Can someone explain this (simple) schematic

[Kasterborus]

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:

http://www.flickr.com/photos/20929480@N03/2037052153

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.

 

[mc]

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.

 

[Anthony Fremont]

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:

http://www.flickr.com/photos/20929480@N03/2037052153

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?

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.

 

[Jamie]

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:

http://www.flickr.com/photos/20929480@N03/2037052153

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.

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.

 

[redbelly]

Yes, you're correct, it should be +

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

 

[redbelly]

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

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