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# Problem switching a 5V mini relay from a low power LED (1,2v 2 mA) input

D

#### [email protected]

Jan 1, 1970
0
I am trying to make a 5V mini relay switch when the HDMI indicator LED
in my DVD player lights up.

I have unsuccesfully tried to trigger the mini relay (5V, 40mA) via an
optocoupler (4N33 - infrared/darlington) "piggybagged" on the
indicator LED (1,2V 2mA), but the relay doesn't switch, only "hums"
slightly. Supplying an input of 1,5V from before the indicator LED
resistor (220ohm) generates the same result. I even tried using the
(normally unused) basis of the optocoupler as input, hoping it was
more "sensitive" - but still only "humming"

The setup works when "piggybacking" a standard 3V LED (>10mA), where
the relay swithes fine. What have I overlooked ?

Any suggestions for (simple) curcuitry I could use to trigger the 5V
relay from the 1,5V 2 mA input (I only have 5V supply power
available) ?

J

#### Jamie

Jan 1, 1970
0
I am trying to make a 5V mini relay switch when the HDMI indicator LED
in my DVD player lights up.

I have unsuccesfully tried to trigger the mini relay (5V, 40mA) via an
optocoupler (4N33 - infrared/darlington) "piggybagged" on the
indicator LED (1,2V 2mA), but the relay doesn't switch, only "hums"
slightly. Supplying an input of 1,5V from before the indicator LED
resistor (220ohm) generates the same result. I even tried using the
(normally unused) basis of the optocoupler as input, hoping it was
more "sensitive" - but still only "humming"

The setup works when "piggybacking" a standard 3V LED (>10mA), where
the relay swithes fine. What have I overlooked ?

Any suggestions for (simple) curcuitry I could use to trigger the 5V
relay from the 1,5V 2 mA input (I only have 5V supply power
available) ?
the opto isn't coming in because 1.2 volts is most likely not enough to
fully.
the LED in the unit is clamping down at a lower voltage.
Disconnect the LED in side and use the same leds to drive
the LED in the opto with no resistor.
if you still need the LED as an indicator, connect the original
LED to the COIL voltage of the relay via a resistor to protect it.

If you don't want to do it that way.
follow the lead wires of the internal LED back on the board to
where there should be a Resistor that limits the current to that
LED when on, connect your option with a Resistor on that side of the
internal resistor..
P.S.
It's possible the internal driver for the LED may not be able to
drive 2 LED's with out damage.
so my first suggestion maybe better, use it's own driver circuit to
drive the opto LED and rig up an indicator from the coil voltage..
also, you need a reverse diode on the coil so that you don't damage the
opto unless it already has one internally.
Also make sure you have sufficient voltage to operate the relay.

D

#### [email protected]

Jan 1, 1970
0
Thank you for your reply.

I think I already tried your second suggestion without success -
"Supplying an input of 1,5V from before the indicator LED
resistor (220ohm) generates the same result."

I followed the lead wires of the internal LED back on the board to
where there was a Resistor (220 ohm) that limits the current to that
LED when on, and connected the optocoupler there. Unfortunately the
result was still no switchin, which unfortunately also indicates to
me, that your first suggestion might also fail.

I think I need another alternative.

C

#### CheapscateDave

Jan 1, 1970
0
I am trying to make a 5V mini relay switch when the HDMI indicator LED
in my DVD player lights up.

I have unsuccesfully tried to trigger the mini relay (5V, 40mA) via an
optocoupler (4N33 - infrared/darlington) "piggybagged" on the
indicator LED (1,2V 2mA), but the relay doesn't switch, only "hums"
slightly. Supplying an input of 1,5V from before the indicator LED
resistor (220ohm) generates the same result. I even tried using the
(normally unused) basis of the optocoupler as input, hoping it was
more "sensitive" - but still only "humming"

The setup works when "piggybacking" a standard 3V LED (>10mA), where
the relay swithes fine. What have I overlooked ?

Any suggestions for (simple) curcuitry I could use to trigger the 5V
relay from the 1,5V 2 mA input (I only have 5V supply power
available) ?
It could be that the LED is being pulsed and is not a steady on
condition. You would need a scopr of DVM with frequency input to
verify it.
Dave

C

#### Chris

Jan 1, 1970
0
I am trying to make a 5V mini relay switch when the HDMI indicator LED
in my DVD player lights up.

I have unsuccesfully tried to trigger the mini relay (5V, 40mA) via an
optocoupler (4N33 - infrared/darlington) "piggybagged" on the
indicator LED (1,2V 2mA), but the relay doesn't switch, only "hums"
slightly. Supplying an input of 1,5V from before the indicator LED
resistor (220ohm) generates the same result. I even tried using the
(normally unused) basis of the optocoupler as input, hoping it was
more "sensitive" - but still only "humming"

The setup works when "piggybacking" a standard 3V LED (>10mA), where
the relay swithes fine. What have I overlooked ?

Any suggestions for (simple) curcuitry I could use to trigger the 5V
relay from the 1,5V 2 mA input (I only have 5V supply power
available) ?

You're almost there. The problem can be seen in the datasheet by
looking at the "Current Transfer Ratio":

http://www.fairchildsemi.com/ds/4N/4N33.pdf

In the middle of page 3, it says I(CTR), or the DC current transfer
ratio, is a minimum of 500%. That means you can't expect more than 5
times the output current from the photodarlington than you're putting
in at the optoLED input, and your 4N33 photodarlington won't be
saturated at that current (which you'll need for driving a relay).

Use a second generic NPN transistor like the 2N3904 to bump up the
gain and give you a clean switch. View this in fixed font or
cut&paste to M$Notepad: | | VCC VCC VCC | + + + | 4N33 | | | | 1 | 5 | | | -----. .----o | | 1N4001 | | | | C| - | | ~ |/ | RY1 C| ^ | V ~ .-| | C| | | - | |> | | | | | | | |/ | | | 2 | | '--| o------' | -----' | |> | | 6| | | | | |4 ___ |/ | | o---|___|---| 2N3904 | .-. | 2.2K |> | 220K| | .-. | | | | 10K| | | | '-' | | | | | '-' | | | | | | === === === | GND GND GND (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Since your 4N33 only has to switch 2mA for the transistor base, you shouldn't have any problems. Good luck Chris J #### Jamie Jan 1, 1970 0 Thank you for your reply. I think I already tried your second suggestion without success - "Supplying an input of 1,5V from before the indicator LED resistor (220ohm) generates the same result." I followed the lead wires of the internal LED back on the board to where there was a Resistor (220 ohm) that limits the current to that LED when on, and connected the optocoupler there. Unfortunately the result was still no switchin, which unfortunately also indicates to me, that your first suggestion might also fail. I think I need another alternative. Did you disconnect the internal LED and connect the Opto to that same line? You must disconnect the internal LED, because it clamps down the voltage. that is where you're getting the 1.5 volts from./ if you disconnect that LED then test the voltage when it's on, you'll find it much higher than 1.5 V most likely at least near what ever supply voltage you have in the unit. You can also place an milliamp meter in series with the OPTO LED to test it's current. You should be getting around 15 ma+ at least. If this is all passing.. Then I would suggest that the output in the opto does not bias low enough to drive the relay, you would most likely need a help transistor.. P #### Paul E. Schoen Jan 1, 1970 0 Chris said: You're almost there. The problem can be seen in the datasheet by looking at the "Current Transfer Ratio": http://www.fairchildsemi.com/ds/4N/4N33.pdf In the middle of page 3, it says I(CTR), or the DC current transfer ratio, is a minimum of 500%. That means you can't expect more than 5 times the output current from the photodarlington than you're putting in at the optoLED input, and your 4N33 photodarlington won't be saturated at that current (which you'll need for driving a relay). Use a second generic NPN transistor like the 2N3904 to bump up the gain and give you a clean switch. View this in fixed font or cut&paste to M$ Notepad:

If you don't need isolation, you can use the simple circuit below, changed
from what was suggested. It should work if the cathode of the LED is at
ground. You might also need about a 10k resistor from the 2N3904 base to
GND to make sure it turns off. Change the 2.2k to 1k if it does not turn on
well enough.
|
| VCC VCC
| + +
| | |
| | |
| -----+---+ | | 1N4001
| | | C| -
| | ~ | RY1 C| ^
| V ~ | C| |
| - | | |
| | | | |
| | | o------'
| | | |
| | | |
| | | ___ |/
| | +------------|___|---| 2N3904
| | 2.2K |>
| | |
| === |
| GND |
| |
| |
| ===
| GND

Paul

D

#### [email protected]

Jan 1, 1970
0
Any easy way to determine if the LED is being "pulsed", before trying
something else ?

Ole

R

#### Rich Grise

Jan 1, 1970
0
Thank you for your reply.

I think I already tried your second suggestion without success -
"Supplying an input of 1,5V from before the indicator LED
resistor (220ohm) generates the same result."

I followed the lead wires of the internal LED back on the board to
where there was a Resistor (220 ohm) that limits the current to that
LED when on, and connected the optocoupler there. Unfortunately the
result was still no switchin, which unfortunately also indicates to
me, that your first suggestion might also fail.

I think I need another alternative.

If you didn't use a series resistor with your opto, you probably blew
its LED.

Either run another 220R resistor from the same spot you mentioned
(switched Vcc), to the anode (probably) of the opto's LED, so there are
two LED/R series sets running in parallel, or to save \$0.003 on a
resistor, unsolder the 220 from the LED and interpose the opto's LED in
series, watching the polarity.

If the opto has a good enough transfer ratio, that should do it.

Another way which doesn't need an opto (i.e., if the grounds are in
common), then take a 1K or so from the LED to the base of any NPN
general-purpose transistor - if you only need 50 mA for your relay,
almost any type should work. Or, better yet, connect the 1K to the
high side of the 220, to be sure to saturate your switch.

Have Fun!
Rich

R

#### Rich Grise

Jan 1, 1970
0
Any easy way to determine if the LED is being "pulsed", before trying
something else ?

Switched + to LED/R o----||-----+---->|---o DC meter
|
_
^
|
GND

That's any ol' cap, and 2x signal-type diodes - I suppose 1n400x would
work. If you use a polar cap, positive goes to the right.

Cheers!
Rich

D

#### [email protected]

Jan 1, 1970
0
Indicator LED was indeed pulsed (which also explains the low power
mesurements)
100 uF capacitor in parallel with relay solved problem

Thanks all,
Ole

D
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