Simple Dusk-Dawn switch

Thread Starter

DC_Kid

Joined Feb 25, 2008
796
if you wish to negate a current vector, the negation will cause the actual vector to flip 180°
thats like marking everything with contrapositive symbols, all the arrows point right but the neg signs mean the actual flow is in opposite direction.

the physical "off" and "on" for a common collector BJT is opposite of 393 input logic.

when input logic is True ( + > -) we get a physical BJT off, no sink. with the series resistors 10k/100k/10k this should leave gate at about 11v
when input logic is False (+ < -) we get a physical BJT on, sink. this pegs fet gate to gnd.

am i wrong, please verify.

sorry to ian, i had swapped out that drive stage to pfet but you quoted me before i could replace the image.


You are going overboard here. The LM393 can sink 16mA. You're only driving an LED. There is no need for external buffering.
i do not yet know what opto SSR i will use, some are like 20mA, but with that last schematic i posted i needed to flip the output logic because input logic is Day-on Night-off (but "393 on" is bjt off, and "393 off" is bjt on-sink), hence use of pfet.
 
Last edited:

MrChips

Joined Oct 2, 2009
19,774
There are more than one ways to skin a cat.

Here are some ways to flip the logic:

1. Interchange the photo sensor and its bias resistor.
2. Interchange the +INPUT and -INPUT.
3. Put LED in series with the load resistor on the output pin of the LM393.

I would opt for option #3 myself.
 

Thread Starter

DC_Kid

Joined Feb 25, 2008
796
There are more than one ways to skin a cat.

Here are some ways to flip the logic:

1. Interchange the photo sensor and its bias resistor.
2. Interchange the +INPUT and -INPUT.
3. Put LED in series with the load resistor on the output pin of the LM393.

I would opt for option #3 myself.
the 393 can handle sinking some of the opto ssr's i am looking at. but with feedback as-is there would be a 0.1mA fwd bias path to gnd via feedback resistor if VO1 led was connected to 393 output directly?
 

MrChips

Joined Oct 2, 2009
19,774
the 393 can handle sinking some of the opto ssr's i am looking at. but with feedback as-is there would be a 0.1mA fwd bias path to gnd via feedback resistor if VO1 led was connected to 393 output directly?
The point is, when you connect the feedback resistor you alter the DC operating conditions of the circuit. By adjusting the photo-sensor bias resistor you restore the desired DC operating point. When the input threshold is exceeded the positive feedback supplies the desire hysteresis, assuming that you choose the appropriate feedback resistor value.
 

ian field

Joined Oct 27, 2012
6,539
You are going overboard here. The LM393 can sink 16mA. You're only driving an LED. There is no need for external buffering.
The datasheet quotes a minimum spec of 6mA - if you don't design around worst case scenario - the circuit will only work with selected parts.

The IRLEDs in optos tend to need more current - 50mA isn't entirely unusual.
 
Top