aware that you can operate a PD in a logarithmic or linear mode. In the linear mode it is usual to reverse-bias the PD by several volts, which lowers the capacitance and causes a "dark current" offset.