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Infrared detector response time

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QuickStrike

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The figure below shows a simple infrared detector circuit.

The response time for a photodiode (T=0.35/2.19RC) is responsible for the fast signal response.

As u can see the formula, if resistance or capacitance or both are made or chosen to be small then the response time will be small (i.e. a fast response time).

Is there any way to reduce the capacitance of the photodiode besides biasing it (when a voltage is applied to the photodiode, capacitance is reduced). i.e. is there some component that I can connect to the circuit shown below to reduce the capacitance, to acquire a fast response time? Maybe connect a capacitor somewhere!!
 

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I'll take a look at that link, thanks

But problem is that when u reduce the resistance then the voltage across R1 will get small.

1) How much current can a photodiode take?...I couldn't find it in the datasheet

2) At 200 ohm the voltage amplitude is 0.1V, when I increase frequency from 0-2MHz. At 2Mhz the signal begins to have ripples, why?

One more question.

I want to calculate the current passing through the photodiode, how can that current be calculated?.....look at figure below
 

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The diode should have a power rating in watts. If it is rated 0.2 watts, then at 5 volts the max DC current is: I = P/E = .2/5 = .04 amps but if the duty cycle is 50%, you could use .08 amps.

Use Ohm's Law to find the current: 0.1 volt across 200 ohms is 500 microamps current. You can get more current by using a stronger light source or more sensitive diode.
 
QuickStrike said:
I want to calculate the current passing through the photodiode, how can that current be calculated?.....look at figure below

I assume you mean the steady-state current; For this, assume C1 is fully charged, hence no current flow through C1/R2. Thus calculate using R1 only; Ignore C1/R2.
 
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