1. Home
  2. Hobby Electronics Basics
  3. Can someone explain this (simple) schematic

Can someone explain this (simple) schematic

Hi,

This is a schematic I've been given for a laser detector circuit. The laser is being pulsed at about 600Hz and the 'PD' (photodiode) is supposed to detect reflected light from this laser:

formatting link

The last stage is a comparator - but not well labeled. I think that the inverting input is taking a copy of the TTL 600Hz clock pulse that is driving the laser, and using it to flip the transistor.

The diode is probably an LED (?) and this represents the output. I think that the LED should be connected to the -5v rail.

Does this make sense? Or have I missed something.

The use of -5v is confusing, why not just +5v? Or is this so that the signal being compared is inverted to the clock pulse...?

This circuit came from a french designer, and I don't get to talk to him often.

Reply to
Kasterborus
Loading thread data ...

From the right bottom:

  • rail (not -5V as you said), 1k resistor, LED, transistor, 1k resistor, comparator or op-amp output.

The negative input of the op-amp or comparator goes to a 10k potentiometer connected between + rail and ground, with a 0.1-uF capacitor from wiper to ground.

At least those are my best guesses of the unspecified part.

Reply to
mc

It's an op-amp thing. You can do the same thing with a single rail supply, but you have to bias the + inputs of the op-amps to half the supply voltage. This is done so that the output idles at a "virtual" ground and can carry an A/C signal. By my estimation, the first two op-amp stages act as high gain amplifiers with a high frequency roll-off characteristic.

The final stage is likely a comparator used to switch the LED on and off at whatever rate the the PD is being modulated. The LED is connected to the

+5V rail since most common comparators have an open collector output and can't source any current. The comparator has one input biased at some positive voltage (with respect to circuit ground) that is adjustable. The rectangular blocks are resistors.

Of course IANEE and I could have this all wrong, so it's worth everything you paid for it. :-)

Reply to
Anthony Fremont

Yes, you're correct, it should be + also be sure to use an OP-AMP and not a comparator since many only have open collector on their output.

--
"I\'m never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
Reply to
Jamie

If they did that then they would also have to reverse the polarity of the photodiode, which needs to be reversed-biased.

With the photodiode in the orientation shown, it does need to be -5V in the upper-left of the diagram.

Mark

Reply to
redbelly

After rereading the OP, it's unclear to me which diode he and you are referring to. Allow me to clarify my response:

The photodiode (PD in the circuit) should be biased with -5V, as shown in the circuit.

The LED (diode at bottom of diagram) should be connected (through the resistor) to a positive voltage. Presumably +5V, though it's not explicitly given in the diagram.

Mark

Reply to
redbelly

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.