Author Topic: Simple remote Loop Switch (Read 28955 times)

duck_arse · « **Reply #20 on:** July 20, 2014, 07:14:57 AM »

exactly. correct. 100%.

I won't type out the led resistor calculation, because I will make my usual long-winded botch/confusion of it. you can probably find an explanation hereabouts if you search "led series resistor calculation", or similar. (sometimes googoo searches this site better than the search button.)

Minty · « **Reply #21 on:** July 21, 2014, 05:18:06 AM »

OK thanks!

I've researched it and understand the theory, and there's plenty of calculators out there, although using this information says that the resistors I have are way over rated (N.B. the LED ratings aren't from the supplier, these I've found elsewhere online. The LEDs I bought are both "Bright Purple/Pink"):

Supply: 9v
Forward Voltage: 3.2v (each)
DC Forward Current: 30mA

The calculator I used tells me I need a total of just 100 ohms resistance, whereas I'm using 2x 4.7K.

Am I missing something obvious?

duck_arse · « **Reply #22 on:** July 21, 2014, 10:56:16 AM »

numbers. ok. first, 30mA is the max value, probably. it will burn the paint on your ceiling at that current.

for a superbright led, you can run at say 5mA andblind the neighbours, attract low flying planes, etc. at 0.5mA, or 500uA, you will have a status indicator instead.

so, V+ = 9V; Vf = 3V2; I(led) = 500uA, R3 = xR.

V+ - Vf = V(R3). VR3 is the voltage we want to "drop". 9 - 3.2 = 5.8. R = V(R3) / I(led) (Mr Ohm). xR = 5.8 / 0.0005 = 11,600 or 11k6. you can use 12k or 10k, and probably not tell the difference.

the led brightness varies with the led current. you must ALWAYS use a current limit resistor with leds (unless they are audio clipping). you can do all the calculations each time if you like, or just use something between 4k7 and 22k for a 9V supply, scale between depending if you want to indicate or room lighting.

you can and should look up the part number and see the manuf datasheet, for parts you know and parts you don't. there will be surprises.

PRR · « **Reply #23 on:** July 22, 2014, 01:37:37 AM »

> DC Forward Current: 30mA
> using 2x 4.7K.

30mA is a burn-up rating. Do you drive your car a constant 113MPH? Usually we poke around at 25-60MPH, which is fast enough for some and too fast for others.

6V drop in 5K is a large milliAmp. Working indoors, especially dim stages, you don't need "113MPH" light output, but a small fraction of a typical LED's MAXimum output. 10%, even 3%. (So a resistor 10 to 30 times bigger than calculated from the MAX rating.)

If it is bright enough for you, do it.

The only reason to know "30mA" is when you need it brighter and brighter and.... don't go (much) over 30mA or the LED becomes a DED (dark-emitting diode).

Minty · « **Reply #24 on:** July 22, 2014, 08:51:29 AM »

Thanks again for your valuable time/knowledge.

I've clearly overestimated even my humble perspective of my grasp of basic electronics

I've got everything but the box to put it in now - just waiting on the delivery. I'm sure I'll have some feedback soon enough, be it excited or with burnt fingers.

Minty · « **Reply #25 on:** July 22, 2014, 10:48:40 AM »

Ok, I think this is making sense now.

xR in this instance is 4,700 ohms, therefore I = 0.001234 [I = V/R]. This is greater than the 0.5mA you recommend....but how much greater in practice?

My lack of appreciation of how much current is actually required here has hampered my initial understanding. PRR's comments on driving at 113mph is an excellent metaphore and I am now grasping the logic.

Will 1.234mA be too much here, or should I be thinking about replacing these with 10k resistors (I have a surplus of them).

PRR · « **Reply #26 on:** July 22, 2014, 10:46:22 PM »

> Will 1.234mA be too much here

31mA is too much for the LED. Like tipping rocket-fuel in your 113MPH car to go 119MPH.

Any lower value is safe for the LED, but you get less light.

Whether 1.234mA is too much or too little pretty much depends on the neighborhood, no? On older LEDs, we "had" to run 5mA to see them in a bright lab. Modern/better LEDs in dimmer rooms, much less may be ample.

Minty · « **Reply #27 on:** July 23, 2014, 04:07:29 AM »

Great, thanks.

Well, I have another issue now:

I had used the internet to determine the power supplied by the Boss GT8 transformer (with an aim to branch off this to the other parts). Vendors on the internet list them as 9v 1000mA.

I've checked mine today, and it's 14v 800mA. So, here's the bigger picture question:

I want to split this supply so that it powers the GT8, The RAT Pedal (which requires 9v) and this switching unit.
GT8 draws 650mA
RAT pedal no more than 5mA (and no more than +120VDC) RAT DATASHEET (point3 = powersupplies
This switching unit? Don't know, but I need to put an inline resistor on the relay now, for sure? HERE IS THE RELAY DATASHEET

Have I totally fudged this, or am I still OK?

Minty · « **Reply #28 on:** July 23, 2014, 07:49:46 AM »

Here's the relevant Data on the Relay:

What do I need to do here to use the 14V supply?

duck_arse · « **Reply #29 on:** July 23, 2014, 08:38:02 AM »

nothing but ohm's law.

the relay max V shows as 21V, so 14V would be safe. seeing as the relay coil is 540R, and you want 9V across it, the other 5V you want to drop, (being near half 9V) suggests a resistor half the value of the relay coil, say 220R. now you get to draw the diagram, add the numbers, and work out what the voltages are and current is.

I think you could put that value resistor between your 14V supply and the point on your switcher marked "+9V", and not worry about anything.

Minty · « **Reply #30 on:** July 23, 2014, 08:56:02 AM »

I'll redraw this as soon as possible. Thanks again!

Any thought on the RAT pedal? That too needs a 9v feed (from a 14v source). I am willing to modify that if needs be.

duck_arse · « **Reply #31 on:** July 23, 2014, 09:15:48 AM »

the rat sounds like the perfect opportunity to learn about three-terminal regulators. 14V is just right as input to a 7809, and the regulator will supply enough 9V to run your relay section as well. no need for the 14V dropper resistor then.

Minty · « **Reply #32 on:** July 23, 2014, 09:19:13 AM »

OK, so, here's the new diagram,

If I'm reading this correctly, then:

V = 9, R = 540, ThereforeI = 0.0166A (16mA)? Sounds a lot for what it is?
Also, the 2 LEDs in there are drawing 1.25mA each. So in total, 18.5mA?

..I'm guessing I'm wrong here?

And so, the RAT pedal too. Apparently 9v and no more than 5mA - so what to do here?

duck_arse · « **Reply #33 on:** July 23, 2014, 09:29:14 AM »

"R6=0.22K" this will confuse you. 0.22k is 220R.

your calcs are correct, mr ohm is very seldom wrong. your rat will be happy and clean with a regulator, as will your relay and switcher, and you will be happy not needing to calculate current draws and voltage drops. as for whether or not you are clean, well .....

Minty · « **Reply #34 on:** July 23, 2014, 09:38:10 AM »

Ok then. 220R it is.

On the regulator, this would be my second choice right now as I've gone quite a lot further than I had planned already, but - if I do go down that route, I'm guessing this is my diagram?

Let's just say I don't do this, and I want the rat to have a 14v feed. What's the process?

Minty · « **Reply #35 on:** July 23, 2014, 11:21:47 AM »

Don't worry about the last enquiry,

I've just bought a 7809 and heatsink, I got the 0.1uf caps anyway.

I also bought a bigger project box to put all of this in

trixdropd · « **Reply #36 on:** July 23, 2014, 02:05:00 PM »

Quote from: Minty on July 23, 2014, 09:19:13 AM

OK, so, here's the new diagram,

If I'm reading this correctly, then:

V = 9, R = 540, ThereforeI = 0.0166A (16mA)? Sounds a lot for what it is?
Also, the 2 LEDs in there are drawing 1.25mA each. So in total, 18.5mA?

..I'm guessing I'm wrong here?

And so, the RAT pedal too. Apparently 9v and no more than 5mA - so what to do here?

I have an alternate wiring suggestion for the relay and jacks;

in: pin 8,11
send: pin 4
ground: pin 6
out: pin 13
return: pin 9

(I'm assuming that pin 4 and 6 are connected by default and when switched pin 4 connects to 8 leaving pin 6 out of circuit)

This properly grounds your effect pedal when NOT in use, which can be very helpful. If you had a high gain pedal in the loop then NOT wiring how I suggest would be like unplugging the cable from the guitar and thowing it on the floor.

Minty · « **Reply #37 on:** July 24, 2014, 06:17:02 AM »

Ok, so there's the updated diagram.

Honestly, I can't tell you all how much I'm enjoying this!!!!

Trix, thanks for your suggestion, I've reconfigured the relay.

Does this all look good to everyone?

duck_arse · « **Reply #38 on:** July 24, 2014, 07:54:53 AM »

nearly. a bit more has come to mind.

the relay will make noise, the rat may well not like it. we seperate the relay supply from the clean, minty-fresh 9V. take the tops of R4, R5 and R3 to the (dirty) 14V (after you've recalculated their value if you like). take the top of the relay and D2 to that 220R you calculated before, and the top of the 220R to the 14V line.

keep C3 and C4 as 100nF, but you'll probably want to add another electro, 47uF~100uF to the 14V line. it will help clean up after the relay and the connecting wires. that should do for today?

Minty · « **Reply #39 on:** July 24, 2014, 09:33:47 AM »

I thought that the whole of the original V supply circuit was to take out the noise when switching?

Or are you talking about hum?

...and yes, that'll do for today!

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Author Topic: Simple remote Loop Switch (Read 28955 times)