Maker Pro
Maker Pro

Home made speaker

Appie.123

Jul 7, 2015
24
Joined
Jul 7, 2015
Messages
24
Hi,

I'm a beginner on electronics, and I'd like to know I bit more about it.
Therefor I decided to make a simple homemade speaker. I followed some tutorials on this on youtube, but still it doesn't work.
I tried it on my phone, laptop and even on the tv but stil it didn't work.
I didn't use a amplifier as I first wanted to try if it works at all, and I thought that when connected to the tv it should at least produce some sound.
I used a coil made of +/- 3 m of insulated copper wire, and a quiet small, but strong, squared neodymium magnet.
I connected my speaker to the wires of a pair of old earphones.
The coil can move freely over the magnet.
The tree rubber bands serve as suspention.

Does anyone know what could possibly be wrong?
20150707_142858.jpg 20150707_142908.jpg
 

Attachments

  • 20150707_143031.jpg
    20150707_143031.jpg
    231.3 KB · Views: 124
Last edited by a moderator:

Gryd3

Jun 25, 2014
4,098
Joined
Jun 25, 2014
Messages
4,098
Hi,

I'm a beginner on electronics, and I'd like to know I bit more about it.
Therefor I decided to make a simple homemade speaker. I followed some tutorials on this on youtube, but still it doesn't work.
I tried it on my phone, laptop and even on the tv but stil it didn't work.
I didn't use a amplifier as I first wanted to try if it works at all, and I thought that when connected to the tv it should at least produce some sound.
I used a coil made of +/- 3 m of insulated copper wire, and a quiet small, but strong, squared neodymium magnet.
I connected my speaker to the wires of a pair of old earphones.
The coil can move freely over the magnet.
The tree rubber bands serve as suspention.

Does anyone know what could possibly be wrong?

View attachment 20803 View attachment 20803 View attachment 20804
Hook it up to a PC and use a function generator app to play various sounds.
Try using a round magnet (or confirm the poled on the square magnet)
Try using a smaller 'plate', as there might be too little current and too much mass to move the speaker.
 

Appie.123

Jul 7, 2015
24
Joined
Jul 7, 2015
Messages
24
I figured out that there simply were some bad connections, now it works

Thanks for your advice :)
 

Appie.123

Jul 7, 2015
24
Joined
Jul 7, 2015
Messages
24
I now would like to make it louder.
What is the best way to do this?
 

Gryd3

Jun 25, 2014
4,098
Joined
Jun 25, 2014
Messages
4,098
I now would like to make it louder.
What is the best way to do this?
Well.. best way to make it lighter is to increase the power to the speaker (up to a point)
You can also use a lighter material for the speaker cone, or build a small enclosure.

Remember that you are using electricity to physically move the speaker cone. The more it can moves, the louder it will be.
 

Appie.123

Jul 7, 2015
24
Joined
Jul 7, 2015
Messages
24
I was also thinking about a amplifier, I wondered if this was realistic to make yourself
 

Gryd3

Jun 25, 2014
4,098
Joined
Jun 25, 2014
Messages
4,098
I was also thinking about a amplifier, I wondered if this was realistic to make yourself
Absolutely is!
The trick here is determining what kind you want to build :p

They can be built with some simple transistors, or you can use mosfets.

Class A, B, and A-B are linear and are pretty easy to build. They will get warm/hot pretty quick depending on what kind of power you are putting out.
Class D is digital and switches the output on/off incredibly rapidly. More efficient, but complicated.

Give this a quick read and see what kind of ideas or questions you may have.
http://www.talkingelectronics.com/projects/TheTransistorAmplifier/TheTransistorAmplifier-P1.html
The author can be a little 'forward' at times, but generally the knowledge is there.
 

Martaine2005

May 12, 2015
4,953
Joined
May 12, 2015
Messages
4,953
I think you should mount your moving coil in this,
and mount your plate on top. Cut the plate slightly smaller than the diameter of the bowl and tape it loosely in place.
You will notice an instant difference.
GENPAKFW032.JPG
 

Appie.123

Jul 7, 2015
24
Joined
Jul 7, 2015
Messages
24
That indeed is a good idea,
thanks.
Nevertheless I think I'd need an amplifier to get a 'decent' volume.
I had a look a the side of the previous post, but I wondered why it could not be made even easier (probably this is a real stupid idea, but as I said I don't know much about electronics):
Schermafbeelding 2015-07-09 om 8.49.25 PM.png

The non amplified signal determens the amount of current comming form the battery and therefore the signal coming form this battery. Would this work?

Thanks for
 

Gryd3

Jun 25, 2014
4,098
Joined
Jun 25, 2014
Messages
4,098
That indeed is a good idea,
thanks.
Nevertheless I think I'd need an amplifier to get a 'decent' volume.
I had a look a the side of the previous post, but I wondered why it could not be made even easier (probably this is a real stupid idea, but as I said I don't know much about electronics):
View attachment 20826

The non amplified signal determens the amount of current comming form the battery and therefore the signal coming form this battery. Would this work?

Thanks for
Almost but not quite... the unamplified signal will swing above and below 0V... So the transistor will only play the 'top' half of the sound.
Additionally, if there is no control over how the transistor is biased, it could very well 'saturate' turning the nice rounded signal into a squarish signal.
That said, it should make noise, even if the noise sounds bad!
 

Appie.123

Jul 7, 2015
24
Joined
Jul 7, 2015
Messages
24
Okay, I think I get it a bit...

So to solve the first problem I could add an NPN transistor as weel as the PNP, so that when the signal is possitive, the PNP transistor works, and when the signal is negative the NPN transisitor will take over.

To bias the transistors, I gues i should simply add a resistor before the base of the transistors.

it would then look something like this:
Schermafbeelding 2015-07-11 om 10.34.23 AM.png
Thanks for your help :)
 

Gryd3

Jun 25, 2014
4,098
Joined
Jun 25, 2014
Messages
4,098
Okay, I think I get it a bit...

So to solve the first problem I could add an NPN transistor as weel as the PNP, so that when the signal is possitive, the PNP transistor works, and when the signal is negative the NPN transisitor will take over.

To bias the transistors, I gues i should simply add a resistor before the base of the transistors.

it would then look something like this:
View attachment 20870
Thanks for your help :)
You're getting it, slowly but surely ;)
I'd like to point you to this page : http://www.electronics-tutorials.ws/amplifier/amp_6.html
It goes over some very simple Class-B circuits and Class-AB.

The most recent drawing you have made will.. make some noise :p but if you want some more clarity, you should eliminate the resistor on the signal input and use a potentiometer instead. Signal in one leg, middle wiper to the amplifier and the last leg to ground. This will give you a volume control that you can adjust to prevent the transistors from 'clipping' when they become saturated.
The other catch here... is that the transistors require a certain amount of voltage before they will even begin to turn on... so there will end up being a flat part in the output amplified signal.
The link I shared will provide some more details about why, how, and what to do about it. Take a look and let me know ;)
 

Appie.123

Jul 7, 2015
24
Joined
Jul 7, 2015
Messages
24
I do get the problem, but I still have some questions:
- why would I use just a potentionmeter? This indeed is usefull to control the volume of the sound, however turning the sound to loud might saturate the transistors. So wouldn't it be better to have one resistor making sure that the transistors will not saturate at the lowest resistence of the potentiometer, and then add a potentiomenter to control the volume?

I did look at the link you send, but I didn't really understand their circuit. This is probably because they use two rails instead of a battery, I'm not really used to this type of notation yet.
However, I did make a new (improved circuit).
20150713_132634.jpg

In this circuit the values of R1 and R3 shoud be the same to give both transistors the same 'lift' in input signal.
I had to create a negative signal to add to the input signal of the NPN. I didn't really know how to do this, so I came up with a quite complicated sollution. I wondered if there's something easer to do to create a negative signal.

Also i realised that this signal has one impurity; The extra voltage I add (or substract) form the imput signal to overcome the lack in voltage in some parts of the curve, is absolute and not relative to the curve, causing the shape of the curve to change. Also it kind of rips apart the curve like shown in the picture below the circuit.
The way to make it relative is to add an other transistor, but this has the same lack in voltage in some parts of the curve.
Meaning this is an infinite problem....:confused:
Is their any way to solve this, or is this a not so significant problem?

And I had one more question:
When connecting your circuit to the ground this is often illustrated with the tree stripes, but what does this look like in real life.
A treat toutching the ground litteraly is namely not so practical .... :p
 

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
You are definitely getting there. A couple of comments, though.

If you use two diodes to replace R2, it will improve the circuit.

And this is a current amplifier with no voltage amplification. Depending on the characteristics of your speaker, that may or may not be the right thing.

Bob
 

Gryd3

Jun 25, 2014
4,098
Joined
Jun 25, 2014
Messages
4,098
I do get the problem, but I still have some questions:
- why would I use just a potentionmeter? This indeed is usefull to control the volume of the sound, however turning the sound to loud might saturate the transistors. So wouldn't it be better to have one resistor making sure that the transistors will not saturate at the lowest resistence of the potentiometer, and then add a potentiomenter to control the volume?
Well.. yes and no. Both are valid solutions. The reason I put a potentiometer here is because it's much easier to adjust the 'volume' knob than to pull out a resistor and put a new one in. Personal preference though, both solutions will work just fine :)

I did look at the link you send, but I didn't really understand their circuit. This is probably because they use two rails instead of a battery, I'm not really used to this type of notation yet.
However, I did make a new (improved circuit).
It's easy enough to get the hang of, and it looks like you have a decent handle on it already.

In this circuit the values of R1 and R3 shoud be the same to give both transistors the same 'lift' in input signal.
I had to create a negative signal to add to the input signal of the NPN. I didn't really know how to do this, so I came up with a quite complicated sollution. I wondered if there's something easer to do to create a negative signal.
Well... two things, R1 and R3 most certainly will be the same, but the value also depends on R2. The goal here isn't to give them the 'same' lift, but a specific lift of roughly 0.7v on each transistor... (So R2 would need to have 1.4v across it)
You don't actually have to do anything here regarding making a signal negative. The signal current is centred at 0v and swings positive and negative by itself. There is a bit of a flaw in your current drawing though... you are missing the capacitors used in the Class B Transformerless and Class AB designs, this is important or you will be shorting out R2 and both transistors will no longer be lifted.

Also i realised that this signal has one impurity; The extra voltage I add (or substract) form the imput signal to overcome the lack in voltage in some parts of the curve, is absolute and not relative to the curve, causing the shape of the curve to change. Also it kind of rips apart the curve like shown in the picture below the circuit.
The way to make it relative is to add an other transistor, but this has the same lack in voltage in some parts of the curve.
Meaning this is an infinite problem....:confused:
Is their any way to solve this, or is this a not so significant problem?
There sure is ;) The thing is, it's already taken care of with the class AB design and a properly tuned Class B on the link I shared.
You see... the voltage you add or subtract is setup so that you add 0.7v to the top transistor, and take 0.7v away from the bottom transistor. This means that both transistors are on the brink of conducting and all it takes is a 'tiny' amount to push it over the edge to conduct. As soon as the signal goes positive what happens?
The top transistor conducts right away, and the bottom transistor stops conducting completely. The problem you are thinking of would only apply if you didn't setup the 'float' voltage to the transistors first... or if you applied the wrong float voltage to the transistors. Remember the voltage is added/subtracted for each transistor... NOT the signal itself.

And I had one more question:
When connecting your circuit to the ground this is often illustrated with the tree stripes, but what does this look like in real life.
A treat toutching the ground litteraly is namely not so practical .... :p
Haha. that little tree icon is commonly called 'ground' but there is nothing so special with it.
The best way to think of 'ground' is how it's used. For example, in a car the 'ground' is attached to the body of the vehicle. This ground is almost always attached to the negative side of the battery, but some cars actually use the positive instead. Both options are fine, and both options use the car body as a 'common' return path. So they only really need to run one wire from the battery to something, then they use the body to go back. In things like the amplifier you are building, the 'ground' is considered to be the 'common' path. You have a pair of batteries making the positive and negative voltages, and as you had drawn the 'ground' is in-between them. To actually build it, the ground is simply a shortcut when drawing the circuit and can be replaced by drawing a line to connect each ground symbol to each other.
Now... that said, some 'ground' connections have more meaning... like in home appliances. The ground it actually connected to 'Earth ground', which is also connected to the body of the appliance. The reason here is that if something goes wrong inside the appliance, the current will go to the ground wire instead of going through a person who touches the appliance. This only works because the power plant also has an 'Earth' ground... so much like the car body being used in the previous example, our Earth can behave the same way as a return path for electricity back to the power plant. This is only used for emergency purposes though because earth is not a perfect conductor.
(Remember that electricity flows in a circle... so a battery operated devices will almost always gain nothing at all from using any kind of 'earth' ground.)
 

Appie.123

Jul 7, 2015
24
Joined
Jul 7, 2015
Messages
24
So should add a capacitor, but according to the link you send, two capacitors are needed, wereas one capacitor will block the way for the 'static' voltage, right. Why would I use two.
To purify the sigal and make sure only the 'waves' are comming through?

There sure is ;) The thing is, it's already taken care of with the class AB design and a properly tuned Class B on the link I shared.
You see... the voltage you add or subtract is setup so that you add 0.7v to the top transistor, and take 0.7v away from the bottom transistor. This means that both transistors are on the brink of conducting and all it takes is a 'tiny' amount to push it over the edge to conduct. As soon as the signal goes positive what happens?
The top transistor conducts right away, and the bottom transistor stops conducting completely. The problem you are thinking of would only apply if you didn't setup the 'float' voltage to the transistors first... or if you applied the wrong float voltage to the transistors. Remember the voltage is added/subtracted for each transistor... NOT the signal itself.
What exactly do you mean by a 'float voltage'?

If you use two diodes to replace R2, it will improve the circuit.
Diodes also have a sertain voltage before they start working, right. This
 

Appie.123

Jul 7, 2015
24
Joined
Jul 7, 2015
Messages
24
would mean there would still be a 'gap' in my signal curve, right?

Also, how do I know what my speaker needs?

I appreciate your help! :)
 

Gryd3

Jun 25, 2014
4,098
Joined
Jun 25, 2014
Messages
4,098
So should add a capacitor, but according to the link you send, two capacitors are needed, wereas one capacitor will block the way for the 'static' voltage, right. Why would I use two.
To purify the sigal and make sure only the 'waves' are comming through?
Nope. Take a look at the designs in the link I shared.
There are two capacitors...
The top transistor will work best if there is +0.7v on it 'before' you add the signal. This way, the signal does not have to go up to 0.7v to trigger the transistor, it's already there!
So a capacitor is used here to couple the AC signal with the DC offset.
... but the bottom transistor needs to be at -0.7v before you add a signal for the same reason. This is where the second capacitor comes into play. The input signal branches off in two directions, to two different transistors that have a different DC offset... If you only use one capacitor, then the transistors will no longer have their own DC offset, and your signal will get chopped off if it's too close to 0v again.
What exactly do you mean by a 'float voltage'?
I may have used the wrong term here... It's actually a DC offset. Currently, a transistor will not turn on until about 0.7v ... so the top transistor will miss the signal while it's between 0v and 0.7v... the solution is to add an offset so that the signals 0v is raised up to 0.7v. Problem solved... but wait, this means that the bottom transistor will miss the signal from 0v to -0.7v ... but because we added that offset, it's now going to miss an additional -0.7v of signal. This is the reason we use two capacitors and branch the signal off... The top transistor has a +0.7v offset, and the bottom transistor had a -0.7v offset.. the pair of capacitors stops the two different offsets from interfering and allows the signal to be coupled with the offset.
Diodes also have a sertain voltage before they start working, right. This
That's right. We are taking advantage of that to get just the right voltage offset for your transistors. This can be done with specially chosen values of resistors too, but a diode is much easier.
 

Appie.123

Jul 7, 2015
24
Joined
Jul 7, 2015
Messages
24
I think I get it now, the circuit in the link you send also makes a lot more sense.
Although the part of the capactors is a little vague to me...
I get that without them there would be now/ very little difference in voltage between the two diodes, but why wouldn't I use diodes on the same place as the capacitors instead?

Besides this I think I can start building my amplifier. :) When this works as well, I might make speaker in full size, for 'real use'. The only thing I would then add would be sth. to split the high and low frequencies so that they can be linked to a bigger and smaller conus.
This however will come later, first I want to built the amplifier, in the mean while I'll try to find out more about it.
I'm on holiday now, making it hard to built my amplifier, in a few weeks I'll be home again, and I'll built the amplifier.
So in a few weeks, I might ask something again.
 

Gryd3

Jun 25, 2014
4,098
Joined
Jun 25, 2014
Messages
4,098
I think I get it now, the circuit in the link you send also makes a lot more sense.
Although the part of the capactors is a little vague to me...
I get that without them there would be now/ very little difference in voltage between the two diodes, but why wouldn't I use diodes on the same place as the capacitors instead?
A capacitor blocks DC voltage, but lets AC voltage through, using this property, we can join almost any DC and AC voltage together. Like 0.7V DC to bias the transistor, and the (most likely) 1V p-p AC audio signal. Now you end up with a signal that goes from 1.7V to -0.3V . This is called a DC offset.
If you use a diode, this won't quite work as planned... for two reasons.
A) A diode has a forward voltage drop like a transistor... so you would end up loosing 0.7V of your signal across a diode :(
B) A diode requires 0.7V 'higher' on one side compared to the other to even conduct... so if you had a 0.7V bias... the signal would get 0.7V chopped off due to the voltage on the other side of the diode, and ANOTHER 0.7V chopped off due to the diode forward voltage :(

Capacitors don't care... there could be 100V DC one one side, and it would still let you couple a 1V AC signal to it. The important thing here is that the voltage rating is higher than the combined voltage of the signal and DC offset, and that the Farad rating is sized for an appropriate cutoff frequency. (Smaller Farad ratings cut off higher frequencies, so if you pick one that's too small, your amplifier output may end up with too much treble)
The link I provided further up also shows a way to build the amplifier with a transformer... the transformer or the capacitor method will both allow you to put an AC and DC together... old phones used the transformer method ;)

Besides this I think I can start building my amplifier. :) When this works as well, I might make speaker in full size, for 'real use'. The only thing I would then add would be sth. to split the high and low frequencies so that they can be linked to a bigger and smaller conus.
This however will come later, first I want to built the amplifier, in the mean while I'll try to find out more about it.
I'm on holiday now, making it hard to built my amplifier, in a few weeks I'll be home again, and I'll built the amplifier.
So in a few weeks, I might ask something again.
Sounds good. Keep asking away and we can see what we can do for you :)
 
Top