Circuit / Electrical Question...

jabaroon

Hi All,

See the circuit diagram below. Im sure some of my terminology is wrong, but please bear with me!

The opto input shown in the diagram is actually a device that detects current (I think its called an Optically Isolated Input), so when it sees a current, its triggered, then it doesn't it isnt.

I have the opto input hooked (indirectly) to a serial port that allows me detect when current is present. So, if I switch on/off the DC power supply, I get the result I expect.

The questions is about the load. Do I need to have something connected in series with the DC+ or - to simulate load to prevent the PSU running a full whack when turned on?

I hope this makes some sort of sense!

Thanks
Jab

fred funk }{

Is that not a relay?

Its not very clear from your post what exactly you're trying to do, tell me what you have connected to it and for what.

Boston

What are you trying to do. Sounds like you're trying to run a PC PSU outside the case. If this is the case, try the modding forum. It shouldn't do anything without a load on it(At least when I've tried it, it mine didn't)

jabaroon

Hi Guys,

Sorry for not being clearer. Basically, I am using one of these:
http://www.electronickits.com/kit/complete/elec/ck1610.htm connected up to the serial port on the PC. When I apply power to the inputs, I can detect the presence of power by sending a query command to the serial port, when in turn returns back a character to indicate whether or not power is present on the input.

I have lots of different ideas in mind for using this device for different types of home automation and gadgets!....but first of all I wanted to make sure that I am not screwing anything up by not having a load on the detection circuit.

At the moment, for testing, I simply connect a 12DV power supply (one of those small universal ones) into the terminals of one of the inputs.

Everything works as I want, however, I am concerned that I have no load on the circuit and am worried that over time with constant use (and the power applied for long periods of time), that I will be running the PSU essentially shorted and therefore at max load?.....hence wondering if I should be putting something in series in the circuit to simulate load?

Thanks,
Jab

Lex Luthor

you only need approx. 0.7V across the base and emitter of the device to trigger it on/off

jabaroon

Lex Luthor wrote:

you only need approx. 0.7V across the base and emitter of the device to trigger it on/off

Ok...how would I achieve this. Given my current setup (eg: where this will be located) I only have access to a 12v DC supply. Whats the simplest way of using this to trigger the input (bearing in mind that the input may need to be in the triggered state indefinatley (eg for days on end!)

Cheers,
Jab

cast_iron

After a quick read of the interface spec, it says that any voltage from 5 to 24 VDC is fine for the isolated inputs. So the unit itself does not need any extra "loads" for anything.

jabaroon

Sorry folks if I am not getting this, but I just want to be 100% sure.

If I wire a 12V power supply (one of those universal ones for example) directly up to the +/- on the input, how much current will the unit draw from the PSU?

The concern I have is that the input is essentially like a short circuit and will draw the max power from the PSU. Hence the question, should I be adding some sort of simulated load into the circuit so as to only be drawing an absolute minimum amount of power from the PSU in order to trigger the input.

Thanks
Jab

cast_iron

jabaroon wrote:

If I wire a 12V power supply (one of those universal ones for example) directly up to the +/- on the input, how much current will the unit draw from the PSU?

It's a voltage source, not a current source - so the circuit will only draw what it was designed for, not a set amount.

It appears fom the circuit diagram (overall one) that the PSU is connected to a regulator, resistors and capacitors and is not in fact a sc. I think you are being misguided by the initial diagram you posted in the OP - it doesn't appear complete.

jabaroon

cast_iron wrote:

It's a voltage source, not a current source - so the circuit will only draw what it was designed for, not a set amount.

It appears fom the circuit diagram (overall one) that the PSU is connected to a regulator, resistors and capacitors and is not in fact a sc. I think you are being misguided by the initial diagram you posted in the OP - it doesn't appear complete.

Thanks a mill for the info. Have taken a look at the schematic for the board and I think I see what you mean. There is a 1k resistor already there so that should mean that I am ok to connect the psu direct to the terminals!

Excellent, thanks a million for the help.

Jab