Recent content by Shikhar

It was mentioned in the Application note of Photodiode that such a configuration helps in minimizing the parallel stray capacitance of the feedback circuit. I don't have any idea why is it so?

THanks to all...finally i have come up with the circuit... Please have a look and let me know of any errors or improvements... ...and for all those who think that i have bad intentions in my mind...i would like to point out that I am a final yr engg student doing my 2 months industrial...

Don't think it in the other way man...actually I am doing my training (a part of our course curriculum) in a defence organisation..we are working on the impact of explosions on a building for which I am using various sensors. Now, in order to trigger my circuitry, i wish to use the photodiode...

sorry to all..actually i wanted to write 20ohms not 20Kohms... o:)

thanks to all there...im really developing a good understanding of my circuit! actually the 5mV signal is from a photodiode which i have used to detect the light during a bomb explosion. When the explosion takes place, the photodiode gives a current which when passed through a small...

its all right sophiecentaur! :smile: but i think u misinterpreted my question...i still haven't started working on the hardware... jst designing the circuit right now! ...u told in your earlier post that i should amplify the signal by a factor of 100 before passing it through the schmitt...so...

i can't understand why the schmitt won't work... if i assume that initially the output is at +V(sat)...the triggering point would be at +V(sat)*R1/(R1+R2). So as u said if R2=100Kohm nd R1=20 Kohm, then triggering voltage would be around 1mv...this is lower than my signal...so why won't the...

The signal Vin is initially open circuited. I want an output of 5V when the signal is applied(Vin would be the range of 5mv)

so how would i know what would be the output at starting? actually, i want the comparator to give a high signal only when the input signal (which is always positive but of very small magnitude) is applied. Before the application of input signal, the output must be low.. can you please suggest...

I have a very basic doubt in schmmit triggers...suppose if I connect the required power supplies to the op-amp (ie. V+ and V-) and make the desired connections(see attached circuit). Now, what would be the output when Vin=0, ie. before the signal is applied.

no current will flow through the circuit as the voltage drop across the reverse biased diode and the 9V battery will be equal to 17V DC source

ohkk...i get the clue! on reverse bias, width of depletion region increases, so capacitance decreases as C is inversely proportional to 'd'. NOw as time constant = RC, so time constant decreases ie. faster operation!

Quiz question -- why should you reverse bias the photodiode instead of running it with zero reversse bias? It's an important concept related to the speed... thanks Berkeman! hmm..i think...it's because photodiodes work on the principle of generation of electron-hole pairs when a light of...

Hi there... i'm working on my college project and wanted to design a high speed photosensor. The intensity of light source is quite high (actually a bomb explosion), so noise signals are not a problem. The photosensor is required to trigger other sensing devices. So which type of photosensor...

thanx every one for ur posts...i really got my answer...