installing led's into amp footswitch

[mr_sooty]

I have a Peavey Classic 50, and I use the reverb and overdrive on it alot. I use the amp footswitch, but there are no LEDs to let me know what's selected. I often have to do little string scratches before the start of a song to hear what channel I'm on.

Surely it must be pretty simple to install on/off led's into this plastic housed footswitch, can someone please give me a run down on how to do it?

[ChrisK]

No, not offhand.

Do you have a schematic of the amp and the footswitch?

I now do, I searched for "Peavey Classic 50 schematics" and found the following
www.prowessamplifiers.com/schematics/misc/Peavey_Classic_50.html

If analog/high voltage signals were being switched/shunted, just adding in LEDS may well not work without implications to both the sound and the LEDs themselves.

A quick look indicates that signals referenced to -27 V are being switched to circuit common (ground).

The channel switching relay is switched by relay in series with a 470 Ohm 1 watt resistor. this indicates that about 30 mA is being switched since the relay is indicated as being a 12 VDC coil. This is a bit much for a typical LED (20 mA max is better).

The reverb is switching a FET gate biased by a 1K5 resistor from -27 V. This results in a current of about 18 mA.

One could try placing an LED in series to the common (ground) terminal in the footswitch from each switch. Since a typical LED has a forward voltage of about 1.6 VDC, for the channel switching relay we could bypass some of the current. I'd put a 150 Ohm resistor in parallel with the LED. This would result in 10 mA thru the resistor and about 20 mA thru the LED.


So, here's what I'd try (I assume NO liability for your actions).

Each switch will have a wire coming from the cable that is unique to that switch, and a common wire that goes to both switches.

You somehow need to intercept the individual switch wires and put an LED in series with it. When the footswitch activates a function, that current flows thru said LED. The additional 1.6 VDC drop introduced by each LED should not matter for proper switching (but it might).


-27 VDC biased Reverb signal-------switch terminals--|<|--common (ground)

-27 VDC biased Channel switch-------switch terminals--|<R|--common (ground)

Footswitch common--------------------------------common

The --|<R|-- is the LED paralleled with the 150 Ohm resistor

Note that if you install the LEDs backwards, they will become toast quickly since the reverse bias voltage of an LED is usually 5 VDC max (they glow nicely, but are crappy diodes). You get to figure that part out. If this is not possible, put a small signal diode (1N914/1N4148) in reverse current flow direction (anti-parallel) with each LED. If they are then wired backwards, the switching will still work, but the LEDs won't get toasted. You could also put an LED in reverse across each LED. In this case, one will glow.

[sumgai]

sooty (and Chris),

I consider 27 volts at 20 to 30 mA to be "low power", and therefore it should be safe to experiment. However, there's always a fly in the ointment, so it would pay to be safer than sorry, IMHO. Since you have to go into the switch anyways, you might as well do it right, and isolate the LEDs from the controlled circuitry. Less chance of screwing things up, in the long run.

The best way (cheapest) is to simply replace the cable in it's entirety. Instead, use a 4 conductor cable - 3 for the 2 switches and common, and 1 more for 14 volt power. You'll also need to replace the switches with DPST units.

Go into the amp, and find the -14 volts in the power supply, at the end of the same line that supplies the -27 volts. It's already DC, so it'll be good to go. Hook it up to one of the four wires in the new cable, and connect the remaining three as they were in the previous cable.

Uninstall the two footswitches, replacing them with the new ones - wire up one side (one pole) as before. The remaining pole will be used to switch on/off the -14 volts to the corresponding LED, which will then go to a 560 Ω resistor, and thence to ground. (The common wire for the other two switches.) If you're not sure how to hook up a LED, see the footnote below.

Voila!, the LEDs will be drawing about 20mA of current, and will turn on only when the effect is active (presuming that you wired them up correctly ). If the LEDs aren't bright enough, decrease the resistance value, but only one step at a time. 390Ω will very likely be too low, and probably burn the LED out - 330Ω will definitely bring doom to the little bugger. If it's almost too bright (it outshines everything else on the stage), you can increase the value as much as you wish, no harm in that.

HTH


sumgai

---

Note:
To keep things straight in your mind, this page shows you how to hook up one or more LEDs:

www.theledlight.com/ledcircuits.html

Pay particular attention to the orientation - the flat side is the cathode, and always goes to the most negative side of the power supply. Ordinarily that would be ground, but in our case, we're using the rather convenient -14 volts. No problem, we'll just hook that to the cathode instead - it's negative with respect to ground. The other side is the anode, and will go to ground, via the resistor. Speaking of which, I found this handy calculator for determining the resistance value:

www.quickar.com/bestledcalc.php?session=

There are other considerations, true, but this is a good first approximation, especially since we're not designing the SS Hoodwink here. ;D

[mr_sooty]

Hmm, this is not going to be as easy as I had imagined. I wonder if one of those Fender foot-switches with led's would work with the Peavey?

[ChrisK]

It's not that things aren't easy, they just involve technical details.

The Fender LED footswitch might work, but then again it might not.


As sumgai insinuated, it might be simpler to just use DPDT switches and use an alternative supply. On could use said switches, add 2 1.5VDC batteries in a holder inside, and wire two series resistors to two LEDs and use the extra switch pole to switch the respective LED. You get "free" battery charge indication if you leave one of them on by mistake.

If I were to go to the trouble of using the internal -14 VDC supply,
I'd go to the trouble of using the -21 VDC supply.

Personally, I'd do what I said, but then I live in

TechnicalDetailLand.

[sumgai]

Chris,

In TechnicalDetailLand, one wouldn't care what the initial voltage was, one would use a voltage regulator, such as a 7905, to make sure that the LEDs received both constant voltage and current. Said regulator would be placed inside the chassis (BIG heatsink), whereupon the 4th wire in the cable would be carrying only -5 vDC, instead of the whole shebang. The resistors, of course, would still be mounted within the footswitch housing.

The batteries are a better, and much safer, idea..... +1 to you for taking the correct way out. ;D


~!~!~!~!~!~
sooty,

Do what Chris says, you'll be much happier for it in the long run (the batteries and DPDT switches, not the -21 volt supply).
(Damn, wish I'd thought of that. )



sumgai

[JohnH]

I think putting batteries in with DPDT switches to run the LEDs is a great idea. Theres nothing extra to run from the amp, and two AA cells will last a long time. When they do go, nothing bad will happen, they will just gradually fade but the switch function will still work.

John

[mr_sooty]

I'd be prepared to give this a go, but I'd need someone to maybe draw up a diagram for me. You guys need to understand that I have very little know;edge of this stuff. I've doen the QTB mos succesfully on my Strat, and done an Indyguitarist mod on my Compressor pedal succesfully, so I can follow instructions easy enough, but I don't really know what I'm doing unless someone tells me.

So if someone would be kind enough to draw up a diagram I would be game enough to give it a shot, although if I wreck my switch I'm stuffed! Haha.

The only other concern is that my Footswitch lives on a pedal board and gets packed up in a padded bag with alot of cables and stuff, so there's a good chance it might get switched on accidently and drain the batteries.

[sumgai]

sooty,

The only other concern is that my Footswitch lives on a pedal board and gets packed up in a padded bag with alot of cables and stuff, so there's a good chance it might get switched on accidently and drain the batteries.

As John notes above, batteries used in this fashion (a simple LED) will tend to last a loooong time. Even if you're in the habit of packing everything up every time you quit playing for the day, then I don't think you're gonna have too much to worry about. (Unless you have little gremlins that come along and mash things down in that padded bag, hoping to cause some extra battery drainage........ ;D) Depending on space available, you can use either two single AA or AAA cells or a common 9vDC radio battery. Alkaline versions of either type should last you at least 200 hours of total usage (lighting one or both LEDs).

Li-Ion, or other rechargeable types, will not last nearly as long, and you'll eventually strip the screw holes from having to open up the casing so often to put in freshly recharged batteries..... not recommended for this purpose.

But as with all other things powered by chemical reaction and transduction, always carry spare batteries!

HTH



sumgai

p.s. It's Chris's turn to be in charge of making drawings, unless he passes the buckdelegates it to someone else. ;D

[mr_sooty]

9 Volt shared would be better, as I have to carry a couple of 9 volts anyway for various things. Wearing out screw holes isn't an issue, becouse the base of the footswitch is kind of rubbery plastic that jut pops off. But yeah, I really would need pictures. I didn't even know what a DPDT switch was!

[ChrisK]

p.s. It's Chris's turn to be in charge of making drawings, unless he passes the buckdelegates it to someone else.

Hmmm, sounds like a call to retaliation action.

Now, If'n those trepidatious a'planet are a'feared to do it my way (-21 VDC), we could use the chemical pile approach. I prefer 1.5 VDC cells since their energy density is higher and their cost is lower.

But, if you wants the 9 VDC battery (the 1.5 VDC CELLS ain't batteries [the term is attributed to Benjamin Franklin describing a brace of Leyden jars that looked like a "battery" of artillery]), I'll draw that up.

Now, since we'll be a'using the DPDT footswitch switch, and we only need 1/2 of a pole (well, SPST) to effect the amp function switching, that leaves at least one whole pole free.

Since I don't like wasting energy and don't like the 9 VDC battery, and, because I'm AFB ???and real dang clever to boot , I'll supply a diagram that switches the LEDs in SERIES thus cutting the power drain in half when both LEDS are a'LEDding (and, yeah, there's no current drawn when they're both off).


I'm unfamiliar with the foot switch switches afoot. Does anyone have pinout info on them?

Can you post some pics of your footswitch (the inside as well.....).
Measure the connections (with the amp not connected) when the switch is operated for all terminals on each switch. If you're lucky, there will already be DPDT switches in use.

Since you've ask for a specific wiring diagram (as opposed to that wonderful API known as a schematic), I'll need exact info to create an exact drawing.

In schematic form...

[mr_sooty]

Thanks for that Chris K. How come the one without the batteries looks easier than the one with the batteries? Obviously I'd prefer to not have to use batteries if it's possible for me to do it that way. Why is is riskier to do it your way (without batteries)?

Here's some pics:


And inside:


Red for reverb. As you guys said, one common wire, and one seperate wire each. The switches have two poles. Something to note maybe, The amp has to be switches to the lead (Overdrive) channel for the switch to work, so the footswitch seems to actually turn the drive off, rather than on, if you know what I mean. I'd prefer to have the light on when it's overdriven, but I could work around the opposite.

[mr_sooty]

Also, I just found this thread on Harmony Central: acapella.harmony-central.com/showthread.php?t=1255486&page=2 which has a schematic that they reckon works, no batteries required. What do you guys reckon?

[sumgai]

sooty,

I can't say anything nice about the posters in that particular thread, so I'll stay quiet. However, 'edster' linked to another H-C thread, here:

acapella.harmony-central.com/forums/showthread.php?t=1717180

Scroll down a little over half way, it's a large diagram by bluesmang. It uses a 9vDC battery, and it's wired with LEDs for both channels, labeled Clean and Dirty. If you want/need a LED for only one channel, then you can simply drop the 3PDT switch, and use the more common DPST switches. (Yours are SPST - they have one pole, not two.) If the idea of one LED per channel strikes you as a good one, then the Channel switch needs to be a DPDT.

The quoted resistor values are approximate, you can experiment up or down a bit in value. However, those values are coupled with the LED colors...... the 150Ω unit is probably a Green LED, the 470Ω part is probably a red LED, and the 100Ω piece is most likely a Yellow LED. Where you put each of the LEDs is your personal choice, of course, but if you start switching them around, be sure to keep the appropriate resistor with it's proper LED.

If the bottom comes off easily (no screws) then your worries about a wasted battery are null and void - simply disconnect it by hand, before stowing it for transit.

HTH


sumgai

[JohnH]

Chris - the version without the batteries feeds the signals for the overdrive relay and the JFET switch biasing through the new diodes, which will involve some voltage drop. Are you sure that that wont mess with the ability of the relay to operate, or the getting of the correct bias to the JFET to turn it off?

Sumgai - in reply 2, you noted using the 14V supply to power the LEDs. This seems to be only a low current reference voltage, since it is created just by two 100k resistors in a voltage divider, so maybe not a good source of LED power.

The 'no battery' design by Chris is cool if it works - subject to the above and if the LEDs dont blow (not a risk to the amp though). The battery version seems risk free.

BTW ive been building into things, the small blue LEDs, and get a nice bright light with just 2mA of current.

John

[mr_sooty]

I'm only after one LED for the Channel switch and one for the Reverb. On for drive and off for clean is fine by me. Red one's are fine too, but then so is any other colour. Green or blue might be quite cool, or it might be cool to have one colour for reverb and one for channel, but really, whatever's going to be easiest.

And I really appreciate the effort you guys are going to to help me find a way of doing this. It's awesome.

[mr_sooty]

I emailed Peavey about this, and they were pretty helpful actually. They said that this footswitch would work:

www.peavey.com/products/browse.cfm/action/detail/item/116308/6505(R)%20Footswitch.cfm

but they also said: "you'll need to reverse the red & black wires going to the LEDS."

Then I asked for a schematic and they sent me this:



Looks pretty basic.

[stratatouille]

M. MrSooty, I am agree - The Peavey customer service people is excellent. Okay.

M. Strat

[ChrisK]

How come the one without the batteries looks easier than the one with the batteries?

Uh, because it is. (It's got less stuff in it.)

Obviously I'd prefer to not have to use batteries if it's possible for me to do it that way.

Free (power) is.

Why is is riskier to do it your way (without batteries)?

Because some folks knees wax a'knocking at the possible involvement of voltages higher that 9 VDC (likely due to the disturbing childhood practice of testing 9 VDC batteries on their tongue - oh , ouch , dang , crap , ouch , dang , shiitake mushrooms , that WAS a good one). (I sincerely hope that it isn't from the grossly disturbing possibility of combining amps AND tongues.....)

From the perspective of UL standards, 27 VDC is low voltage since it is under 30 VAC/42.2 VDC and both of these signals are inherently current limited.


I proposed this initially since I actually know what I'm doing ..........(unfortunately, retribution will also be a three-way street )

(Now, you DO have to mind the LED polarity (in either my way or the battery cop-out). If you don't, they may/will rapidly "go to be with Jesus".......... Either an anti-parallel one (as I show) to limit the application of excessive reverse voltage, or a series one (to limit the application of excessive reverse voltage) is in order. I wouldn't call this a double-bagger, but I would "use protection" since you may not know where that amp has been.

The 'no battery' design by Chris is cool if it works

- subject to the above

and if the LEDs dont blow (not a risk to the amp though).

1. Yep, that's what I've been a'saying.................

(This is confirmed by the diagram link that mr_sooty posted.)

Chris - the version without the batteries feeds the signals for the overdrive relay and the JFET switch biasing through the new diodes, which will involve some voltage drop. Are you sure that that wont mess with the ability of the relay to operate, or the getting of the correct bias to the JFET to turn it off?

Sure as in absolute, no. Sure as in real confident, yep.

(This is confirmed by the diagram link that mr_sooty posted.)

The possibility that an additional 1.5 to 2.0 VDC of additional drop will cause a loss of function is slim since the gate threshold on the FET has a way wider range of tolerance, and to allow for this, Peavey would have designed significant margin in.

(This is confirmed by the diagram link that mr_sooty posted.)


The relay is a 12 VDC device in series with a resistor. the net effect will be the reduction of the applied coil voltage of about 1/2 of the diode drop (1 VDC total). Again, since tolerances come into play, relays will pick up at 70% or less of their rated coil voltage (one DOES have to mind the current ratings under reduced effected mechanical contact pressure, but this is out of scope here).

(Again, this is confirmed...................)

I routinely design OEM controls that apply reduced coil voltages to relays since the limiting factor for a relay (within contact rating scope) is the insulation temperature rating. One can reduce the heat generated in the coil to extend the ambient operational range of a product, quite handy when products must operate on rooftops in the tropics in the summer, in missile defense systems during the end of days, or on the surfaces of other planets. 100,000's of products shipped (single ones in some cases, and all long-lasting) that are based on such designs have proven the concept.

Red for reverb. [Black for channel] As you guys said, one common wire, and one separate wire each.

Yeppers.

The switches have two poles.

No. The switches have two terminals.
Each switch has one pole as in SPST.

The amp has to be switched to the lead (Overdrive) channel for the footswitch to work, so the footswitch seems to actually turn the drive off, rather than on. I'd prefer to have the light on when it's overdriven, but I could work around the opposite.

Simple is.

Since JohnH has alluded to low current LEDS (which are also available in red) a solution readily exists;

means

The clean/overdrive is switched by the relay.

The relay has a 12 VDC coil.

The coil is in series with a 470 Ohm resistor from -27 VDC.

When the relay is switched on, there is -27 VDC across the relay and resistor.

Therefore, the relay has aboot a 375'ish Ohm coil for 12 VDC across the coil.

Therefore, the current thru the relay coil and resistor is aboot 32 mA.

Did I mention that relays will turn on with 70% or less of rated coil voltage?

Relays won't turn on if the current thru the coil is real low.

Ergo'ski, at 2 mA thru the coil, the voltage developed across the coil will be about 0.75 VDC.

Turn on, it ain't gonna.


Soooooooooooo,

-27VDC/0.002 A = 13,500 Ohms

470+375 = 845 Ohms

13,500 - 845 = ~~ 13K Ohms

Connect the overdrive LED in series with a 1/4 Watt 13K Ohm resistor.

This is from the Black wire to the Tan common wire. The LED/resistor circuit can go across the two terminals of the clean switch.

This current (2 mA) will ALWAYS flow thru the relay coil and the 470 Ohm resistor in the amp.

To turn the clean channel on, the foot switch will short across the LED/resistor circuit.


Now, for the reverb channel, since 20 mA needs to flow thru the LED, you will have to use a regular LED rated for 20 mA continuous or add a parallel resistor to enable this LED to "glow long and prosper". (They used to say that LED's lasted forever. Then we discovered the optical semiconductor wearout effect. Now they say 100 years. Well, Sparky, LEDs ain't been around for 100 years, so..........................wishful is.)

/means

[ChrisK]

And I really appreciate the effort you guys are going to to help me find a way of doing this

Well, since we actually ONLY exist in netspace and hence have no real lives ourselves, we're always looking to poach upon humans (if someone doesn't respond, do we still exist?).







You ARE keeping an eye on your credit report as you converse with us....................................

[mr_sooty]

The main problem I'm having Chris is that your explanations are waaayyy over my head, and you lose me. Could you do a 'for dummies' version? Imagine its a recipe for fudge (I'm good at baking).

Ingredients: 2 x such and such an LED, blah blah

Directions: melt LEDs and add sugar etc...

I actually don't read the schematics very well, I don't know what IN4002 means for example. And I don't know what that 1/4 w 13K stuff means really. I need a 'do this, then do that.' 'cut this, then solder this to this' Then I can do it. But I'm game to try the non battery version. How can I get toasted from a footswitch that isn't plugged in to anything?

I'm sure I can put it back how it was if I stuff it up. also, can you tell which way around the led's are supposed to go by looking at them?

[ChrisK]

I don't know what IN4002 means for example

It's the part number for the diode, the symbol that it is nearest.

And I don't know what that 1/4 w 13K stuff means really.

It's a 13,000 (as in kilo) Ohm 1/4 watt resistor.

How can I get toasted from a footswitch that isn't plugged in to anything?

You won't get toasted, but the LEDs might if they're in backwards.

also, can you tell which way around the led's are supposed to go by looking at them?

Sure, if I could see them.
While the vast majority of same-sized LEDs share a common footprint, I have seen ones that were different or reversed over the years.

A Typical T1 3/4 size LED
The left lead is most probably the cathode.

A Typical LED Datasheet


Since you live on the upside-down side a'planet, can you tell me where you would get the electronic components needed?

If you post a link to their web site, I can help you select the components (and possibly get pics of them as well).

For instance, in the U.S. we have Radio Shack and such.


Also, do you have a digital multi-meter?

[mr_sooty]

I'd probaby go here: www.jaycar.co.nz/ or here: www.dicksmith.co.nz/cgi-bin/dse.storefront/. And no, I don't have one of them whojimmywhatsits. (cue banjos playing).

[ChrisK]

Note that in the first picture, both protection diodes (1N4002) are in series with the LEDs. If the diodes ARE in correctly (based on the polarity band on them) even if the LEDs are wired in backwards, they won't be damaged. The LEDs can then be rewired and all should work well.






Here's a resistor that is close enough in value, larger than needed, but in stock
Buy a few of these (spares are).

Here's a 1N4004 diode that's higher in rated voltage, but better, and in stock
Buy 5 of these (spares are).

This might work for the channel LED since it's bright and may work at 2mA
It may/will be too bright for the reverb channel. Buy a few of these (spares are).

Here is a relatively not bright yellow LED that might match (at 20mA) the high efficiency red LED (at 2 mA)
Buy a few of these (spares are).

LED bezel
Buy 5 of these (spares are).

Unfortunately, both sites are void of meaningful technical specs relating to forward current on LEDs. As a result I can't exactly choose an LED.

[mr_sooty]

Wow, that's awesome man, now we're really getting somewhere! A couple of questions:

Why again do we need the resistors and diodes? I just ask because the 5150 switch doesn't seem to have any.

What does 'cathode polarity band mean? Is this a marking on the diode that indicated which way around it goes?

While you say you 'can't exactly choose the LEDs', are you fairly confident the two you've suggested would work?

Lastly (for now) your drawing seems to suggest I can just use the existing switches, is that right?

Thanks again. (and thanks for picking the electronics with a branch in my area!)

[ChrisK]

Why again do we need the resistors and diodes? I just ask because the 5150 switch doesn't seem to have any.

Yes, the 5150 switch doesn't have these functions. It's just switches.
While this may seem like a lot,
This is what I consider to be complex, it fits into a guitar and doesn't even have the tone and volume controls shown.

We need the LEDs because.........................

We need the 13K Ohm resistor to limit the current thru the relay coil so it doesn't turn on when the switch is open. The LED will have about 2 mA flowing thru it when the switch is open, thereby providing illumination until the switch is closed, thereby switching channels as you requested. We could have just used the LED directly, but then the switch would only turn the LED on or off without ever "changing the channel".

We need the diodes to isolate the LED in the event that they are wired in backwards. LEDs are crappy diodes that glow. The 1N4002 diode can block reverse voltages up to 100 VDC, The 1N4004 blocks up to ?300 or ?400 VDC. The LED per typical data sheet can tolerate a reverse bias of only 3 to 5 VDC.

If you're absolutely certain that you can get the LEDs in properly (remember, I'm looking at the same tiny web picture as you and don't have any real data sheets), you can leave the diodes out and just bypass them with wire. I certainly could (because I have several of them there whojimmywhatsits and know how to use them), but would still use the diodes since I prefer to practice
"safe dioding" (something about LEDs being somewhat crappy diodes).

What does 'cathode polarity band mean? Is this a marking on the diode that indicated which way around it goes?

Yep. When the voltage at the cathode end is 0.7 VDC more negative than the voltage at the unbanded end (anode), current flows thru the diode. Otherwise it doesn't. It's like a flow check valve that only works one way. The water (electricity) flows from a positive source to a negativer (less positive - downhill) point. The 0.7 VDC is analogous to the need for some "water pressure" to "push the ball valve off of it's seat".


Now, before sumgai reminds that ALL of the electricity flow symbols are really backwards (and have been forever), we follow the convention that the current follows the arrows (that's why the diode schematic symbol is one). Electricity, the flow of electrons (why we call it electricity and electronics) actually goes from the more negative point to the more positive point. In vacuum tubes, the electrons "coaxed" from a heated cathode by the heater, have their flow to the (much) more positive plate modulated by the grid, thereby achieving current flow. This is why the Brits, who have a tendency to name things by what they do, call them "valves" (while I see the validity of round'abouts, ladders in stockings, and dual-carriageways, I have more trouble with "loo" [pronouncing a number - public buildings standardized on numbering the public toilet as room 100], and don't get "flats" as apartments and while the charge on a battery can be exhausted, only tires can actually go "flat". In the U.S. we have a "straw pole" while they have the "Aunt Sally").

While you say you 'can't exactly choose the LEDs', are you fairly confident the two you've suggested would work?

Yes, because the vendor sells LEDs sans actual manufacturer's part numbers, no one can know in advance what can be minimally expected from them. (This might be why we invented part numbers. )

Lastly (for now) your drawing seems to suggest I can just use the existing switches, is that right?

Now, that would've been clever of me, eh.

Thanks again. (and thanks for picking the electronics with a branch in my area!)

Sure, it's cheaper to ship "bins of bits" than "bits of kit".

[mr_sooty]

chrisk said:

your drawing seems to suggest I can just use the existing switches, is that right?

Now, that would've been clever of me, eh.

Bah! Riddles! So that's a 'yes' right? I don't need these DSP..D..er...AB... other switches you guys were talking about before.

[JohnH]

From Chris' battery-less version, it appears to be using your existing switches.

Good luck finding a 13k resistor at Dick Smith. !2k or 15k are more common.

John

[mr_sooty]

D'oh, those Dick Smith's links have stopped working properly, they're not linking to the products anymore. Chris....please......pretty please.....is it possible you could find the parts again and give me the shops item codes or something? Pleeeeaaasssssseeeee..........?

[sumgai]

sooty,

Just checking in late for a change (or rather, back to my usual habits ), I see that the DSE site is worthless, all of Chris's links go to the home page. Sigh, the miracle of modern programming, as envisioned by management that thinks monkeys grow in trees.

Anyway, here are the part numbers, just plug them into the Search field, towards the top of the site's home page:


13K resistor - R0206

1N4004 - Z3204

LED - Z4074

Hi-efficiency LED - Z4089

LED Bezel - H2030

HTH




sumgai