Hi,
I am trying to detect the DLP white Pulse through photodiode (#PD438C/S46) but DLP signals are not detected when ambient light is ON.
Here is the circuit that i am using
Here is graph when the input data is in black/green colour with some 3D content.
We got continuous 5V signal when the ambient light is ON so can anyone tell us how to detect the DLP signal and remove the unwanted noise?
Thanks,
Yamini
Yamini,
I suspect that the problem is that either R1 feedback resistor value is too high for the maximum photodiode current (what is Iimax? it should NOT be higher than 2.15uA - see below) and/or the actual current pulse frequency in your application is above the circuit bandwidth, fp = 1/(6.28*2e6*2.7e-12) = 29,488Hz you created by using 2M||2.7pF feedback network (what is fp?)
Please determine max R1 based of Iimax and than use it to determine maximum C1 based on the frequency of current pulses, fp - see below.
You do NOT need to recalculate 3) above because 18MHz GBW of OPA381 should be more than enough to meet the requirements of your application.
As far as stability is concern, in its current form the circuit is very stable with almost no overshoot - see below.
Hi Marek,
The value of maximum photo-diode current is not given in datasheet so we have taken the typical value (PD438C/S46 - 18uA) of it and the pulse width of the DLP signal is around 50µs(which may vary with respect to projector). we recalculated the Value of Resistor and capacitor. After changing the components, we are not able to detect the DLP signal in both ambient light ON/OFF.
We got the DLP signal in previous circuit when the ambient light is off.
Here are the results which we got:
We are Getting DLP signal ONLY when we place the photo diode circuit Infront of the projector but we want to detect those signal even when the photo-diode is far from the projector.
Can you please suggest, what changes are required to get the DLP signal when the the photo-diode is far away from the projector in both condition (ambient light off/on).
Thank You,
Yamini
Your calculation of R1 based on typical photodiode current of 18uA will result in the output railing around 4.4V for any higher photodiode currents level.
Other than that, your calculations are correct. DC sweep shows the output voltage range from 100mV to 4.24V for 0<IG1<18uA - see below.
-3dB bandwidth is around 20kHz (Gain=0.707 from DC value), thus to avoid much attenuation the input signal would need to be below 10kHz.
Pulsing the output with a current source shows a very stable circuit with no overshoot - see below.
Applying 10kHz input current source (50us pulse width) shows proper operation of the circuit - see below.
Thus, I believe the photodiode current pulse must be much narrower than 50us - it must be much shorter and for the reason the circuit is too slow.
Looking at the below scope picture makes me think the actual pulse has much higher frequency component, which would be attenuated or outright filtered out by the circuit with corner frequency around 20kHz.
See what happens when the photodiode pulses width is 5us - see below
All in all, I believe you must increase the bandwidth of the circuit by lowering the value of C1 by at least 10x - see below.
