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I'm designing a new frontend photodiode for a four quadrant4456.tr12.zips system.My requirements are the following :
Photodiode capacitance : 16pF / quadrantDC blocking (from DC to around 5kHz)DC current level : from 0 (system used in the dark) up to 10mABandwidth of the measurement : useful from 1MHz to 20MHz (30MHz would be a plus)Sampling with a 12bits system (to be defined)Photodiode current range : from 1uA to 20mA
I made a design (see attached file) which has a very flat transimpedance (at -6dB) from 20kHz to 20Mhz.Transimpedance is 50kOhm (95dBOhm) and total noise level is 1.77mV.With a 1uA pulse (half-sine 20Mhz), it gives a 25mV output pulse, which is very comfortable for sampling and analyzing.
The main problem that I'm facing is that my transimpedance is too high for the high photodiode current that I may receive.
I have changed my transimpedance 50kOhm with a tee system (so I can connect or disconnect the foot resistance using a photocoupler) and tried to add a resistance divider just after the AC-coupling capacitance, without any success (I add too much noise).
I also tried using a 100Ohm transimpedance and a VGA, but i add too much noise into the input of the VGA. I get bad SNR results.
Do you have any ideas on this subject? How can I improve the dynamic of my measurement system?
One possibility that comes to my mind would be to divert a major portion of the diode current from flowing through the feedback resistor once 'high current' condition is sensed. You could:
Sense the diode current after your AC coupling cap (e.g. through a series 50ohm resistor), convert it to a current source, tied to amplifier inverting input, that you turn on only when the 10mA condition occurs. This current source, designed to source 99.8% of the incoming current of 10mA, would allow only 20uA to flow through the 50kohm feedback to produce 1V at amplifier output. So, with the switched current source turned on, once a pre-determined diode current level is reached, the transimpedance gain is dialed down so the amplifier output won't saturate.
There may be challenges;
1. Not loading the summing junction
2. Not causing instability with this current steering scheme (need a fast current source?)
If speed is not very important in "high current" mode, one could use an analg switch across the feedback resistor to reduce RF value when turned on to reduce gain. A series resistor to inverting input could keep the analog switch capacitance isolated. Analog switch in off state could be in low pF value range.
Hope this helps.
The approach that you are showing with the T-network by itself is certainly an option. Note that T-network will add noise in your system as they are inherently higher noise solution than a straight transimpedance amplifier solution.
I do not quite understand how you intended to use the resistance divider, but 100kohm is certainly too high and will add even more noise than the T-network.
A lower noise alternative to the T-network you are suggesting is to have 2 transimpedance amplifiers connected to a high speed multiplexer such as the OPA4872 (4:1 MUX). This way you always have access to a transimpedance amplifier that is not saturating. The noise of the MUX will not dominate the system as the noise will be dominated by the transimpedance amplifier.
Each transimpedance stage can use a different amplifier to maximize the dynamic range for each stage. In most cases the noise will be dominated by the noise gain peaking happening due the the compensation capacitance and the source capacitance. As the gain is reduce, an amplifier having a lower GBWP as opposed to decompensated amplifier should have better performance. Also some gain can be added in the last OPA4872 stage, allowing you to reduce the transimpedance gain, reducing your noise, improving your dynamic range and increasing your bandwidth some.
For unity gain stable low noise amplifier, the OPA842 is a good choice,
For gain >3V/V, the OPA843.
For G>7V/V, the OPA846.
and for gain greater than 12V/V, the OPA847 you are using.
Note that I am not recommending FET input amplifier for your application as you are targeting low transimpedance gain and bipolar input will have better overall noise.
In reply to Xavier Ramus:
Thank you both for your ideas.
Here are the results of my investigations :
- About my system with a tee feedback : I don't see an increase in the noise level. Please look at the attached model, with in one case a 60kOhm resistor and in the other case a 10k+100/500 tee. I get a similar (1.75m and 1.77mV) total noise level and the same (95dBOhm) transimpedance resistor.
Here is the model : 0882.tr12.zip
- About my foot 100k resistor : Here is a picture of a simulation (with a 5k series resistor in the -input of the op amp)
with a varying foot resistor. By making this resistance variable (various resistances with optocoupler or fet resistance) i can make a variable transimpedance. The problem is that the 5K series resistor lower my overall transimpedance gain and decrease my SNR : Without the 5K series resistor, I have 95dBOhm of transimpedance and 1.75mV of total output noise)With the 5K series resistor, I have 75dBOhm of transimpedance and 1.35mV of total output noise).
If I accept the degradation of SNR, I then can change the foot resistance to have a variable gain transimpedance.
I should make many simulations with current source or 4:1 MUX (possible still with the photodiode current source?).
Thanks for your help and do not hesitate to comment
(i'll be unavailable until Thursday, please keep my topic open ;-)
In reply to Piwi Bel:
I have made differents trials on this setup :4477.tr16.zip
My idea is to split the transimpedance resistor with different values in series and senses the voltage using a voltage follower. Afterwards i'll select the right channel using a 4:1 multiplexer (OPA4872 i guess).
I have the followings problems with my model :(see attached file) : - This system doesn't work if I remove U5 or U6. I really don't understand what's going on, because these two opamps are of the same use. I reached this schematics by trials/errors- The noise performance (from 1K to 1G) is really amazing : 182uV of total output noise with 71.9dB of transimpedance. With a simple transimpedance model of 4K (with 100f in parallel) i also get 72dB of transimpedance gain but with a total output noise of 500uV.
Therefore, i guess that something is wrong. Why can't I remove U5 or U6? Why is the noise performance (while keeping U5 and U6) much better that a very simple 4k transimpedance? Why is the noise performance much worse when I delete them?
Please give me feedback ;-)
thanks for your help
If I interpret your question correctly about the difference in noise between a simple 4k feedback resistor and the T-network (R21, R27, R24) of similar Transimpedance gain, I'd say that the reason might be that the simple feedback results in a much higher bandwidth (110MHz simulated compared with 20MHz with the T-network ) and hence the higher noise. The loop gain is very different between the two cases.
About the effect of U5, and U5 on noise: The U5 and U6 modeled input capacitance might be a strong factor in the simulated response at VF1. You could use an R in series with U5, U6 input, for isolation, and see if that removes the variance on VF1 noise. Frequency response peaking could especially cause large noise readings. Also watch out for the bias current of the bipolar amplifiers you have on the summing junction (U2, U5, and U6)- all bias currents need to flow through R21 and may cause a significant shift in operating point at VF1.
Hope this makes sense.
In reply to Hooman Hashemi:
My mistake, yes, I forgot to check the required bandwidth.
I try to use OPA4872 but I have troubles in splitting the current of the photodiode between the differents inputs. I tried with mirror currents without any success.
Can anyone send me a schematics that would work using this 4:1 MUX? That would be wonderful.
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