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# simple intercom ?

Discussion in 'General Electronics Chat' started by Mathematics!, Apr 16, 2009.

1. ### Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
4
I am makeing a simple intercom from a microphone to a 8 ohm 20 watt speaker. All I want to do is talk into the microphone and hear it out the speaker.

I am using a 9 volt battery , the speaker , the microphone (condensor) , and some 20 gage wire. I know I need also a capacitor to transformer the DC to AC for the speaker.

Correct me if I am wrong
I believe since the speaker is 8ohm 20 watts. I need sqrt( 200 /8 ) = 5 amps to the speaker to have it operate.

I also need to transform the DC supply voltage to AC voltage to the speaker.
The problem is How do I bump up the current and what capacitor do I use with the microphone to produce the AC voltage to the speaker?

I am a little shake on how I would go about bumping up the current.
I know you can put a ton of batteries/capacitors in parrell .
But is their a better way other then using an amplifier chip or transistor's.

I want it to be the simplest thing.

I have tried a 9 volt battery with 2 old telephones and it worked fine Ofcourse Those speaker and microphone pair had a ampifier circuit made for bumping up the current and makeing the AC to the speaker ,...etc

Thanks for any help

Apr 5, 2008
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3. ### Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
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Yes , but my problem is I want to understand the circuit completly so I can design my own.

What I am not getting is why you need all the capacitors , resistors ,...etc.

Shouldn't the microphone produce varying DC electrical current with variations in the millamps. If this is true then all that would have to be done in theory is convert the varying DC current into varying AC current that the speaker can play?

So I would think the 9 volt battery and the microphone already produces the electrical sound signal all that is need is to boost it up so the speaker can move enough to produce the sound loud enough to hear.

But I don't know how much I should amplify the current and I don't know what the best way to produce AC for the speaker. I am assuming with capacitors but how do you know which kind and how do you know how much amlification is need to hear the sound thru the speaker?

Thanks

4. ### mbohuntr Senior Member

Apr 6, 2009
430
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You might wish to look up coupling and de coupling theory. Basically your voice is creating a sinusoidal wave that is moved through the transistors. The transistors need DC to be correctly biased, The speakers are isolated from the DC component by the capacitor., the sinusoidal signal is carried through and perhaps amplified by the BJTs and then de-coupled by the other capacitor, giving the other speaker a clean signal. Everyone...Feel free to add or correct anything I missed, I am still learning.

5. ### Audioguru Expert

Dec 20, 2007
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The output of a microphone is 0.01V with a current of 0.0000005A.
A speaker is fairly loud with a power of only 0.5W then an 8 ohm speaker needs 2V at 0.13A.
A capacitor is not an amplifier. A transformer is also not an amplifier and will not work with the DC from a battery.

You need an amplifier to "bump up" the 0.01V to 2V and "bump up" the 0.0000005A of current to 0.13A.

6. ### Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
4
Wouldn't it be P = I^2 * R which implies I = sqrt( P /R )

This gives I = sqrt( 0.5W/8ohm ) = 0.25 amps
And since V = I* R imples 0.25 amps * 8 ohms = 2 Volts

So the current should be 0.25 amps not 0.13 amps?

I understand that capacitors are not amplifiers but I would think their purpose would be to create AC current for the speaker. I believe speakers need AC to operate but correct me if I am wrong? Because I believe speakers work on the change in alternating current induces a magnetic field which drives the speaker back and forth to produce the vibration of air
molecule which is the sound we here. But maybe a speaker can operate on only pulseating DC?

How do you know the microphone is 0.01V and 0.0000005A is this some kind of standard?

Anyway with all this said I should be able to use just an battery , microphone , wire , speaker and something to amplify the current so the speaker can play it and you can here it? From this I am having a hard time figuring out why we need so many capacitors , resistor , and transistors to do it.

I know 2 or more transistors can be used to amplify but why so many resistors and capacators . And why 10kohm resistor ,..etc as opposed to 100kohms or .000003 ohms whatever. I don't understand what math determines the values of the components to use and why we need some of the components to amplify?

Thanks for any help.
My many purpose is to understand the design anybody can build it on a solderless breadboard but it takes more skill to understand the purpose of each component.

I am fully aware of capacitor's in parrell and in series calculations maybe this stuff has something to do with the values used ? I cann't see it though

Last edited: Apr 16, 2009
7. ### Audioguru Expert

Dec 20, 2007
11,251
1,350
You are correct and I was wrong. I wrote 0.5W but calculated the current for 0.25W.
0.5W is only a little louder than 0.25W. An intercom is used within arm's reach so you don't need much power.

A capacitor blocks DC and passes (does npt amplify) AC. An amplifier is needed.

Look at the spec's for an electret microphone. Its output voltage is 10mV and into a 10k load is only 1uA. Again my math was wrong before.

What is the battery for? To power the missing amplifier. A battery is not an amplifier.

It is an intercom amplifier. The LM386 IC has many transistors inside. The circuit shown is missing 3 important parts.

You need to learn about electronics to see how an amplifier circuit or any other circuit is designed.

8. ### Wendy Moderator

Mar 24, 2008
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I've built a few myself, when I was beginning. AG is right, you don't need much power, this isn't a bullhorn. The biggest problem I had was radio, those long wires make good antennas (or were you planning on going wireless?).

Mine used the speaker as a mic, with a simple spring loaded toggle switch to select to transmit.

A master station can have the amplifier, with the satellite stations are plain speakers. This is what schools use (or used to, at least).

9. ### Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
4
Ok , I will study up on op-amplifier circuits but in the mean time could I use a LM386 IC from radioshack and just connect the microphone to the IC amplifier chip then connect the speaker to the output of the chip?

I thought that you need a battery to power the microphone as well as my missing amplifer? Or will the microphone produce a variation in electrical current not hooked to anything down it's to legs?

Also does the speaker work on pulseating DC or AC ?
I beleive the capacitors are used to create either the pulseating DC or AC need to power the speaker. But I am unsure why you would choose one capacitor over another what determines what capacitance to use in the circuit ? Obviously smaller capacitance will full charge quicker then larger.
But how do you know how fast you want it to charge at?

As for using amplifier chip's from a radioshack how do you know how much gain you will have from the chip some chips don't specify this info they give all this differential characteristics stuff?

10. ### Wendy Moderator

Mar 24, 2008
21,838
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A speaker and mic next to each other? Sounds like a recipe for acoustic squeal. There is a reason I used the speaker for both mic and speaker, they would never crosstalk that way.

11. ### Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
4
Ok ,but is the mic no different then a speaker?

Do they both work in both directions.

Like a mic works by when you talk into it it vibrates a coil of wire around a magnet to produce a changing current in the wire. ( i.e from physics a change in magnetic field produces a current in a wire )

So you need electrical current to flow thru the coil to produce the magnetic field which moves back and forth from the magnet it is around these to magnets changing produces the extra current in the wire. When it produces current in a wire I would think this current is AC since the magnets are moving back and forth.

Similarly for the speaker you need a changing current to go thru the coils of the speaker to induce a magnetic field which is attracted to the magnet in the back of the speaker. This will move the speaker back and forth producing sound. (i.e it uses the principle of a changing electric field produces a magnetic field , the opposite principle of the mic).

So when I connect the mic ---wire--- speaker the only thing that is missing is an amplifier and a battery to power it. So I would think if I got a amplifier chip and hooked it up between the mic and speakers. It should boost it up enough so I can hear it. Does amplifiers boost both voltage and current or just current? And if this is true then why all the capacitors ,transistor , resistor, etc Could I by pass using all these just by getting amplifier chip instead of making an amplifier circuit?

12. ### Wendy Moderator

Mar 24, 2008
21,838
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An electret mic isn't a speaker. A speaker is bidirectional however, it can be both (though not at the same time).

Even amplifier chips need support components. We're only talking complexity here.

13. ### italo New Member

Nov 20, 2005
205
2
start with lm386 1w 20w will drive me out of the house use 100 Ω speaker for input- output no microphone this is an intercome not a hi-fi one DPDT switch and the design is done. if you use 9v battery one " hello" will eat the battery up with 20w. simplicity is the mother of all virtue. get with the program.

14. ### Audioguru Expert

Dec 20, 2007
11,251
1,350
An electret mic is the most popular one today. 40 years ago the dynamic mic was the most popular. It has a coil and magnet like a tiny speaker and generates a very small voltage and current.

Yes, for a dynamic mic or for a speaker that is used as a dynamic mic.

No.
A dynamic mic generates a small voltage and current. You do not give it current.

No.
The coil is moving back and forth. The magnet does not move.

An amplifier boosts the signal voltage and the signal current.
The amplifier chip needs some extra parts for it to work.

15. ### Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
4
Ok , I am studying the op-amp's but I don't get why you would use the inverted terminal or non-inverted terminal.

In the book they say you can think of the input terminals as having 0 volts and zero current. (open circuit) Because the internal resistors are megohms
But I don't see why you want a large internal resistance in an op-amp.
Thought the whole point was to speed up/bump up the current/voltage.
Resistors only drop voltage or current?

And they say A( v+ - v- ) can be assumed to be inf * ( v+ - v -)
How is this possible if the open-loop gain was assumed inf then we would have an inf voltage/current gain unless we are talking about the undetermined form inf * 0 but this isn't very helpful in analysis?
Plus I would believe that an op-amp cann't amplifier voltage more then the supply battery can supply. Conservation of energy still always must apply.

Also they talk about the output resistance of the amplifer to be as large as 75 ohm or more and so may not be negligibly small. Where is this resistance coming from and what the heck are they talking about when they say they can reduce it to be negligibly using negative feedback?
Why would you even want to provide an output resistance with a resistor internal or something.

For example I am having a hard time getting the inverting amplifier.
They have it with Ri , Rf , RL , vo , vi as varibles.
The + input terminal is connected to the - op-amp input
with Rf being a resistor runing from the input to the output of the op-amp.
The rest is grounded and the output also has a RL resistor before ground.

I know how to use KCL but how do you know which way current is flowing in Rf is it back to the input or forward to the output terminal of the op-amp.

Obviously I know how they derived that vo = -Rf/Ri * vi but this is assuming current is flowing into the - terminal of the op-amp from both the Rf line and the Ri line. How do they know which way current is flowing in the Rf line it is attached to both input and output of the op-amp. So I would think path of least resistance. As the current pass the Ri resistor won't it flow initially into the Rf line faster then it would flow thru the internal megohm resistor. If this was true KCL would give
vi/Ri - vo/Rf = 0 not vi/Ri + vo/Rf .

Hopefully somebody can clear my issue up.
I am using the schaum's Basic Circuit Analysis book if that helps chapter 6.

As well once current flows out the output of the op-amo won't it try to flow back into the input so I am confused on which way it goes. AHhhhhhhhhhhhhhhhhhh

Last edited: Apr 17, 2009
16. ### Wendy Moderator

Mar 24, 2008
21,838
3,047
Their really isn't any difference between the + and - inputs in the op amp, they are primarily used for feedback purposes, ie, controlling gain. Other uses for feedback is frequency response.

You can also vary the input impedance of an op amp by picking the input, though this can be done different ways for each input. Overall, it is a nonissue, I wouldn't worry about it too much. Just pick a design and go with it.

Working on another thread, we came up with what I thought was an interesting driver for an op amp.

If you're using a ± power supply you can eliminate the capacitors, if an op amp is feeding this driver you can eliminate C1. I've never built this circuit, but I have pretty good confidence in it. Adjust R2 until the circuit is barely on, say 5ma or so. This circuit has no voltage gain (which is why you would need an op amp), but tons of current gain. R6 and R7 can be dropped even lower for more current, but then Q3 and Q4 can get hot (but only when driving a speaker).

Last edited: Apr 18, 2009
17. ### Audioguru Expert

Dec 20, 2007
11,251
1,350
It will probably have "thermal runaway" unless the diodes are connected to the heatsink of the power transistors.
Usually it is made like this for some negative feedback from the 1 ohm resistors:

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18. ### Wendy Moderator

Mar 24, 2008
21,838
3,047
You'll get a lot more negitive feedback from the emitters, as shown on my print. The idea is for less than an ohm of output impedance. The regulator is there to prevent the diodes from shifting around, also Q1 and Q2 are going to be carrying a few ma, not the main load.

19. ### Audioguru Expert

Dec 20, 2007
11,251
1,350
Audio amplifiers have used the circuit I posted for about 35 years. The output impedance is almost zero when the circuit is included in the negative feedback loop of an amplifier.