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The attachment is the design of a photodiode preamplifier.The signal generated by the photodiode I=I(dc)+I(ac), which means DC current and AC current respectively. However, only the ac part is needed. So a capacitor is used to block the DC current from going through the opamp.
Can I design in this way? If so, the path for DC cannot be provided actually.If not, how can I get rid of the dc current? Thanks in advance.
The ac signal needed is 1kHz-2kHz sine wave. The total power received by the photo-diode ranges from -10dBm to -20dBm. The ac part, which is the part needed, is only 0.024%-5% of the total power. That is the reason why I want to block the dc current.
Thanks in advance
Capacitive coupling in this way will cause C1 to charge to a DC voltage that will forward bias the photodiode. It would not produce your desired result. It would be better to perform the AC coupling after the transimpedance amplifier. Are you able to AC couple the signal after the transimpedance amplifier or is the DC component too large?
In reply to Bruce Trump:
Dear Bruce, are you suggesting a way like the one above?
I was wondering if it will introduce more noise since two opamps are used.
Furthermore, I need to change the sine wave into all positive, which means V=Vdc+Vac. It enables the ADC to work properly, since the ADC I am using cannot detect negative voltage. As a result, another opamp (maybe instrumentational amplifier) is needed to add a DC voltage to the output.. If just like the first design, the dc part could be added at the only opamp used.
If the first plan is used, can I just add an reverse bias voltage to the photodiode to ensure that the photodiode is not forward biased?
Thank you Bruce.
In reply to Richie Chen:
This approach will introduce more noise. This is because the first amplifier must be reduced in gain so that it does not saturate from the DC level of light. Noise performance of a TIA is best when all the gain is achieved in the TIA stage. Still, this approach may work for you. The question is whether more noise is too much noise.
It's not clear exactly what you want to do with the resulting AC signal. Do you want to accurately digitize the shape of the AC component? If so, then the second amplifier could be offset with a DC voltage applied to the non-inverting input of the second amplifier.
Thank you very much!
1.For the first plan: Can I replace the end of the photodiode from ground to a voltage? It may cancel the voltage generated by capacitor and make the photodiode reverse biased.
2.If a voltage(Vdc) is added at the non-inverting end of the second opamp, the output will be:
Vout=-(R16/R17)*Vin+Vdc*(1+R16/R17); am I correct? It seems that the offset is amplified more than the signal. If the offset voltage is not quite stable, maybe more noise will be
introduced. What do you think?
Yes, I hope to digitalize the exact shape of the signal.
Thank you very much.
By the way, for the first plan, it seems that the current will flow from c1->photodiod->ground.
It means that negative charge will accumulate at the capacitor. So the voltage at ground will be higher than the voltage at capacitor, the photodiode is in reverse-biased actually...
Am I wrong somewhere?
Thanks a lot.
The capacitive coupling will cause the photodiode to forward bias, even if you connect a positive voltage bias on the cathode. Consider a simple model of the photodiode as a current source in parallel with a diode. The capacitor cannot conduct DC current. Thus the continuous DC current produced by the photodiode will flow through the forward bias direction of the diode. Think of this current as flowing in a circle, from the current source, through the diode and back to the current source. This causes the photodiode to forward bias and enter the photovoltaic region of operation.
When the capacitor is removed, the op amp forces a constant voltage across the photodiode. Photo current then is forced to flow into the summing junction of the op amp and through the feedback resistor.
Regarding offsetting the second amplifier: The AC gain is as you describe but the DC gain at the non-inverting input is unity (due to the coupling capacitors on the input). Thus the voltage you apply to this input should be equal to the desired output offset voltage. This could be accomplished, for example, with a voltage divider from the reference voltage of the a/d converter. A capacitor on this divided-down voltage can reduce the noise.
Sorry I dont quite understand what do you mean.
For the first: what is the direction of the current source? Is it from anode to cathode? If so, the current will flow from cathode to anode as reverse-biased.
For the offset voltage: thank you very much! You mean the dc current is blocked!
Your help is really appreciated.
Shown below is another possible implementation for your circuit. It uses an integrator, U2, to control a transistor current sink. This balances the DC component of light current from the photodiode with the current in the collector of T1. The average output voltage at Vout is equal to the voltage applied to the non-inverting input of U2, 2.5V. This control loop allows the gain of the TIA to be set to a high value without being overloaded by the constant DC light level.
The diode and current source that model the photodiode are shown. Note the polarity of the diode and light-generated current. If the photodiode were AC coupled with a capacitor as in your circuit, the DC current would flow counter-clockwise in loop formed by IG1 and the diode. This would forward bias the diode.
If the photodiode is forward biased, what will happen?
Furthermore, what is the problem of operating in photovoltaic mode?
I connected the cathode to ground actually so it was in photovoltaic mode..
Is photovoltaic less noisy? The signal needed in this system does not need to have large bandwidth; but high precision of the wave form shape is needed.
The response of the photodiode becomes highly nonlinear in photovoltaic mode. It also becomes slower because capacitance increases. I don't know about the noise properties in this mode.
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