Circuit to measure DC and AC from a photodiode at low-frequencies

Discussion in 'General Electronics Chat' started by MailerDaemon, Jul 17, 2015.

  1. MailerDaemon

    Thread Starter New Member

    Jul 17, 2015
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    Hey all,

    like the topic says I am looking for a solution to measure the DC and AC signal from a photodiode simultaneously at a relative low frequency range starting around 100Hz up to maybe 1MHz. I think duo to the need of a very high inductance at this low frequencies I didn't find a suitable bias tee for this application to seperate AC and DC parts.

    I found a very interesting circuit which uses a second op-amp to remove the dc contributions. The basic idea of the circuit is described here:
    http://electronicdesign.com/power/photodiode-amp-nulls-ambient-light

    or in the book

    Photodetecion and measurement - maximizing performance in optical systems on page 146.

    Here the circuit is used to remove unwanted dc part from ambient light. Further I found this publication

    "A low-noise transimpedance amplifier for the detection of “Violin-Mode” resonances in advanced Laser Interferometer Gravitational wave Observatory suspensions" by N. A. Lockerbie and K. V. Tokmakov, Rev. Sci. Instrum. 85, 114705 (2014)

    http://scitation.aip.org/content/aip/journal/rsi/85/11/10.1063/1.4900955

    Here they used this idea to seperate DC and AC parts.

    Nevertheless I am not very familiar with electrical circuits and I am trying to understand the principle behind this idea but can not find much additional information for some deeper understanding.
    As I understand it would be possible to modify this circuit by removing for example the low-pass filter which is formed by R3 and C3 in Fig.1 b) to extend this circuit to higher frequencies?
    But what are the limitations of this circuit and how to design it for the proposed problem?

    I hope that someone has some further input and ideas for me.

    Thanks in advance!
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    What is it about the detector and the photosource that makes you think there is useful information in the MHz. range?
    In order to process signals in this range the device needs to be capable of operation at about ten times the frequency of interest. This amplifier might have a GBW of 10 MHz. or so.
     
  3. MailerDaemon

    Thread Starter New Member

    Jul 17, 2015
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    As a source I use LEDs. I want to measure the relative intensity noise (RIN) of LEDs at this low frequency range in order to see a 1/f or 1/f² behavior in the RIN at low frequencies, which should go over to the shot noise limit for higher frequencies. Therefore it would be perfect to cover this large frequency range.
     
  4. Papabravo

    Expert

    Feb 24, 2006
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    I'm not entirely sure you can make an LED turn on and off at that rate. I think you get a kind of mushy blur at that speed.
     
  5. MailerDaemon

    Thread Starter New Member

    Jul 17, 2015
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    I don't modulate the LED I want to measure the intrinsic noise characteristics of LEDs in the named frequency range. The LED is operated in continous wave but it's intensity noise is limited by the shot noise limit for high frequencies and limited by 1/f or 1/f² for low frequencies.
     
  6. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    A standard red or green LED has on and off times below 50 ns, so its output should be "square" enough. Let's try for a bit more clarity - what do you mean by "continuous wave", something truly continuous (a sine wave) or something continuing but discontinuous (a square wave)? If square, is ti symmetrical or a pulse train? If pulse, what are the min/max pulse widths?

    In post #5 it sounds like you are using one photo diode system to measure several different LEDs. yes/no

    Back to the LED driving waveform, is there a DC component to the continuous wave, such that the LED never is completely off? If so, is this the DC component you want to separate from the AC in the photodiode signal, or is that something else?

    ak
     
  7. OBW0549

    Well-Known Member

    Mar 2, 2015
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    The circuit of Fig. 1 (b) in the Lockerbie and Tokmakov paper is actually a fairly common scheme used to remove an unwanted DC component from a composite signal, leaving only the AC component as an output. The data sheet for Linear Technology's LT1793 operation amplifier (http://cds.linear.com/docs/en/datasheet/1793f.pdf) shows a slightly different implementation of this method for processing a photodiode signal (at the bottom of page 1), and application to a piezoelectric accelerometer (bottom of page 12).

    In all versions, one opamp (IC3 in the Lockerbie/Tokmakov paper) is used as a transimpedance amplifier to translate the sensor's output current to a voltage, and another opamp (IC2) functions as a non-inverting integrator which integrates the output of the transimpedance amplifier and sends a compensating current back to the transimpedance amplifier's input so as to keep the amplifier's output near zero in the long-term. Short-term fluctuations in IC3's output (i.e., the desired AC component of the output signal) do not affect IC2's output, and therefore are not nulled out.

    Yes, removing that filter will raise the upper frequency limit of this circuit somewhat.

    The main limitation on this circuit's frequency response is the gain-bandwidth product of opamp IC3 which, for an AD743, is 4.5 MHz. To get a usable 1 MHz bandwidth from this circuit, the opamp will have to be a lot faster than the AD743, while still remaining low-noise. It may take a lot of searching to find a suitable part.

    Hope this helps a bit...
     
  8. OBW0549

    Well-Known Member

    Mar 2, 2015
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    As I understand it, what the TS is trying to do is measure the luminance noise of an LED when operated at constant DC current; unless I'm missing something, the LED current is not being modulated in any way.

    At least, that's what I think he's doing...
     
  9. #12

    Expert

    Nov 30, 2010
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    That is the impression I get. If true, then this is a problem for frequency filters. AC couple the signal from the LED, bandwidth limit it, amplify to a comfortable level, measure the voltage. A capacitor separates the DC very nicely, the bottom end is audio frequency, and the top end is even easier to filter.
     
  10. OBW0549

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    Mar 2, 2015
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    Ummm... there is no "signal from the LED." He's not trying to measure the noise in the LED's forward voltage drop, he's trying to measure the noise in the LED's light output. Hence the photodiode, and the need for some sort of transimpedance amplifier to process its output.
     
  11. #12

    Expert

    Nov 30, 2010
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    So...apply the amplifiers and filters to the output of the photodiode. Either that or send the information from the photodiode to a spectrum analyzer. Maybe there are other ways to find noise in defined frequency bands.
     
  12. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    I yield to y'all. So, with a lowest frequency of interest of 100 Hz (post #1), my first thought was an AC coupled wideband mic preamp-ish. 100 Hz isn't a very low low end, but the coupling capacitor has to behave over a 4 decade frequency range. Hmmm, maybe a 1.6 Hz servo (or 16 Hz in the second link) is the way to go. It seems to be a popular choice.

    To the TS question - both circuits you linked to function in the same way, so we'll stick to the first link. The opt201 is a fully integrated photodiode and transconductance preamp. Steady state light on the sensor translates to DC output voltage, and the device has only so much range for output swing. R1 and C1 lowpass filter the output to leave only the DC component and the very low frequencies. A1 and C2 integrate this to produce what is an corrective offset voltage that represents the DC level at the output. Because the other end of A1 is tied to GND, the A1 output represents how far from ground the sensor output is. This is an active version of a coupling capacitor - it removes the DC component and leaves the AC stiff. The corrective voltage is converted to a very low current by R3 and injected back into the input of the sensor. In the high-end audio world this is called a DC servo, used to eliminate evil coupling capacitors.

    ak
     
  13. MailerDaemon

    Thread Starter New Member

    Jul 17, 2015
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    1
    Thank you all for your replies!!!

    OBW0549 you are completly right. The LED is supplied with a constant current and I want to measure the fluctuations of optical output power which is radiated by the LED.

    To quantify this fluctuations I want to measure the relaive intensity noise, which can be understand more or less as the quotient of I_AC(f) / I_DC.

    I_AC(f) = AC signal of the photodiode used to detect the emitted light of the LED (Measured with a electric spectrum analyzer)
    I_DC = DC signal of the photodiode used to detect the emitted light of the LED

    Because of this I need AC and DC part simultaneous.

    In the MHz frequency regime this is no big issue since one can use an bias tee (e.g. http://194.75.38.69/products/BiasTees.shtml) to sperate AC and DC.

    Since I want to measure the noise in the low frequency regime I need to think of an other way to seperate AC and DC. Therefore I thought that the circuit described in the publication may be useful.

    As far as I now understand the freuqency limits are give by the filters, which can be modified/removed with suitable capacitors and resistors, AND the gain-bandwidth product of the amplifiers.

    But I am a little bit concerned about the right choise of amplifiers. If I got it right the IC2 (Fig 1b in the publication) sends some sort of low frequency feedback back into the circuit to cancle out the dc? But how high should the gain be of IC2? Is the frequency behavoir given bei the two R=100k resistors and the two C=100n capacitors? Why are the capacitors and resistors selected like this?
    Does the IC2 also need this high Gain Bandwidth Product? Or since it only amplifies the low frequencies it only needs low noise?

    For the IC3 I found a simular amplifier as the AD743, the AD745 which has a 20 MHz Gain Bandwidth Product at the same noise properties. Is this one suitable? (http://www.analog.com/en/products/amplifiers/operational-amplifiers/jfet-input-amplifiers/ad745.html)
     
  14. OBW0549

    Well-Known Member

    Mar 2, 2015
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    Correct. IC2 acts through resistor Ri to sink the current being passed by the photodiode to servo IC3's output toward zero volts; but since IC2 is configured as an integrator (i.e., its voltage gain decreases with increasing frequency), this corrective action takes place only slowly, therefore the output of IC2 represents the DC component of the photodiode output.

    IC2 gain doesn't matter much; just about any operational amplifier will have more than enough gain to do the job.

    Those resistors and capacitors determine AC low-frequency cutoff point (i.e., the frequency below which the AC gain of the circuit begins to decline). Those values were probably selected to give a low-frequency cutoff somewhere below 100 Hz. They have no effect on the high-frequency performance.

    No, IC2 does not need a high GBW product. It only needs low input voltage (and current) noise, especially with regard to 1/f noise and 1/f noise corner frequency, and low input offset current given the high value of the resistors R. Although the LT1028 has excellent noise performance, it has rather large input bias currents and offset current; it is also not guaranteed to be unity-gain stable (meaning that in this circuit, it might oscillate at some high frequency up in the MHz region). My own choice for IC2 would have been a MAX44241 (http://www.maximintegrated.com/en/products/analog/amplifiers/MAX44241.html).

    Yes. The AD745 would be my first choice to get a higher frequency response from this circuit, without compromising its noise performance.
     
  15. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    A1 or IC2 has one of the pins is connected to GND, so the integrator is a form of a comparator, showing the accumulated time and low-frequency average energy of the input signal away from GND. Because of this, A1/IC2 does not need very high gain-bandwidth, but it does need very low input offset errors. Any DC offset at its input shows up in the Vdc output signal. Three error sources are the opamp's inherent input offset voltage, differences in its input bias current times each of the two input resistors, and leakage current and dielectric absorption in the integrator capacitor.

    ak
     
  16. MailerDaemon

    Thread Starter New Member

    Jul 17, 2015
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    Hey thank you for your suggestions.

    As I am really not that deep in the whole op-amp thing, its a great benefit to know which parameters are important.

    Now I am wondering if low input voltage (and current) noise, low input offset current and low input bias current are the important properties, why shouldn't I just use the same amplifiers for IC2 and IC3?

    The AD745 has

    input voltage noise: 5.5nV/sqrt(Hz) @ 10Hz // 3.6nV/sqrt(Hz) @ 100Hz ...
    input offset current: 20 - 75pA
    input bias current: 250 - 400pA

    Am I getting something wrong or why not choosing also this amplifier?
     
  17. OBW0549

    Well-Known Member

    Mar 2, 2015
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    No reason at all why you couldn't use an AD745 for both IC2 and IC3. The main thing the MAX44241 has going for it in the role of IC2 is that, because it is a chopper-stabilized opamp, it has essentially no 1/f noise. But since you're planning on ignoring frequencies below about 100 Hz anyway, this may not be very important. On the other hand, if you were to want to extend the frequency response of this rig down to 1 Hz or below, I'd say definitely use the MAX44241 for IC2.
     
  18. MailerDaemon

    Thread Starter New Member

    Jul 17, 2015
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    Do the chopper-stabilization and the nonpresent 1/f noise have any positiv influence on my dc signal, which I am measuring from the upper port of fig 1 b) ?
     
    OBW0549 likes this.
  19. OBW0549

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    Good catch! Now that you mention it, the MAX44241 would indeed give you some benefit there, as it would result in less low-frequency noise on your DC output than with the AD745. Whether that benefit would be significant or not, I couldn't say; the difference in noise level would be only a few microvolts, I expect.
     
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