Bit of electrical help please

[AnthonyL]

I'm involved in a small project which needs to respond to the output from a photodiode.

A Maplin L67BQ power adapter provides the 9v needed to power a LED which then reflects via a moving mirror to activated the photodiode, which in turn provides 9v.

This works fine but the signal has to be taken over about 30m and into a computer adapter which should detect the voltage but doesn't.

On checking, the Maplin adapter appears to be isolated from mains earth and neutral. So I can register the 9v between signal and the adapter neutral but not between signal and mains neutral.

I assume the computer is checking the voltage with reference to it's own neutral, which would be power supply neutral or signal ground.

How do I get round this?

Thanks

[The Natural Philosopher]

I think understanding electronics would be a start.

I do, and none of the above makes sense..

1/. What has the Maplin PSU got to do with the photdiode output whih is entirely independent and separate.?

2/. What is it 'in(to) the computer' that measures voltage? Nothing in my computer measures voltage apart from a few embedded sensors, and they don't get to the outside world.

[Paul Herber]

You'll only get a maximum of 0.6V or so from a photodiode, not 9V. This 0.6V is relative to the diode's anode and cathode, nothing else.

[John Williamson]

Are you using two wire cores to take the signal from the LED/ photodiode module to the computer or one? As described, your LED/ photodiode system may be floating with respect to the computer input, so will not register a voltage.

If you're not using two cores for the interconnect, then you should be. You can't rely on using mains neutral as a return in any system of this sort.

When you put a meter across the input to the computer module at the computer end of the cable, do you get a clean 9 volt reading?

It is also possible, but unlikely, that you're picking up mains interference which is causing the fault. You may need to use shielded cable, depending on the circuit impedances.

[The Natural Philosopher]

+1.

Need an opamp to boost that a bit.LM724 is good.

Or some of them can be used reverse biased FROM a 9v supply with a resistor in series..when illuminated they leak like a wimpey home.And the voltage drops..

[Baz]

Hi Ant. What Computer Adaptor are you using and do you have the drivers for it? Baz.

[Jules Richardson]

Just to be clear, you have a power supply, LED, mirror, and *something* (I suspect not a photodiode - or at least not in isolation) which is putting out +9V with respect to the power supply's ground when the light falls on it?

What is this computer adapter?

If you're feeding +9V into an old PC parallel port, you'll probably break something (they're only designed for TTL levels).

If you're using one of the lines on a serial port, that'll probably work (although some PCs can be a bit picky, so I'm not sure that +9V is guaranteed to work despite what the spec might say).

But "computer adapter" could mean all manner of things...

That confuses me. Typically these "wall wart" supplies are plastic-cased and have no earth, true - but they'll still be connected to live and neutral on the AC side; they won't work otherwise.

You're not running a single wire back from the detector to the computer, are you? You need to connect ground on the computer to ground on the detector (and forget about anything to do with "neutral" or the mains AC side of either the computer or the Maplin supply)

30m is quite a distance, too; using good quality shielded cable is probably a must...

cheers

Jules

[Paul D Smith]

Sorry if I'm being dumb here, but are you only providing a single wire to the PC and expecting the PC to measure this relative to some floating datum?

[Martin Brown]

The LED lifetime on 9v will be exceedingly short unless there is a current limiting resistor in series with it - depending on the colour an LED will drop anything between 2v (red) to 4v (blue & white).

This seems unlikely. The signal might be a few volts at negligible current depending on how you wire it up with a series resistance and how well the detector is screened from ambient light and filtered. It will not travel well down a long cable though...

Test it with a shorter cable. What you say doesn't make a lot of sense. you seem to be complaining that the Maplin PSU is floating wrt mains earth (which seems quite likely). You just have to run a pair of cables or better a screened signal cable with earth on the screen to the PC.

You probably need to buffer the photodiode with an opamp so that it can drive the cable to the PC with suitable signal levels and frequency response. How are you measuring it at the PC end? Line in ?

You would probably be better off asking on sci.electronics.basics

Daqarta will give you a quick realtime display of the audio in channels.

[John Rumm]

9V is well in spec for RS232C which is what a traditional PC serial port works to, however it only reads digital levels, and does not perform analogue to digital conversion as such. Fine if all you need is an illuminated, not illuminated indication - not much use for an indication of the level of illumination.

(a standard joystick port can do A to D if you scale the voltage a bit first)

[Andy Burns]

When was the last time you saw a PC with a joystick port?

[dennis]

Buy one of these

and modify as required.

[John Rumm]

about a second ago...

(you can also do analogue input on the line in - but even more restrictive on range)

[Andy Burns]

Yeah, I was half-expecting that, but other than one you own?

[Robin]

FWIW I can see one too (if I stick my head under the desk).

[cynic]

Are you explaining this correctly? Photo-diodes are not generating devices and do not produce a voltage output on their own.

On what do you base this statement?

[Andy Burns]

Don't tell Harry ...

[cynic]

Hmmm I had in mind photo-resistive response devices. I normally think of the devices producing a voltage/current output as photovoltaic. - Point taken!

[dennis]

I have a USB joystick interface somewhere, it was used for a flight yoke.

[Tim Watts]

Did anyone say VDR (voltage dependent resistor)?