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Isolated pulse detector

S

Sylvain Munaut

Jan 1, 1970
0
Hi,

I have two wire from which can came a pulse. The pulse voltage might be anywhere from 6V DC -> 50V DC (about ...).
I'd like to get from that a nice 3.3V pulse, and be isolated from input.

Wire 1
o
|
---
| |
|R|
|1|
---
|
|-----,
| |
--- |
| | |
|R| |
|2| \_|_
--- ^ \
| / \ Zener
| ---
--- |
\ / |
v |
--- |
| |
| |
|-----'
|
|
o
Wire 2


With the led being a optocoupler input.

Should that work ? Is this common, or a better way ?



Sylvain
 
J

Joerg

Jan 1, 1970
0
Hello Sylvain,

The zener and second resistor is often not necessary. An opto coupler
with a plain diode input characteristic limits the voltage across it by
itself. You might want to include a reverse diode across the opto
coupler input if you suspect that ringing might occur on this line.
That's always a good idea and cheap.

If the coupler has too narrow a current range you may have to replace
the resistor with a current source but that can be easily achieved with
a couple of transistors and resistors.

Regards, Joerg
 
N

Nicholas O. Lindan

Jan 1, 1970
0
Sylvain Munaut said:
3.3 -> 50V
o
|
---
| |
|R|
|1|
---
|
|-----,
| |
--- |
| | |
|R| |
|2| \_|_
--- ^ \
| / \ Zener
| ---
--- |
\ / |
v LED |
--- |
| |
| |
|-----'
|
|
Gnd

ILED = 10mA
Should that work ? Is this common, or a better way ?

It will work.

At 3.3 volts input and Iled min = 10mA the quantity
R1 + R2 = 230 ohms max. where VLED = 1V.

This sets Vz = .01 * (230 - R1) + 1, or Vz ranges from 3.3V to 1V as R1
ranges from 0(not a good idea, BTW) to 230 ohms (also not a good idea).

Minimum power loss will be with the lowest acceptable value for R2. This
requires Vz be close to 1 volt. You will now have to trade off zener
regulation, led voltage variation and led allowable current swing.

Power in the zener will be about .25-.50W and in R2 it will be
10-15 watts at 50v continuous input, depending on the R1 and R2 values
selected.
 
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