LED Pushbutton hookup to chipamp problem

[JohnnyP00]

Hi,

I'm trying to attach an LED switch to a Brian GT chipamp and have a few questions. First off I wanted to attach the LED through the V+ and V- of one of the power supplies and using a resistor to knock down the voltage to meet the LED's requirements. Is this correct?
From the pdf I read it requires up to 24vDC. I'll be supplying it with around 72vDC and will have a 7.5K resistor in series on the positive leg. Is this enough resistance for the LED? I tried some calculations but the manual doesn't give any current limitations so I couldn't follow the calculations I know.
Also if I hook it up to one of the power supplies and not the other, will this cause noise to rise to one channel and will a cap in parallel help this?

I attached some photos and the link for the pdf of the switch in question.

http://www.schurterinc.com/pdf/engli..._MSM_19_LA.pdf
It's the ring version.

Thanks for your help JohnnyP.

[jaycee]

Just put it across V+ and ground. the LED draws so little current it will not affect the sound in teh slightest.

The formula is (Voltage - LED forward voltage drop) / current = resistor to use. The current flowing will affect the brightness. Most LED's are rated for 20mA max but this will be way too bright. For just across one rail, 2K2 would be enough.

[tomchr]

72 V DC?! That sounds rather high. Is there a lower voltage available in your amp? You only need a couple of volts (slightly higher than the forward voltage drop of the LED) to make the LED light up.

Anyway. As the previous writer said, the equation to use is Rseries = (Vsupply - Vf)/Iled, where Rseries is the series resistor in ohms, Vsupply is the supply voltage in volts, Vf is the LED forward voltage in volts, and Iled is the LED current.

The best way to find Vf and Iled is to look in the data sheet for the switch. You don't want to go with the maximum Iled as this will limit LED lifetime and (more importantly) be way, way too bright. If nothing is stated, I suggest starting at 2 mA (0.002 A). If this isn't bright enough, increase the current to 5 mA by decreasing the resistor. You shouldn't have to go beyond 10~20 mA. When calculating Rseries, round up to the nearest standard value.

With 72 V, you will likely dissipate quite a bit of power in the series resistor (hence, my suggestion to use a lower voltage if available). You calculate the power dissipated in the resistor as: Presistor = Iled*(Vsupply-Vf), where Presistor is the power dissipated in the resistor in watts. When specifying power resistors you really have to be conservative. Let's say you use 5 mA and Vf is 2 V. So Presistor = 0.005*(72-2) = 0.35 W. The nearest higher standard value is probably a 0.5 W resistor. You could use this resistor if you can guarantee that the ambient temperature around the resistor will not get beyond 25 deg C (that's where the resistor power rating is specified at). The insides of your chassis will be hotter than that so the resistor will likely fry. Even if it doesn't, it'll be screaming hot -- like, 200 deg C! So multiply the calculated power by a factor of 4~5 and get a resistor with a power rating above that. So you'd need a 0.35 * 5 = 1.75 W resistor. 2 W is the nearest higher value. A 5 W would be even better.

Hope this helps.

~Tom

PS: The forward voltage of the LED in your switch should be listed in the data sheet for the switch/LED combo. Vf does depend on the color of the light the LED is emitting. Should it not be in the data sheet, use the following as a guideline:

RED, YELLOW, AMBER: Vf = 2 V
GREEN: Vf = 2.4 V
BLUE: Vf = 4 V

The blue ones tend to vary all over the place as a lot of different materials are used for the blue ones. But with your 72 V it's not likely to matter greatly as the voltage drop across the series resistor is much, much greater than the variation in Vf between different color LEDs.

[lanchile]

Originally Posted by JohnnyP00

Hi,

I'm trying to attach an LED switch to a Brian GT chipamp and have a few questions. First off I wanted to attach the LED through the V+ and V- of one of the power supplies and using a resistor to knock down the voltage to meet the LED's requirements. Is this correct?
From the pdf I read it requires up to 24vDC. I'll be supplying it with around 72vDC and will have a 7.5K resistor in series on the positive leg. Is this enough resistance for the LED? I tried some calculations but the manual doesn't give any current limitations so I couldn't follow the calculations I know.
Also if I hook it up to one of the power supplies and not the other, will this cause noise to rise to one channel and will a cap in parallel help this?

I attached some photos and the link for the pdf of the switch in question.

http://www.schurterinc.com/pdf/engli..._MSM_19_LA.pdf
It's the ring version.

Thanks for your help JohnnyP.

Why don't you use some of these lamps, they are small and they run on 120v so you do not need to share voltage from power supply.They come in many colors. RED, GREEN, ORANGE

[tomchr]

Originally Posted by lanchile

Why don't you use some of these lamps, they are small and they run on 120v so you do not need to share voltage from power supply.They come in many colors. RED, GREEN, ORANGE

<sarcasm> Great advice! Especially now that he's done a nice job of drilling holes in the chassis and mounting the pesky switch... </sarcasm> Sorry. Couldn't resist....

Actually, LEDs can run on 120/230V as well. Just use a capacitor to limit the current and mount a diode (1N4007 or similar) in anti-parallel with the LED so the LED is protected against excessive reverse voltage.

~Tom

[JohnnyP00]

Hi,

I ended up being a little confused because if you look at the spec sheet it doesn't state any amp maximum for the ring type LED. It only states max 24 vDC. I'll use the common 20mAmps to do the calculation.
If I use Jaycee's suggestion it would be R=(35-24)/0.02 Amps giving me 600 ohms. I would probably use a 2k resistor to limit the current. Should I put anything on the ground leg for protection?
For Tomchr suggestion it seems very interesting except I don't seem to have any AC caps on hand and how would you calculate the amount of capacitance needed? I do have some 1N4004 kicking around and also some Murh860 220 AB 3 leg doides somewhere.

Thanks for everyones help... JohnnyP.

[Th3 uN1Qu3]

If you put the LED on one of the DC rails (ie after the diode bridge and filter capacitors) you don't need a protection diode.

[JohnnyP00]

I just attempted it and got nothing. I put V+ to 2.7k resistior to LED + and put PG+ straight to the negative. I think I'm reducing the voltage too much or the negative leg needs to be hooked up to V-. I'm going to try changing the ground leg to V-

Hope this goes well... JohnnyP.

Got it working!! I still would like to see an AC version circuit for future projects

[tomchr]

This is one way of doing it:



0.47 uF cap in series with the LED, 1N4007 (or eqv) in anti-parallel with the LED. Use 0.22 uF for 230 V circuits. Use caps rated for use on the AC mains - i.e. caps that fail safely. They are typically labeled X1 or X2. I would use 250 V caps on 120 V circuits and 400 V caps on 230 V circuits minimum!

~Tom