installing led's into amp footswitch

[ChrisK]

You want them to "glow"" not "go" (away).

LEDs are really crappy diodes (they glow after all). They have a very poor reverse voltage rating. I suspect that if it isn't an over current issue, it's that a reverse voltage more than perhaps 5 VDC is being applied.


BTW, are the internal circuits of the Classic 50 and Classic 30 the same?

[newey]

I found this little applet a while back and posted a link to our "Links Page-Sort of", under Calculators:

LED Drop-Down Resistor Calculator

Don't know if that helps any (or, for that matter, whether it's accurate or not)

But FWIW.

[mr_sooty]

Dec 4, 2008 19:17:58 GMT -5 ChrisK said:

\
BTW, are the internal circuits of the Classic 50 and Classic 30 the same?

No, I don't think so. The Classic 30 seems to be a slightly different beast. I figured maybe the Classic 30 was shooting more ma's through the switch, except that my friend has been using my footswitch while I've been modding his, and he so far hasn't blown the LEDs on my switch with his amp (Classic 30). So that's odd.

Anyway, today I chucked a 200ohm resistor across the LED instead of the 150ohm one, and it seemed to work OK, except that I'm only testing on my amp because he has his back. But the LED was really bright (I have the same issue in mine). So I started experimenting with resistors to drop the brightness a little. I tried an extra 150 ohm one across it (parralel to the 200ohm), and it dulled it, but slightly too much. So I tried a 51 ohm one, and it cut out.

That's when I realised that maybe I really wasn't getting the idea of resistors at all. I decided to try going up. I tried another 200ohm one (in parralel with the original 200ohm one) and it was better, but still a little dull. I ended up with something like a 510ohm one wired in series with a 51ohm one, both wired in parralel with the 200ohm one. This seems just right. The LED glows bright enough, but not too bright.

So this is all very confusing for me. I thought increasing the ratings of the resistors would reduce the current flowing from the LED making it duller or stopping it going. But increasing the Ohms reduced the amount of 'dulling', giving me a brighter LED.

It seems I have no clue about resistors and ohms law. I understand that adding extra speakers to an amp will usually reduce the ohm rating, reducing the resistance and causing the more power to come from the amp - or something. That's about it.

I saw a website today that was teaching about this stuff. It asked 'if you take two 100ohm resitors and wire them in series, how many ohms do you have? I answered 200ohms, which was apparently correct. Then it asked the same question, but in parallel. Obviously 200ohms was not the right answer, and I have no idea what was the right answer (it didn't tell me, it just told me I was wrong).

So now I'm wondering (among other things), does a higher number on a resistor allow more current through or less current through?

[JohnH]

sooty - are you putting your resistors in parallel with the LED? They should be in series and usually the resistor value is alot higher than 150 Ohms - see the calculator that newey posted. It would explain why its so bright and often pops!

John

[mr_sooty]

Dec 5, 2008 4:00:07 GMT -5 JohnH said:

sooty - are you putting your resistors in parallel with the LED? They should be in series and usually the resistor value is alot higher than 150 Ohms - see the calculator that newey posted. It would explain why its so bright and often pops!

John

Hmm, well Chris K told me to wire it across the LED to shunt some current, hus diagram even has it in parralel. So yes, all my Resistors are going across the legs of the LED, in parralel.

I was looking at this equation last tonight, which is supposedly how you calculate resistors in parralel.

R1xR2
R1+R2

I have a 510ohm and a 51ohm in series, making 561ohms. They're in parralel with a 200ohm one. If I use the above equation, I come out with 148ohms, about where I started. Except that the LED has dulled nicely, so now I'm really confused.

[newey]

The calculator I posted presupposes that the resistor is in series with the LED (see the diagram thereon). The recent loop box I built had its LED in series with the resistor, and that's the one and only time I've ever wired an LED, so I don't know about wiring 'em in parallel.

In any event, if you're now happy with the result and it works, leave them there sleepy dogs lazing about.

[ChrisK]

I have a 510ohm and a 51ohm in series, making 561ohms. They're in parallel with a 200ohm one. If I use the above equation, I come out with 148ohms, about where I started.

Except that the LED has dulled nicely, so now I'm really confused.

Why, isn't this your amp and not his now? They likely have different currents flowing thru their foot switch circuits.

Remember, this foot switch design was for your specific amp. All bets are off when another amp (of unknown circuitry and values) is used.

There is a series resistor or relay coil in the amp that forms the major total circuit current limitation value.

Sheet 1, lower right-hand corner.
www.prowessamplifiers.com/schematics/misc/Peavey_Classic_50.html

Sheet 1, left side middle
www.prowessamplifiers.com/schematics/misc/Peavey_Classic_30.html

Remember, this foot switch design was for your specific amp.


Looking at the two schematics linked above, one will clearly see that the internal circuitry AND VOLTAGES are different for the two amps. Mechanical foot switches won't care, but electronic circuitry (a.k.a. LEDs and balancing resistors) WILL.

All bets are off when another amp (of unknown circuitry and values) is used.


In this case we are putting a current shunting resistor in parallel with the LED to bypass some of the current in order to reduce the brightness.

The higher the value of the parallel resistance, the less the value of bypassed current. The lower the value of the parallel resistance, the greater the value of bypassed current.

Assuming that the LED has a constant voltage drop of about 1.8 VDC, a 180 Ohm resistor will have 10 mA flowing thru it. This means that whatever the current flowing thru the LED is, it will be 10 mA less.

However, if the current flowing thru a small resistor value is insufficient to generate a voltage drop at least equal to the LED ON voltage drop, no light will be emitted.

It's a balancing act in that you are adjusting the bypass current for illumination at a lower desired level. The total circuit current needs to remain constant to effect proper amp operation.


And yes, this is indeed how you calculate two resistances in parallel.

R1xR2
R1+R2

For two or more resistances in parallel, it's;

1/(1/R1 + 1/R2 + ..... 1/Rn)

[mr_sooty]

So, in order to protect the LED, would I be better to wire a resistor in front of it, rather that parralel to it?

But yeah, unfortunately I don't have this guys amp to test on. But given that I've succesfully lowered the brightness, is it safe to assume that less current is going into the LED now, and therefore it is less likely to blow?

[ChrisK]

The likely culprit has been identified.


In the Classic 50;

The reverb channel has a maximum current of -27 VDC/ 1,600 Ohms = -16.9 mA.

The channel select is a 12 VDC relay coil in series with 470 Ohms. Assuming that the relay coil is operated at 12 VDC, there is -15 VDC / 470 Ohms = 31 mA.


In the Classic 30;

The reverb channel has a maximum current of -36 VDC/ 1,600 Ohms = -23 mA.

The channel select is a 12 VDC relay coil in series with 1,500 Ohms. Assuming that the relay coil is operated at 12 VDC, there is -24 VDC / 1,500 Ohms = 16 mA.


The reverb currents are fairly similar.


However, it is unlikely that the same relay is operated at a current twice that in one amp vs the other (which implies an effective coil voltage that is twice that as well).

Based on a quick analysis of simultaneous linear equations, with the assumption that one resistor is 500 Ohms and the other is 1,500 Ohms, for equal coil current 22.5 VDC would be developed across the coil in both cases [27 / (X + 500) = 36 / (X + 1,500)], which is at 9 Ma. This implies a coil resistance of 2,500 Ohms, which is quite believable for a 24 VDC coil in a small relay.

For the LED being too bright at only 9 mA, it must be a very high efficiency LED.


The culprit:

The Classic 30 has an additional 10 uF cap (C55) across the channel select foot switch input. When the channel select switch is not on, this cap WILL charge to -36 VDC.

Since there is no series resistance between this cap and the foot switch, the peak discharge current from the cap can easily exceed one amp when the switch makes connection, and may be as high as 5 amps. This surge current goes thru the LED and turns it into a DE(a)D.

(Darkness Emitting Diode.)

[mr_sooty]

Right, so I guess I should probably ask my friend to stop using my footswitch with his amp! Otherwise I'm gonna get a fried LED.

Is there anything I can do to protect the LED in the Classic 30?

[ChrisK]

Yeah, but let me run a simulation to characterize things.

[ChrisK]

The three extra diodes form a 2.1 volt clamp. The 51 Ohm resistor in series with the LED limits the surge current thru it.

[mr_sooty]

Cool, thanks. Now I just have to find some more of those Diodes.

[ChrisK]

You can use;
1N4002
1N4003
1N4004
1N4005
1N4006
1N4007

They're just higher reverse voltage versions.

[mr_sooty]

OK, so I've wired this all up. Doesn't really work on my amp (Classic 50 - Should it?) but I don't know yet if it will work on his. Currently the channel switches, but the LED just makes a faint flash.

I was trying to retain the drop in brightness that I cooked up before, so instead of the 150ohm resistor, I have a 200ohm one in parallel with a 510ohm one. Should I switch back to just the 150ohm one?

[ChrisK]

No, 200 Ohms in parallel with 510 Ohms is 143.66 Ohms. This is within the tolerance on your 150 Ohm resistor, and would be virtually the same.

See how bright the LED is with no parallel resistor. Then try a 510, and then a 200 Ohm one. If you have a 500 pot, you could dial the resistance that you want for the brightness that you want.

I have no way to predetermine the brightness since I can't see that far (and have no data on the LED).


Did you indeed use 1N400X series diodes?

[mr_sooty]

Yeah, I used 1N4004's. The only differences between what I've done and what you drew are:

- I didn't use the original LED protection diodes, as we pretty much found out last time that we had the LED's in the right way around. The only diodes are the three in series in the new design.

- The extra parralel resistors.

Would the brightness theoretically be the same in both amps?

[mr_sooty]

OK, so I tried just about every value resistor I had, as well as no resistor at all (this is just in the parralel application accross the LEDs legs), and everything gives me the same result, a quick little faint flick, and then nothing. The LED's still good, I checked it with a battery. I'm thinking there's something wrong here.

Are we sure the Diodes are around the right way? They confuse me, because when you buy them they put a big red stripe of tape at the end the polarity band is at, which would seem to suggest that end goes to positive, but the bands in this design are going to the same end as the negative end of the LED. And ever other time I've used diodes the polarity band seems to go to the negative end.

[ChrisK]

Are we sure the Diodes are around the right way?

Yes.

They confuse me, because when you buy them they put a big red stripe of tape at the end the polarity band is at, which would seem to suggest that end goes to positive,

No.

but the bands in this design are going to the same end as the negative end of the LED. And ever other time I've used diodes the polarity band seems to go to the negative end.

These two sentences are in agreement.


There is no positive and negative end to a diode. Depending on the application, either end might be connected to either.

The band marks the cathode. The other end is the anode.

Anode-----|>||-----Cathode

For current flow (which is unfortunately defined as the opposite direction of electron flow), if the anode is more positive than the cathode by the voltage (conduction) drop of the diode, current will flow in the direction of the arrow (>).

In both of these amps, the foot switch common wire is at amp ground potential, and the signals that are switched are negative voltages. So, in our case, the amp ground is the most positive voltage.

This is what the new circuit does.

|---|LED>||---|51 Ohms|---- to channel select input
|
|---|>||-----|>||-----|>||----- to channel select input

Each 1N4004 has a forward voltage drop of about 0.7 VDC. Three of them should be about 2.1 VDC. If the current thru them is very small (except for the surge current), they may not develop enough of a drop to allow the LED to turn on (it has a forward voltage drop of perhaps 1.8 VDC.

The fact that the LED flashes seems to indicate that except for the surge (turn-on) current event, insufficient voltage is being developed across the LED.

The simplest thing to do is to add a fourth diode in the three diode series string. This will develop 2.8 VDC max (or less as we seem to see).

|---|>||-----|>||-----|>||-----|>||------

The series string forms a voltage clamp in that the drop across all four is limited to perhaps 3.5 VDC at the surge current of an amp or more, limiting the maximum current thru the LED. Once the surge current is dissipated, the LED and 51 Ohm resistor conduct the current.


A little more fishing and trial and error are in order. Remember, I'm doing this with posted schematics of unknown vintage that may well NOT match either amp, and have no data regarding the LED, internal amp channel select relay, or actual currents occurring.

[mr_sooty]

Dec 7, 2008 23:09:10 GMT -5 ChrisK said:

but the bands in this design are going to the same end as the negative end of the LED. And ever other time I've used diodes the polarity band seems to go to the negative end.

These two sentences are in agreement. .

Yes, but both are in disagreement with my misguided perception of what the red tape meant.

Dec 7, 2008 23:09:10 GMT -5 ChrisK said:

Remember, I'm doing this with posted schematics of unknown vintage that may well NOT match either amp, and have no data regarding the LED, internal amp channel select relay, or actual currents occurring.

I know. I know it would be a lot easier for you if I knew what the heck I was talking about and could explain things in proper electronic talk. But the truth is, it's your ability to decipher my information, which could only be described as 'vague hints' and come up with working solutions based on next to no information that I admire the most.

Honestly, your posts are like gold to me. I sit at my computer when you're online (sometimes) eagerly awaiting another drop of wisdom. If I wasn't already a Christian, and if I was in the market for a new God, you would be in serious contention.

Anyway, success! Fortunately I had one diode left over, and the forth diode in the chain seems to have done the trick. Initially I had put back the parallel 150ohm resistor, and it didn't work, so I cut that out and got a strong, blinding glow.

I tried a few other parallel resistors and settled on a 750ohm one, which 'took the edge off' the ultra bightness of the LED, so it's bright without being obnoxiously so. I still don't understand why a higher rated resistor has less of an affect that a lower rated one, why it doesn't work with a 150ohm one, but is just slightly dulled with a higher rated one. But, oh well. It does, and that's that.

So anyway, hopefully this will go well in the Classic 30 and the LED will survive. We'll just have to wait and see. Hopefully I can let you know within a couple of days.

Thanks so much for your help as always!

[ChrisK]

Fortunately I had one diode left over, and the forth diode in the chain seems to have done the trick.

Good. We had to empirically determine things. Always buy extra parts, they're usually cheaper than the petrol to go get more.

Essentially we have created a hard voltage clamp (as hard as the power diodes can tolerate). This limits the voltage under current surge to no more than 3.5 to 4 VDC. Although rated at 1 amp of average forward current, for a single half cycle surge (8.33 mS) the 1N400X series can handle 30 Amps.

www.fairchildsemi.com/ds/1N/1N4004.pdf
If you look at page 2, Non-Repetitive Surge Current, you'll see that it's a fairly beefy little sucker. Just don't use it for back EMF protection for relay coil drivers, it's much too slow in reverse recovery (use a fast recovery diode).


As a result, the LED (did I mention that they are crappy diodes? - they do make good fuses tho', eh) is protected from having to handle the current surge.

[mr_sooty]

Dec 8, 2008 19:10:21 GMT -5 ChrisK said:

Just don't use it for back EMF protection for relay coil drivers, it's much too slow in reverse recovery (use a fast recovery diode).

Right, well I'll keep that in mind...


...( )

Also, I was giving an electritician friend a guitar lesson last night and he explained the resistor thing to me, how the signal follows the path of least resistance. Made things make sense to me slightly more. So if the resistor value is low, the signal will go through it, bypassing the LED (= LED doesn't go) and if it's too high, the signal will bypass the resistor completely and go straight to the LED (= resistor has no effect). So you get a range in between there where it's going to both, and the higher the resistor value, the more current is going through the LED (= brighter LED), and vice versa (lower, through resistor = duller LED).

Sound right? Makes sense based on my experiences with this.

[hellseeker]

UPDATE!

I've had success with my Classic 50 410 by just wiring the ring and sleeve from the footswitch cable directly to LEDs, then to the switches. I rehoused everything for a smaller foot print.
Thoughts?

[newey]

hs-

Hello and Welcome to G-Nutz2!

You've resurrected a very old thread on this, I don't know that we've heard back from Mr. Sooty in a while. But hey, you've got pertinent experience, so it's all good!

It looks like your footswitch is a bit different than the one Sooty was modifying, as the photos of his don't show a TRS jack- the footswitch was hard-wired to the cable.

Also, you have used the LEDs which come complete with a housing which contains the step-down resistor internally, eliminating the need to calculate the appropriate step-down value (assuming the other parameters are met). Did you calculate the voltage to the LEDs, or did you just get lucky?

Your solution is equivalent to the "no battery" solution diagrammed by chrisk back on the first page of this discussion, but with the difference of the jack being on yours.

[hellseeker]

Yeah, I cut the original cable and soldered on a TRS plug, then put a stereo jack into the enclosure. I like being able to unplug it if necessary. Right now, there are no resistors in the enclosure, the LEDs are wired in series from the jack to the footswitches. They are a tad bright, so I'm going to add some resistors at some point to tame them a little, probably 1k-1.5k based on what I've heard.
I actually bought and drilled the enclosure myself, adding the LEDs and bezels and transplanting the switches from the original Peavey footswitch that came with the amp. All the materials only cost me about $10 (including the great powder coating - I suck at painting...)
The size is great because I can click both buttons with my foot simultaneously, allowing my to go from overdrive with no reverb to clean with reverb, without extra tapping.

[dan]

Hello,

I realize this is an old thread, but my goal is to do the same job Sooty has and then post the actual part values used so that others will not have the same confusion I am having thus making a more complete thread on this project (that is making it stupid proof for those like me who can solder but aren't sure what they are soldering).

I need to install LED's as Sooty has done.  I have the same foot switch and a Classic 50 head. 

I have read this entire thread many time but I just am not certain as to what values I should use for the parts.  Being the links no longer work ((6 years, 42 in dog years) I'm getting confused on all the exchanges of info.

I decided to measure the volts and amps at the switch.  My digital died so I used my analog.  I have 23 v at both switches.  The line for the reverb is 14 mA and the one for the channel switch is 20 mA.   From reading ChrisK's responses I gathered that the voltage could be around 27 and that the mA for the channel switch could run 32mA.

ChrisK's first drawing on the switch shows the diode, LED, 13K resistor wired in series but connected parallel to the switch for the channels. Later he shows it wired in series with the switch with a 150 parallel resistor to the LED.  Sooty's drawing shows it this way also.

My confusion is that when I talked to my local electronics parts store (a real one with all sorts of goodies) he asked about what voltage and amps thus what resister did I need.  Googling all this suggests I need a resistor in series to get the voltage down to around 2v and Chrisk notes 20 mA.  The calculators all come up with 1.2K resistor using my measured voltage.

For the reverb, all the drawings and pictures show the LED with the diode in series but no step down resistor. 

One calculator notes: Voltage drop is usually 1.9~2.1V for AlGaInP, 3.1~3.5 for InGaN and 1.2V for Infrared.
Current is usualy 20mA, for UFO LEDs current is: 30mA for InGaN and 50mA for AlGaInP.

Not knowing what these different LED's are it looks like I could get a couple that will tolerate the mA's even at the level ChrisK has calculated but that the voltage is still the issue?

So, do I need a step down resistor in series?  I understand the diode is to protect my potential miss wiring of the LED but does it act to step down the voltage?  Is the 150 Ohm resistor paralleled in the channel circuit acting as the calculated 1.2K resistor?

Or should I just put the cover back on the pedal and forget about it?

Thanks in advance for the help.

 

[dan]

Well, I've done some more reading and I see where my mistake was as to using the calculators for the resister.  I needed to subtract the LED voltage from the present voltage and enter that as the voltage drop.

I purchased some LEDs rated at 3 volts, 20mA (yellow and green) and wired them in series based on the last set of pictures with just the resistor parallel on the channel switch and no resistor on the reverb switch.  I used the diodes also. 

I still don't get the 27 volts vs the 3 volts of the LED as to how this does not blow the LED if there is still enough voltage to do the switching.  I don't see how the voltage won't kill the reverb LED with no resistor.

I could have read my analog meter incorrectly and be off by a decimal?

Anyway, I hope someone responds and helps me understand what I just wired. 

[newey]

Dan-

Hello and Welcome to G-Nutz2!

I don't have an answer for your question, but someone will be along. You might consider re-posting this into the amp section. I don't really know how Sooty's original thread was left here, but no point to moving it now.

[dan]

Nov 17, 2013 22:54:35 GMT -5 newey said:

Dan-

Hello and Welcome to G-Nutz2!

I don't have an answer for your question, but someone will be along. You might consider re-posting this into the amp section. I don't really know how Sooty's original thread was left here, but no point to moving it now.

Thank you for the welcome.  I will post this to the amp section as you suggest and link back to this topic.

Just to be complete here, the switch does work with the LED's I purchased.  The components were from Radio Shack (I could not get to my favorite store and figured it's not like I'm doing the tone stack). 

I rechecked the voltages. For the channel switch on (closed) there is 22.5 V across the polls.  In this position the LED is off and the amp is running in the normal channel.  There is no voltage measured across the LED.  With the switch off (open) there is no voltage across the switch, the LED is on and there is 2 V across the LED.  In this position the amp is running through the high gain channel (pre and post amp volumes).  I used the green LED.  If I try to measure the mA, the LED goes out.

The reverb channel works the same.  With the switch off (open) the LED is on and there is reverb.  2.8 volts across the LED.  Zero across the switch.  When the switch is one (closed) there is 20 V across the switch, nothing across the LED.

I just don't get this as to how there can be power in the LED with the switches open thus no connection to the common ground.  When the switch is closed how is it that the LED is not popping with 20+ volts in the circuit?