Newbie LED question

[Jayco]

Hey all...

I have a question that I hope someone can help me with:

I have an Orange amplifier head that provides ~ 7.6VDC on the jack that you plug the channel switch into.

The amp doesn't come with a switch, so I wanted to build my own.

Closing the switch is obviously easy to do (and works fine), but I am adding an LED for BOTH states of the switch (closed and open) and want to use the provided 7.6VDC to power the LEDs.

Right now I come off of the tip of the input jack (which carries the + from the amp) and hit a DPDT switch. 

The circuit looks like this:



The 7.6VDC comes from the amp and hits a DPDT with the two center pins tied together (to provide voltage to the amp side and the LED side).

When the DPDT is open for the amp, LED 1 lights without any problem.

When the DPDT is closed for the amp, the amp switches (as it should), but LED 2 doesn't light.

I am assuming that the LED is providing enough resistance so that the electricity just grounds out on the amp side, but I'm not sure how to correct it.

Any ideas?

I'd like to avoid putting a 9-volt source in there if possible.

Also, I am currently using two LEDs I just grabbed from RadioShack... one is a 1.7v 20 mA and the other is a 5v 30mA.  Both are PLENTY bright with the 1k resistor, but am I hurting anything with those voltage of LEDs?

Thanks a bunch!

Jim Coates

[smccusker]

just going for the most obvious answer here, but is led2 oriented correctly?

[Jayco]

Yes... if I remove the ground on the amp side, the LED works correctly.

Jim

[smccusker]

Pure speculation since i'm out of my depth here, but is it possible that since you're providing a path to ground with no resistance, all your current is going there instead of through the LED?

[Jayco]

Pure speculation since i'm out of my depth here, but is it possible that since you're providing a path to ground with no resistance, all your current is going there instead of through the LED?

That's my guess too... that's what I was trying to convey when I said:

I am assuming that the LED is providing enough resistance so that the electricity just grounds out on the amp side, but I'm not sure how to correct it.

I've tried adding resistance to the amp ground side, but then the relay that switches the channels doesn't seem to have enough voltage... not sure how to handle it.

Jim

[Jayco]

Ok.. new piece of information.

If I put the 1.7 volt LED DIRECTLY across the amp side (no resistor), it will light and switch the amp.  However, I assume the voltage would quickly damage the LED.

The LED itself drops the voltage from ~7.6VDC to ~1.8VDC.

If I use the 5 volt LED directly, it won't change the channel, but will light.

Jim

[spudulike]

[Jayco]

Too much resistance... as I stated a few messages ago, adding resistance to the amp ground side the relay that switches the channels doesn't seem to have enough voltage.

I checked on some resistance calculators and it looks like for the 1.7v LED, I can go down to about a 300ohm resistor... so I'm going to try and see if that will work with enough voltage to change the channel as well.

Jim

[Jayco]

Ok... so since I know that the circuit will trigger and the LED will light if I don't use a resistor on the LED, will that cause problems?

I know its a good idea to use one to keep the LED from burning out, but will I harm things by not using one?

I had it on without the resistor for a while and the LED didn't burn out...  so are we just talking about a matter of a shorter LED light or is it more a matter of it might give out in the next several hours?

btw - I tried using a 330 Ohm resistor, but it still was too much resistance to trigger the relay that changes the channels.  The only way I've gotten it to work thus far is by either not using an LED or by wiring the LED direct (with not resistor)

Thanks!
Jim

[niftydog]

since the normal operation of the circuit is to simply ground the 7.6VDC, then the source of that voltage must be current limited because otherwise the short circuit would destroy components from excess current. So, it's possible that the LED will be fine, provided that current limit is appropriately low.

The issue of the circuit not switching is not due to too much resistance, rather that the LED is a diode and as such it has a voltage drop across it when it's biased into operation (ie; when it's working!) So, one solution might be to find a low voltage drop LED. The resistor only exacerbates this problem, but a low value resistor will not protect the LED so it may not be worth putting it in at all. The same theory applies to both LEDs as they are both being fed by the current limited source.

If you're really curious, you could hook up a current meter across the circuit. A current meter is practically a short circuit, so you should be able to get an idea of the maximum current that flows when the terminals are shorted. You can then decide if it's likely to damage your LED, or purchase an appropriately rated LED to suit the current that's flowing.

[Jayco]

Niftydog..

Thanks for the response.

I put a meter across the amps jack and this is what I found:

0.07 amps (70mA) measuring current based on a 10A scale and 63.5mA based on a 300mA scale.

Trying the same measurements on a 9-volt battery I get the following (just to compare):

2.7 amps.

The LED that I am using at the moment is a 1.7v 40mA 3mm LED... so I'm still 23.5 mA in descrepency.

Is that acceptable, or do you think I need to find something with a little better match?

Sorry for the newbie questions...  I'm just having trouble getting my hands wrapped around whether this is a 'bad thing' or not.

Jim

[niftydog]

ok, so the max current is about 65mA, and your LED is rated to 40mA. That's probably ok-ish, but not great.

Try the same test again, but this time with the LED in series. That is to say, put it back the way that it works (with no resistor) and measure the current through the LED. So, just to be clear, the current path will be out of the amp, through the LED, through the multimeter then to ground.

The LED voltage drop might help things... perhaps. Otherwise, a small value resistor might provide just enough of a voltage drop to bring the current down... try that 2nd measurement first and get back to the forum.

[Jayco]

Measuring from amp.. through LED... through multimeter to ground = 43.4mA

So that is measured AFTER it hits the LED with the forward voltage of 1.7 VDC.

Jim