Photodiode/Transimpedance Amplifier Design
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- čas přidán 3. 08. 2016
- I recently designed an infrared sensor board (dubbed "IRis") for my friend's Defcon talk (defcon-wireless-village.com/sp.... This video walks through the photodiode amplifier circuit design, and discusses some of the pitfalls associated with photodiode amplifier design. References and additional reading: www.analogzoo.com/2016/08/phot...
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You don’t lose sensitivity when reverse biasing the photodiode. The responsively actually increase when reverse biased because the depletion region grows so sensing volume increase in the diode. However as the bias is increased so as the dark current, thus noise equivalent power (NEP) also increase. This cause the signal to noise to reduce. So optimal biasing depends on the NEP requirement and the bandwidth that is needed. Biasing also increase the linearity of the photodiode since it operates in photoconductive mode instead of photovoltaic mode at higher input light.
devttys0 - Excellent description - I was with you until 7:42 when you stated reverse bias decreases sensitivity. I designed beta particle detectors (basically specialized photodiodes) and we reverse biased them with 80v for the specific purpose of increased sensitivity. Reverse bias (aka photoconductive mode) increases the depletion region of a PN junction, and the depletion region is where beta particles (and photons) are 'trapped' and thus detected. Larger depletion region, more particles detected. Downside of reverse bias is increase in dark current and thus shot noise, and drift. For what you are doing here (photovoltaic mode) the gain in sensitivity is probably not worth the extra noise. For most photodiodes this increase in noise can swamp the increase in signal and thus lowers overall signal to noise - if that is what you mean by reverse bias reducing sensitivity then ..... nvm. (BTW - it takes a VERY specialized diode to handle 80v reverse bias ..... don't try at home ....)
Reverse bias decreases sensitivity compared with zero bias because the leakage current in reverse bias increases the noise floor. I think that "sensitivity" in this case means the lowest signal that can be detected against the background noise, rather than the responsivity (the current in amps per watt of incident energy) which is what you seem to be taking it to mean. In all of the photodiode applications I've made, the only reason to increase the reverse bias is to decrease the junction capacitance and therefore improve the response time.
IR light = about 1 EV, beta rays = 10 kEv and up. That might have something to do with the choice of reverse bias, I'm guessing.
If you want to detect over a wide range, one or more diode+resistor networks in the feedback can be useful, to give a stepped gain charactristic, reducing gain as signal exceeds the Vf of the diode(s)
Man, if I could put a thousand likes, I would and more! Excellent breakdown and description. Thanks a lot!
Thank you for going into depth about how the parts work and which parts you ended up purchasing. Very helpful.
Fascinating glimpse into designing a real IR amp! Really high-quality stuff here man, thanks!
Fantastic! You solved the dilema that I recently experienced due to poor sensitivity IR Photo Diodes. I will definitely try your brilliant highlighted techniques. Thank you PROFUSELY!!!
Super well-explained! I’ve never worked on a circuit that uses LEDs in this way and it was very easy to understand and fun to watch. Thank you!
I was just trying to figure out the compensation capacitor the other day, and then you post this video. THANKS
Fantastic description of the circuit.. I'm not very experienced in analog electronics, but you made it very easy to follow..
This is a quite useful video on designing a good photodetector. Thank you!
Woah! This is awesome!
I have been trying to design a hyper-sensitive IR receiver using an LMC660. Of course I am guarding everything. This video brought up some points I would have not thought of. Thanks devttys! I give you internet hugs.
Thanks and congratulations for this very well explained PIN+TIA configuration. Cheers!
WOAH man! One of the top explanation designing op amps. Thanks a lot.
Did you consider using an OOK circuit for the comparator. In this the DC coupled signal is applied to both the positive and negative inputs of the comparator. At the positive pin, feedback is applied to turn this input into a sensitive Schmitt trigger input. At the negative input the signal is low passed filtered with a high value bias resistor added across the capacitor and the divider formed referenced to the positive rail. This has the effect of putting the Schmitt trigger point slightly above the average noise voltage and any DC bias on the signal. Provided the Schmitt trigger threshold voltage is designed to be slightly higher than the bias voltage, the circuit will trigger on the signal as it just appears above the noise. I have used this circuit to decode OOK modulated on the 433.92 MHz RF data channel, where the background noise can be high and variable, which makes the choice of a simple AC coupled comparitor threshold difficult. The detector can be DC coupled to the signal, picking out only the peaks which exceed the Schmitt trigger threshold above the variable background noise. Effectively the tiny trigger threshold floats above the average noise floor and any DC bias, no matter what that might be. This works well where the channel is dominated by band limited white noise. In my case this solution proved to be better than just AC coupling the signal onto a compatible with a fixed offset bias. The Schmitt action gives well defined rail to rail output pulses, it is even possible to modify the Schmitt recover time to ensure a minimum recovered data pulse width.
Yea, Ill have to watch this afew times just to let this sink in. Thanks for the great explanations.
Irda receivers can be useful for receiving baseband IR
Such a good vid makes me wanna get into wireless signals
Very well explained as usual!
Very nice! I ordered the PCB on OSH Park, I'm looking forward to trying it out. Thanks.
The links to the PCB are so old they are gone... do you know where I can get the PCB and other info of the old links? Thanks
This is amazing! Thank you for sharing with us.
You explained the circuit verry good. Thanks!
This is a fantastic breakdown and analysis. I'm starting a project that is similar. I'm wanting to track an IR beacon from a few hundred feet away (the beacon can be much much brighter than a phone output). So this is super helpful, and has given me confidence that it might actually be possible. @7:40 you mention that reverse biasing photo diodes reduces their sensitivity. But looking at the datasheet, figure 4 on page 3 shows "Reverse Light Current vs. Reverse Voltage" actually is extremely linear. figure 5 shows the capacitance drops quite a bit with higher reverse voltage. This would indicate to me that a higher reverse voltage would actually be a good thing for high speed applications, and may make it easer to filter out high frequency oscillations.
Yes, higher reverse bias will increase the response speed of the diode, but at a cost of greater noise.
If you're looking for small signals (and you don't need very fast speeds), then keeping the reverse bias as close to zero as possible to reduce the noise is a better strategy.
Thanks! The explanation was awesome!!
This was a great explanation. thanks!!
Amazing design tutorial. I wish I can like it many times. How did you solder the SMT components? Did OSH park do it for you? For how much?
Great explanation!
Thanks! Very informative video!
Hi. Thanks for the video. What kind of schottky diode did you use to remove undershoot? Thanks!
Top work on this
sweet photo diode video, i would like to make an analog isolation 1:1 ratio 10vdc op amp pcb using the vishay IL300 photo-diode chip. any circuit suggestions before having the pcb made? thanks :)
Good video! Subscribed. Have you ever looked at using photodiodes for detecting radiation either directly, or by use of a scintillation crystal?
Very informative, thank you!
Great video, thanks.
one thing that always pops into my mind with low current applications and "the diode drops 0.7V " is that for low currents the drop is much lower, and often this bites people when their nanoamps start flowing away at .2V instead
Yes, that's important to consider. I'd like to know what professionals do about that, since I've faced that problem a couple of times and I don't know if my solutions were optimal.
@@guillep2k Reducing a diode's forward current by a factor of e (2.718) decreases its voltage drop by approximately 25-26mV at room temperature. So reducing the current from 1mA to 1nA implies a reduction of its forward voltage by about 350mV or so. That's probably as much as you can do to estimate the effect.
Well explained.. Thanks!
Interestingly enough the circuit you designed is almost identical to the charge-sensitive preamplifiers used for radiation detectors (gas-filled tubes, photomultipliers, HPGe photodiodes, etc.). The main difference is that the DC blocking cap in those applications would be put next to the detector rather than on the amplifier's output, mainly because a bias voltage of hundreds or thousands of volts are put on the detectors and you want to keep that away from the amplifier.
So I'm thinking of borrowing a preamp from my lab and hooking an IR photodiode to it to see what it thinks about my phone...
Ah, very cool! Actually, come to think of it, putting the DC blocking cap next to the detector might be better for this application, since any constant light source is unwanted here. It would cause a light-source-dependent voltage build up across the photodiode which would change some of the operating properties of the photodiode, but that downside might be offset by the fact that the unwanted ambient light wouldn't be amplified by the op-amp. I'll have to try it out!
That could be interesting; just be advised that the amplifier's operation changes quite a bit when you give it pulses instead of a current. It won't preserve the shape of the square wave and the output voltage will peak at V=q/C and decay exponentially with a time constant of t=RC (q is the charge produced by the detector, R and C are the values of the op amp's feedback components). However, if all you care about is the timing of the leading edge of the square wave it should work.
Yeah, decay time might be a show-stopper. It'll be fun to experiment with though. :)
Very cool vid. I usually test the step response of this kind of amplifier and tweak it using the twisted wire capacitor trick. Also, this amp requires shielding, otherwise it picks up tons of crap, especially with a high gain like this one.
I know enough to appreciate this, but not enough to fully understand everything. BTW im still in HS. Any sources I can use to learn more about this online?
This is really cool, tank you!
AMAZING VIDEO
Please make a video on RF amps
great help man ! thanks !
can you explain lockin amplifiers using LED and PD?
Hello, nice work! One question. Why using decoupling caps 0.1uF at 22k and 10k whe your supply is a battery? thanks
Noise does not only come from the main voltage source.
Great video! Is there a reason you wouldn't shunt the current from the photodiode across a resistance and measure the voltage instead of inputting the current directly into the amplifier?
Good question! The answer is yes, you generally don't want to do that because you don't want to develop a signal voltage across the photodiode, since changing the photodiode's voltage bias will change its operational properties. For example, changing the voltage bias across a photodiode will change the photodiode's sensitivity (aka, "flux responsivity"). You usually don't want the signal that you're measuring to affect the sensitivity of the sensor that you're using to measure it with, so keeping the signal voltage off the photodiode is important. By using a transimpedance amplifier, the op amp varies its output voltage to ensure that the voltage at its inverting input - and hence, the voltage across the photodiode - remains constant. Hence the signal voltage appears at the op amp's output, and not across the photodiode.
@@Analogzoo Beside's amplitude errors you describe looks like you don't care about it too much in your case, but one more obvious problem in that case is maximum voltage output from terminated PD in voltage in photovoltaic mode is ~0.5V and relative linear operation is up to 300-400mV, then it saturates in weird way producing all sorts of artifacts, so that way is more bumming... Plus to preserve square wave fidelity (reasonable fall&rise t) you have to terminate PD well enough which leads to low gain (in the beginning of video for that case detection range discussed was few")... You could employ discrete+hybrid opamp (2 stages) to lower noise and increase bandwidth, also PD bootstrapping might offer a lots of Cpd cancellation that adds more BW and gain but overall nice informative vid.
Thank you.!
Hi, Very nice video to clear all points.....How about if my transmitter is transmitting data at 9600 baud rate....how same circuit works....
You need to increase the bandwidth by a factor of 10, so you'll need to reduce the feedback resistor down to 1 megaohm. That will reduce the gain, and hence the range by a factor of 10. Perhaps you will need a multi-stage circuit, rather than this simple one.
What is the frequency response of this circuit? Can it see the 38khz modulation in typical IR signals from remotes? It would seem that the 10M feedback, while providing high gain, would also degrade the frequency response considerably.
I'd like to know that too.
You’re correct. If it’s designed correctly, the dominant pole will be governed by the RC in the feedback path. In this case, BW = 1/(2*pi*R*C) = 7.2kHz. Remember how tiny 2.2pF is - and how easily board parasitics could swamp it. Reducing this capacitance would not be a good way of widening the bandwidth
Is there a recording/transcript of your friend's defcon talk? It sounds really interesting
Unfortunately not all the talks were recorded, and his wasn't. :(
Cool video. What kind of EDA tool do you use for PCB design?
KiCAD. It has its idiosyncrasies (for example, at least on Mac, the standard Command+C copy shortcut key *fricking closes kicad windows*...ugh...), but then I'd venture to say that all CAD programs are like that.
This is in the OSX bug database bugs.launchpad.net/kicad/+bug/1420714 - if some intrepid soul wants to fix it.
Nice video 👍
dear host, I really enjoyed this cirquit. but instead of 3 mosfets isn't it more practical to use one zenerdiode in reverse polarisation?
You could definitely use a zener, or three normal diodes. The advantage of using JFETs as diodes is that they have very very low reverse leakage current. In the original design, the circuit was powered from a split supply, so there was a possibility that the diodes could become reverse biased; later, the design was switched to run off a single supply and reverse biasing should never happen now, so the JFETs really don't give much of an advantage, they're just a hold-over from the original design.
Hey devtyso ,can you give me your Email ID please.i have some queries regarding my own same amplifier circuit!?!??? Waiting for your reply
Nice video, I subbed!
very good !!
How would you add in AGC in this circuit in order to give the comparator a predictable signal?
I would low pass filter the signal and use this to compare against the signal itself. Brigther pulses will rise the filtered signal, causing the comparator to be less sensitive.
Doesn't the circuit get swamped by noise from fluorescent lighting, which produce an IR carrier component modulated by low frequency noise? In the case of compact florescent lights there is a strong modulation from the control electronics.
I haven't looked at CFLs specifically, but I do have fluorescent lighting above the workbench which generates a 60Hz modulated signal that is picked up by the detector. The high pass filter formed by the coupling capacitor between the amplifier and comparator helps to reduce the effects of lower frequency modulation like this, but the circuit is still fundamentally susceptible to it.
Hello,
I do not understand why the circuit is not stable. So I tried some things :
First I replaced the photodiode by a perfect current generator in parallel with the capacitance. Then I transformed the Norton generator by a Thevenin generator delivering the voltage : e = id/jCw. id is what I called the photodiode current, e being the entry for transfer function. By using Milmann on the node with the potential V- (entry minus of the TIA), I found H(p) = -Rf*Cp.
Now this doesn't look like any transfer function I've ever seen. I would like to analyse the stability but it has no denominator so I don't know how. Maybe that in this simple case the stability is just by analysing H(p) when p tends towards +infinity? Then I would get H(+inf) = +inf, can this alone let me conclude that the system is not stable?
Also I first considered the resistance Rs of the substrat of the photodiode, following this method I found H1(p) = -RfCp/(1+RsCp) but then this transfer function is stable so it contradicts what you said.
Give me details of highly recommended books on noise
awesome
I wonder if an avalanche photodiode suited for the wavelength in question could be used with a strong optical IR bandpass filter to create an extremely sensitive detector precisely for IR. The high gain would perhaps compensate for the filter loss. I've been reading about APD applications when it comes to laser rangefinders, which is a fascinating topic in itself. Though while ~950nm APDs exist (like the First Sensor 9.5) they're pretty expensive afaik. From some rough research they should have maybe ~100x more sensitivity on average compared to a photodiode, I absolutely stand to be corrected though since I'm new to this area myself.
I've never used them myself, but yes, avalanche photodiodes are much more sensitive than the more common PIN photodiodes. I didn't really consider them as an option in this case due to cost (as you noted) and AFAIK they typically require large voltage biases. Would be interesting to play with!
devttys0 Yep, but the large voltage biases is where the gain of APDs comes from. The drive circuits would be similar to a geiger counter like some other people mentioned. I'll advertise a cool bunch of posts for you in case you'd like some geeky reading - google "Designing a Laser Range Finder", should be first link on hackaday. The poster is a South African manufacturer of laser rangefinders. :>
devttys0 Also your channel to me is as fantastic for pure learning as W2AEW's. Thank you so very much for the videos that you've made.
At 4:26 "... at some frequency that phase shift is going to be 180 degrees". No, it isn't. The phase shift in a single RC filter never exceeds 90°. You'll have to think again about what's causing oscillations. You also need to consider the phase shift introduced by the opamp itself at very high frequencies. Probably the layout is going to be critical in these sort of cases.
For modern rail-to-rail opamps, you have a CMOS output stage and that allows output voltages very close to the rails, especially with small current drawn. In addition MOSFETs don't suffer from the same problems of recovery from saturation that BJTs do, so you don't have to worry about that effect. I might have considered an MCP6002 for both the transimpedance amp and the comparator. You don't need a huge bandwidth to detect the 1KHz pulses and the MCP6002 is cheap as well as needing a supply current of just 100µA. Nice project.
So coin cell batteries are usually rated for
Most of the ones I've seen are tested at
Cool. Thanks. Also what 3D renderer do you use?
It's the one built into KiCAD.
Schweet. Thanks.
10 meg feedback resistor seems too high, noisy and maybe unstable?(I think it's resistance will vary a bit depending even on air humidity). wouldn't it be better if there was two amplifier stages(more stable, less noise)? One transimpedance amplifier with say 100 kOm fixed resistor, and second stage - voltage amplifier with another 100 kOm feedback resistor, maybe variable so you can adjust it for best performance. And then comparator.
It is quite a large resistor, and cascading two amplifiers would likely be perfectly acceptable here; in fact, what you suggest is commonly done when high gain and wide bandwidth are both required. However, wide bandwidth was not a requirement in this case.
Making the feedback resistor larger actually increases stability, and even the compensation capacitor in the schematic can be removed without adverse affect; even a few pF of parasitic capacitance in parallel with such a large resistance forms a low pass filter with a pretty low-frequency cutoff, effectively preventing high frequency oscillation/instability.
And yes, the resistor value will change over temperature, humidity, etc…even putting a voltage across a resistor causes undesirable effects such as non-linearity (practical resistors are not perfectly linear). I don’t expect any significant negative effects in this case, although to be fair I have not performed any extensive temperature testing.
A larger resistor will indeed have more inherent Johnson noise, but noise is not really a concern; what’s really important is the signal-to-noise ratio. 10mVpp of noise isn’t much a problem if your signal is 10Vpp, but it is a problem if your signal is only 11mVpp! The transimpedance amplifier’s gain increases directly with the feedback resistor value, but the noise increases with the square root of the resistor. All else being equal, this means that a larger feedback resistor results in a better SNR. In addition, the narrow bandwidth formed by the large resistance and its parallel capacitance is very effective at reducing noise in the system. It’s all a bit tricky though, because proper noise analysis is rather complex when dealing with photodiode amplifiers.
Bottom line is that I would expect both approaches to work acceptably, but I’d highly recommend Jerald Graeme’s Photodiode Amplifiers book, which deals extensively with noise, cascading amplifier stages, stability, etc.
Thanx, that was very informative answer. BTW I had such problem while designing similar circuit where high value op amp gain resistors changed they'r values and part of circuit tended to be unstable. But circuit was used in marine application, and it took some time to find the cause. :)
@@Analogzoo Late reply, buy hey.
A higher TIA feedback resistor would increase the SNR of the electrical signal itself, for the reasons you mention - but wouldn't it also make it more susceptible to radiated noise, though? Noise pickup would be higher on a higher impedance node than on a low impedance node, so (as always is the case with noise analysis) it would all depend on what your main noise sources are.
Just curious, you picked the JFets as diodes, for their low inverse leakage current - when does this come into play? IMHO the diodes (JFets here) will never be reverse biased at all (even if there is 0 current flowing through the IR diode, the output of the opamp will always be ever so slightly above the V- pin of the opamp), so no current will be able to flow in the reverse direction.
~
Heh, you're right of course, I'm surprised you're the only one to mention that. :) Leakage was more of a concern in a previous iteration of the design which ran off a split supply and the op amp's output was nominally 0v; in that case, a fast-moving input signal could briefly reverse bias the diodes (or they could possibly be reverse biased in a quiescent state due to op amp offset errors). Still, the JFET-as-a-super-low-leakage-diode trick is one of my favorite parts of the circuit, even if unecessary. :P
Yeah, if the opamp was powered from a split supply, then they will make sense. That crossed my mind, but when I saw the single coin cell battery, I started scratching my head :-) I totally agree that for low-current nodes, JFets are way better then a simple diode!
Regards,
Aleks
0
I have been searching for this video. trying to make long-range IR things..
Did you succeed?
@@hoplahey haha not really. I made a prototype that worked in daylight but only 6ft or so
How exactly are the JFETS wired up? Anyone please help :(
nice!
how does one RC provide 180 phase shift ?? at the max 90 degrees!
Beyond its dominant pole, the OpAmp can contribute an additional 90deg
Add another op-amp amplifier stage, and keep the first stage gain low ?
You'll get swamped by noise.
Hi. Can anyone explain how to configure this MMBF4117 N channel JFET into a diode case? I dont understand how to connect g s d terminal in this video so that it will become like a diode
Connect drain to source; as it's an N-channel, that's the diode's cathode. The gate is then the anode.
@@RexxSchneider thx bro. Can u briefly explain why is that?
@@wenhaoye8773 An N-channel JFET can be thought of as a length of N-type silicon with P-type doping around half-way along that length. So if you connect the drain and source together, you have a connection to N-type silicon. The gate is then a connection to the P-type silicon and that's what you need for a diode.
The construction of JFETs is such that the diode you get has low reverse leakage current.
@@RexxSchneider thx bruh, ur explaination is super useful and i understand now. But the final question comes up: in the video, the youtuber mentioned that configure jfet in diode case will hv lower leakage current compared to normal nmos. Do u hv any clue why is that?
@@wenhaoye8773 To be accurate, connecting a JFET as a diode tends to have lower reverse leakage current than a normal silicon diode. That's a complex issue, but in simplistic terms, because the JFET is bigger than a normal diode, the charge carriers have a greater distance to go, and the levels of doping will normally be different from those used in a silicon p-n diode.
A MOSFET doesn't have a junction to reverse bias, and the gate is insulated by a metal oxide layer. Leakage through the gate of a MOSFET is a completely different effect.
For anyone who is looking for more information, check out the book by Jerald Graeme Photodiode Amplifiers: op amp solutions
You are wrong about the photodiode getting hit by more light equates to to a higher gain. In your circuit, more current will flow thru the PD and lessen the amplification. I think you mean to add a +Vcc to the PD in your circuit
Darlington pnp is better than a op amp her i belive .. put it in a sim !!
Best circuit is a 2N7000 mosfet and connect photodiode between D and G
Transistor as a source follower... With 3 IR leds and small reflector works up to 1000m
lol who disliked this video?
3 dorks who quit class
1k likes
This video had a high yawn factor.
An Avalanche Photodiode driver would have been much more sensitive and exciting.
Yea...yea, then also you can also step the game to MPPC's but all the cost for this mundane application would be overkill! i think it is pretty appropriate to the task.
You are wrong about the photodiode getting hit by more light equates to to a higher gain. In your circuit, more current will flow thru the PD and lessen the amplification. I think you mean to add a +Vcc to the PD in your circuit