EMRFD Message Archive 10907

Message Date From Subject
10907 2015-03-20 22:51:10 Kirk Kleinschmidt Optoisolator Question
Hi, gang,

I'm trying to make a simple optoisolated CW keying interface so my logging software can key my transmitter. The reference circuit provided (www.n3fjp.com/cwschematic.html) shows a standard-looking optoisolator with the usual output transistor, emitter-collector, to key the downstream circuit, with the base of the output transistor floating.

The optoisolators in my junkbox (Sharp PC900V) have the same 6-pin package, the same LED input pinouts, and an output transistor (emitter-collector) that has the same output configuration, but with slightly different pin numbers.

The big difference is that the PC900V, instead of having pin 6 connect to the base of the output transistor, pin 6 is listed as "VCC," and connects to an internal voltage regulator (and an internal amplifier between the LED and the phototransistor thingy).

This suggests that the PC900V needs some kind of dc supply to function, and that it's probably not a standard optoisolator that switches the output transistor on when the "light is shining."

With no VCC supplied to pin 6, when I energize the input with a weak 9-v battery (through a 1.2k series resistor), which I'm substituting as a test signal for my PC's serial port RTS signal, which I think is +12 V, I get -0.4 V dc on the output transistor, with no indication at all on the resistance scale of the DMM.

My first "cheap and dirty" thought was to simply supply the signaling voltage to the signal input AND the VCC pin. But then I thought that might destroy the isolation I was seeking in the first place.

I haven't yet connected a separate battery as VCC.

Do some optoisolators require VCC to function, or am I doing something wrong?

Building a battery into my keying interface isn't the end of the world, but if I don't need one, or if the input signal itself can also supply VCC (the output only changes state when the input signal is supplied...), I'd leave it out.

Or, I could get a more suitable optoisolator...

Anyone know the answer off the top?


--Kirk, NT0Z

My book, "Stealth Amateur Radio," is now available from
www.stealthamateur.com and on the Amazon Kindle (soon)
10910 2015-03-21 01:12:14 Eamon Egan Re: Optoisolator Question

Hi Kirk.

Looking at the datasheet, this part does indeed need a VCC supply. If your rig has a positive supply on its Mic or utility connector it will probably be suitable (the part will be happy with up to 15 volts). The open collector output should do fine to ground the PTT.

In case there's any doubt, yes, using the signaling voltage to power the output side would indeed defeat the whole purpose! :)


10911 2015-03-21 01:18:59 Eamon Egan Re: Optoisolator Question

Hi again Kirk.

Rereading your message I see you want to key CW, not PTT. My answer may still apply, except it won't be as natural to get the VCC from the rig if you hope to simply interface directly to the CW contact closure circuit.


10912 2015-03-21 01:29:19 Kirk Kleinschmidt Re: Optoisolator Question

Leave it to me to have scrounged some oddball parts... :)


--Kirk, NT0Z

My book, "Stealth Amateur Radio," is now available from
www.stealthamateur.com and on the Amazon Kindle (soon)
10913 2015-03-21 04:24:35 Kirk Kleinschmidt Re: Optoisolator Question
It looks like a battery will suffice until I can find a more standard optoisolator that's "solar powered."

When VCC is 9 V, idle current is about 2 mA, and "switched on" current is 5 mA (with the output transistor driving only my DMM). Don't know if the current will increase when the output transistor is loaded "for real." Probably will...

On-state resistance is about 15 ohms between the collector and emitter of the output transistor, which should be okay for solid-state keying.

Although simple, this exercise was a sweet little victory for my "inner experimenter," which I have been mostly suppressing for some time. I found a simple circuit that performed a useful function. I thought I had the right IC, but discovered that mine was slightly different. The datasheet showed that external power was required. I surmised that a simple battery would suffice, but I asked the group for input before I risked letting the smoke out (and while I watched a movie :). My guestimate affirmed by Eamon, I connected the battery and the circuit functioned. I measured various parameters and noted that they were within reason. And I then reversed the polarity of the input signal to test my protective diode. Everything worked, and it's time to test with the actual RTS signal instead of the 9-V battery and move the pieces and parts to a little ugly-style board and then into service.

The proto board had been packed in a box since 1994! I have built things since then, but never with a proto board :)

If it were only that easy with RF power amplifiers or Arduino code... :)

73, and thanks again for the help,

--Kirk, NT0Z

My book, "Stealth Amateur Radio," is now available from
www.stealthamateur.com and on the Amazon Kindle (soon)
10915 2015-03-21 06:00:32 Ed - K9EW Re: Optoisolator Question
Hi Kirk,

Maybe you're thinking of a 4N25... no Vcc, no base connection required.
You turn on the LED with your RTS signal, and that turns on the
photo-transistor to key your circuit. I've attached a data sheet.

Two things to keep in mind are:

1) The voltage on the RTS line when you're in the 'key-up' condition. It
may be negative, so you'll have to protect the photo-diode in the reverse
direction with a clamping diode.

2) Depending on the current available on your RTS line, you may not be
able to saturate the photo-transistor enough to key your rig.

It may turn out that neither of these are issues, but I just wanted to you
aware of them.

I'll send you a couple to play with.

ed - k9ew

10916 2015-03-21 16:59:49 iam74@rocketmail.... Re: Optoisolator Question
The difference is the Sharp unit has a schmidt trigger inside, so it is a true TTL device. Connect it as per the data sheet with ~300 ohms between the Vcc and the Vout. Take your signal from Vout. It will act pretty much as any isolator, but will have a very sharp transition...and probably will require 5v.

Note that the output signal is the inverse of the input signal to the led.

As K9EW says, 4N35 and the like are more common, but yours will work if you mind the data sheet.