Recent content by bob987

So this is the circuit. the 100k resistors in the photodiode circuit are necessary to produce a .1V change when a light shines through the finger onto the diode. I want my v3 to simply be the difference between the output of the photodiode circuit minus the 4.5v bias. So ideally it would be...

I don't have a scanner near me atm. But for my photodiode circuit there is a voltage divider, the 4.5 from that divider feeds into the - terminal of the opamp. from the + terminal there is a resistor going to the output of the opamp, and the photodiode's - terminal is attatched to the + feed...

Sorry, I was just picking two numbers that were .1v apart. i should have said 4.5 and 4.6. I have actually been running the voltage subtractor at the moment off a DC supply, but just using the voltages i would normally see, since i didn't want to drain my battery and it allowed me to change...

Could I increase the resistors for the voltage divider to reduce that amps running through it? or would that throw off the circuit? Also, what if i used a voltage regulator before the voltage divider, and put a voltage follower after it leading into the voltage divider, would that help to keep...

I don't quite get how this works. You say it will give me 5.5 V out? how is 4.5-4.6 * 10=5.5 shouldn't i be seeing an output voltage of 1V? or is the 4.5V subtracted from that 5 since it is inverted?

i'm also using a 9V battery as a source, so putting in a +- 5 volts isn't doable.

what I am saying is if i put 1.7 into v1 and 1.8 into v2 i get an output of one and some change or 2 and some change. not the .1V difference it should be. I can't figure out why it stops at some random voltage and won't find the difference of the input. I am aware the gain is 1, which is why...

Im using the set up on this page http://www.electronics-tutorials.ws/opamp/opamp_5.html with all the resistors set = to each other. I have tried 100, 500, 1K 10K but i will always get a diference of 1.x v or 2.x v and if i check the voltages going into the v1 and v2 the difference is only...

I am building a pulse oximeter, and there is a photo diode that is reversed biased by 4.5 volts. When there is a light shining on the diode the voltage changes between .1-.3V. I need to cubtract off the 4.5 volt bias so the change can be multiplied and analyzed. I tried using a basic...

hoever, I only have a single 9v dc source. wouldn't I need negative voltages from the oscillator to switch between the diodes?

I found the following circuit http://www.555-timer-circuits.com/bi-polar-led-driver.html that suits the task I am trying to achieve. My only issue is I don't know what the frequency would be or how to change it. Is the frequency determined by the cpacitor and the 33K resistor or is it more...

Thanks for your reply. So I found this circuit setup that seems to do what I am looking for. here is the link. http://www.555-timer-circuits.com/bi-polar-led-driver.html It uses the LM555 timer, the only thing I'm not sur of is what the freuency of the switching light would be. Would that...

I don't know a lot about PLD's so hopefully someone here can help me. Is there a relatively cheap PLD that can switch between 2 inputs every given amount of time, like every second or every 5 seconds? I'm just trying to find what I am looking for and can look up and read through the data...

I am trying to take two lights and usign a 9v battery switch between the LED's. My thought was if i could get a square wave to go from positive to negative, since the LED's are polarity sensative I could just put them in parallel with the positive of one and the negative of the other being fed...

Is it possible to use a 9v battery and run it through some circuit or system to get a squarewave output that has a positive and negative voltage. Ideally the two voltages would be the same but the negative of the other.