liltimmy

Hi, im fairly new to the whole electronics thing but I have a couple of questions.

This is to do with two electronic bass pedals:

This schematic shows the circuit needed. Note that it requires its own 5V power supply - one based on the simple LM7805 should be sufficient and can be powered from a 9V battery.

You should be at least somewhat familiar with electronics to assemble this. If you don't know what a 5V power supply is or why you would need one, you're going to have trouble if you try to make one of these.


"Make sure you get the 74HC14 Schmitt-Trigger Hex Inverter. The 7404 will not work. All of the diodes can be pretty much any kind. The 5V relay should be normally-open and should be non-latching. The transistor is a 2N3904, but any garden-variety NPN BJT should do.

What happens is that the closing of either input switch (i.e. either pedal) causes the output (the relay) to close for a short time, then re-open - regardless of whether the input pedal is still down or not.

This allows pedal #1 to be pushed down to trigger a Bass Drum hit, then while pedal #1 is still down, pedal #2 can be hit to trigger another Bass Drum hit! When working both pedals at once one after theother, this allows for some very fast bass drumming."



That's the circuit diagram and here are my questions:

1. What should I use where it says "jack"? I know I need two female jacks. The pulses are sent via a regular 3.5mm mono jack but im not sure how I would connect the jack to the leads leading into the breadboard.

2. What is a 5v relay? and where would i buy 1 (bearing in mind it has to be non-latching and normally-open)?

3. Similar to question 1, how would I wire the male audio jack that would be connected to the output.

DonaldG

Here is the relay

This is the male jack that is on the output side (right of the cct)

This is the female jack that you need 2 off

These are the jacks if you want to mount the input jacks to the circuit board.

All other components can be ordered from Maplins here


EDIT: it will become obvious how to connect the jacks - there are only 2 connections.

Cheers.

liltimmy

I have bought some female jacks off ebay, similar to this http://www.maplin.co.uk/Module.aspx?ModuleNo=1148, but it is plastic instead.

I can see two contacts (i think one branches into 2 so i counted that as one) if i look deep in the hole, but how do i solder into the jack without cutting it open?

I presume that I want to buy this product (out of the available) on your link:

1A DPDT 5Vdc N05AW £1.91

seeing as it is 5Vdc?

And I forgot to ask, next to the box labelled 5v relay, I see there is a diode in parallel, do I add that? or is that part of the relay?

DonaldG

The barrel on the jack will probably just unscrew.

Yes, you will need to buy the diode.

If you look at the specification of the relay, you will note that it is a 5 VDC one.

liltimmy

Ok thanks a lot for your help!

2 last questions :) :

1) On a breadboard to put something in parallel do I put the two ends of the diode on the same numbered line as the 5v relay?

2) Finally, which end do i put the black end of the diode, negative or positive?

Seeing as this is my first project on a breadboard that I hoping will work, I will post a photo up if I cant get the project to work.

DonaldG

This link will help you get everything sorted out.

Not only will you be able to identify components, you will also be able to understand a wee bit more on how it works too.

liltimmy

Thanks very much!

Will post back my results when I recieve the relay

liltimmy

http://www.maplin.co.uk/Module.aspx?ModuleNo=37494
1A DPDT 5Vdc N05AW £1.9

I'm confused by the markings on the underside of the 5v relay.

2 pins are clearly related to the coil.

the lower 6 look like this:

DonaldG

The relay that you have is fine for what you need. Ignore your drawing for a moment.

A relay is an electrically controlled switch. When electricity flows through the coil, the switch goes on or off depending on how you wire it in. The relay that you have has two such switches and are independent from each other. It is known as a double pole, double throw type. (DPDT) You will only need one of them. This means that you can ignore half of the pins.

For the purpose of explaining, I will only refer to the pins on the left side of your drawing. (The pins on the right act in exactly the same way and at the same time.)

TL = Top Left
ML = Middle Left
BL = Bottom Left
(Look carefully, you should actually see that the pins are numbered but TL, ML & BL will put you on the right track)

The ML pin is the common pin, sometimes called the 'wiper'. It is this pin that either connects to TL or BL depending on if voltage is flowing through the coil.

My guess, from your drawing, is that when there is no voltage, ML and BL is open circuit (switched off) but ML and TL are closed circuit (Switched on)

when voltage is applied to the coil, ML and TL become open circuit (switched off ) and ML and BL are closed circuit (switched on)

Therefore for you application, I suggest that you use only the two pins ML & BL. Alternatively you could use the right hand pins MR & BR...

Cheers

liltimmy

First of all, I want to say that I greatly appreciate the time your taking to explain this stuff to me.



Because I'm not sure what contact on the jack I have to connect each wire too heres an image of one I have.

If you look carefully there there is a dual contact. Lets call this *A* so I know what you are referring too.

and

then a longer single contact that curves to make a circle. Lets call this *B* so I know what you are referring too.

Sorry if this seems really obvious but I'm not sure which is which.

So using *A* and *B* can you please explain which contact I have to connect to the 5V and the 10K.

Oh and also could I use a 100K or a 1K or 270 ohm resistor instead of the indicated 10K?

Thank you very much!

DonaldG

The jack that you have is a stereo one. The 'dual' contacts are two seperate contacts. In effect you have three contacts and must use the correct two. Using the wrong two, you could/will blow or burn out the 5v power supply.

The long arm is the common or 'Ground' contact. I suggest conecting he 10K resistor and IC to that. The other connectrion should be on one of the two stubby cionnection. The one to use is the one that has the contact NEAREST to the BACK of the barrel. You will need to look inside the barrel to try and figure out which one.

If the circuit calls for a 10K resistor, then that is what you must use.

liltimmy

Then I am very glad I had the foresight to buy a second supply, knowing I would do something like this.

I dont understand, both contacts are the same height.

liltimmy

Oh! I understand what you mean now, There would be three contacts for the ring tip and sleeve, and the two smaller contacts are ring and sleeve. And because the ring is further away from the barrell when you put it in, I want to avoid that and use the other one.

I've noticed that when I was soldering I managed to dislodge the contact and move its position hence why I couldn't see the difference in height.

Will post back if I have any more problems. Thank you!

liltimmy

I've been re-reading what you said and I realize I don't really understand.

Does that mean I should connect the bottom left pin and the middle left pin to the jack?

And if that is the case...

Which pin ML or BL would go to the sleeve/tip?

DonaldG

YES - the ML & BL

It doesn't really matter which pin goes to the tip. It is only acting as a switch. Polarisation does not matter

liltimmy

Ok.

I hooked it all up and plugged it in but I made two fatal errors.

My diode was facing the incorrect way in parallel to the relay and I smelt burning (maybe the transistor)?

I plugged 9V in instead of the maximum 6v.

When I completed the circuit when I put in the battery, I heard a single click as it touched the breadboard. Would that be related to the voltage?

DonaldG

You have learnt a major lesson that we all go through when starting to experiment with electronics: Always, without exception, check that all components are the right value, in the right place and the right way round.. Then check the wiring/breadboard. Check all solder joints for dry joints or for too much solder that can short out to a nearby component or track.

Failure to do that will result in a failed unit.



The diode and transistor will probably be shot. Replace them.

The click you heard was probably the relay switching as the diode blew up and the current went through the relay coil instead of a diode the wrong way round.

The next worry is that you applied 9v that also went to the input of a device that should only see 5v - that means that the IC 74HC14 could also be shot too. I suggest replacing that too

When you do replace them, remember - Check everything twice...