The most simple amplifier

First, make sure that your ampliifer (transistor) is properly biased and your circuit is working correctly. Second, double check the electrical specfications of the components and make sure that you have connected the mic in a manner that is compatible with your circuit. Keep in mind the "loading" concept and make sure that you don't have a divider circuit that prevents a signal from appearing at the "input" to your ampliifer.

A speaker will respond to the AC component of the signal, which is why a DC blocking capacitor is typically used in the circuit. As far as 0

10V, working with your speaker, this will depend on the characteristics of the speaker (i.e. impedance, power disipation capability, etc).

If you are still having problems, a simpler way to start may be either an opamp or an audio amp (basically an opamp geared towards the audio band).

Thanks for the reply!

What is the difference between the different amps? Seems like there are at least 3 kinds of amps.

1. Whichever one you were discussing.
2. As an alternative: an opamp; or
3. audio amp.

I don't know thw specifications of the mic in my computer headset. Is there ANY way I can test the signal with my voltmeter/ampmeter? since the voice on the mic will be a wave, I suppose the voltage generated will change between -x and x. So how do I determine the range of voltage for the mic? I know what all the other components are e.g. the multiplication of the transistor etc, so if I know what I get in, I can calculate what I get out.

You probably have an electret, which needs a bias resistor of say 4.7k from

+5 or +10V. Then take the signal off with a coupling capacitor and have fun.

Tim

-- Deep Fryer: a very philosophical monk. Website:

Also, be aware that your single-transistor amplifier probably does not have much gain - depending on the circuit you're using, maybe you have a gain of

1. The signal from the microphone (even when it has appropriate bias voltage, using Tim's suggestion) is only going to be a few millivolts. So the voltage to the speaker will be only a few tens of millivolts. That is not very much.

I would start by using a signal source that is stronger, and easier to work with - for instance, the output of a sound card or an iPod, which will be a few hundred millivolts, and which have the advantage that when you're listening you're not also talking.

And also, you need to think about impedance matching. If you're driving a speaker, it probably has 8 ohms or so of impedance. You need to think about that in the context of your circuit, because it is effectively in parallel with whatever is driving it. Suppose you're using a circuit like this:

Vcc -----o-----. | | | .-. | | |Rc | | | .-. '-' | | | || | | o---||-. '-' | || | | | | __ /| || | |/ '--| | | IN -||--o---| .--|__| | || | |> | \\| .-. | | | | | | | | | '-' .-. | | | | | | | |Re | | '-' | | | | GND -----o-----o------'

Then the speaker impedance is effectively in parallel with Rc (for AC signals; you have to remember that the AC difference between Vcc and GND is zero). If you've been assuming gain = Rc/Re, then that will throw off your calculations by quite a lot. A simpler way to think about that is that in the above circuit, the only thing the transistor can do is let more or less current through itself. Even if it's letting zero current through, so that all the current flows through the speaker, the amount of current is limited by Rc; if Rc is, let's say, 10k ohms and Vcc is 9V, then 0.9mA is the most that can flow. That 0.9mA, through the speaker's 8 ohms, equals 0.72mV - probably even less than the mic is putting out, and that's the MOST the speaker voltage can ever be.

So, to get from a mic element with impedance around 1k ohms, to a speaker with impedance 8 ohms, you need impedance "gain" as well. One way to get this is by using a transformer, say a 1k:8 ohms transformer.

All that said... if I were you, depending on what your goal is, you might consider using an LM386 as your amplifier. If you do, make sure to read about "bypass capacitors".

Whao Walter, thanks for the reply (and the schematic)

I understand your recommendation of using a line-out rather than a mic as input. However, if I use the soundcard's out, that will already be a amplified signal, so I won't know if the transister did any amplification? Same for the iPod. Or am I missing something?

How will my experiment be affected if I use a class A amp? (Ie. I want to get rid of Re and remove the other resistor and the wire that is next to Re) The reason I'm asking is that I can understand the class A better and it will make experimentation better for me.

So, turn the volume on the iPod down to the point where if you plugged the speaker directly into it, it wouldn't be very loud.

The schematic I drew is class A. Any transistor amplifier where the transistor is biased into conduction for the entire signal wavecycle, is class A; that includes virtually any audio-frequency single-transistor amp. Classes AB and B refer to circuits where there is typically more than one transistor, and where each transistor is conducting for only part of the wavecycle.

If what you mean is that you want to tie the emitter directly to ground, that will increase the gain slightly, at the expense of making it impossible to control how much gain you have and more difficult to bias properly. But it will not solve the basic problem you have, which is that you can't easily drive an 8 ohm speaker with a common-emitter transistor circuit (that's one where the load is connected to the collector).

You'll have an easier time driving 8 ohms with a common-collector amp (that's one where the load is connected to the emitter), but that won't give you any voltage gain. So, what most practical circuits do is they have a common-emitter stage to give voltage gain, followed by a common-collector stage (aka an "emitter follower") to give current gain. As I mentioned before, you need both, to get between a microphone (low voltage, low current) and a speaker (higher voltage, higher current). Amplifier chips such as the LM386 contain several gain stages of various appropriate sorts, conveniently packaged up into a blob of plastic.

Horowitz and Hill's "Art of Electronics", 2ed., contains an excellent discussion of the issues you're trying to understand.

[...]

A tad impractical. He'd learn more and it would be easier just to add an emitter- follower stage to the output from his amp. The first stage could then concentrate solely on voltage gain; the second on current/power.

He'll learn Jack about amplifiers by copping-out with a 'black box solution' like that.

--

"What is now proved was once only imagin\'d" - William Blake

usually they are electret mics - this is a good thing, but you'll need to send some DC its way for it to produce any output

they need more amplification

They will work with a one-sided signal, they work better with an signal that goes positive and negative.

Bye. Jasen

You can think of a "0 to 10V" signal as being the same as a -5V to +5V AC signal, plus a constant 5V DC signal. If you feed a speaker a signal that contains a DC component, the speaker can't turn that into acoustic energy (in other words, its frequency response does not include DC); so instead, it turns it into heat, in the voice coil.

If your power levels are high enough, or your speaker is small enough, the result is that the voice coil warps or melts, destroying the speaker. On the other hand, as Jasen suggests, if the speaker is rated for enough power, it just gets a little warm, no problem.

A fellow sound guy once told me the story of a power amplifier failing with a short to the DC power supply, at a concert he was working in the '70s. This was a big amp and a big speaker, so when the speaker failed, there was apparently enough AC component left to eject the flaming voice-coil ring out of the speaker cabinet and into the audience. Pretty cool effect. Your amp is unlikely to do that, though :-)

To get rid of the AC component, all you need to do is put a capacitor in series with the speaker, as I showed in my schematic. The capacitor needs to have a big enough value that it lets enough low frequencies through. C =

1/2*pi*R*f at the rolloff point, so for 8 ohms and 80Hz (which is lower than any little speaker is going to reproduce, there's no point in going to 20Hz), you get C = 1/2*pi*8*80 = 248uF. You can use a 220uF capacitor, it's a more common value. Make sure that you aim the '+' pin of the capacitor the right way, towards the amp rather than towards the speaker. The capacitor needs to be rated for the DC voltage it's going to withstand: in your 0-10V example it sees 5V, but I would play it safe and get one rated for at least 10V.

Note that for simple testing purposes a headphone element is effectively the same thing as a dynamic mic element, though certainly not optimized for that use. So if you have an old pair of headphones, you might want to give them a try for this use.

Best regards,

Bob Masta dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis

Home of DaqGen, the FREEWARE signal generator

Hi Jasen

Thanks for your brilliant description of 0 to 10V question I had! One more thing about the capacitors. Why do you have one on the MIC side. You know you'll get both positive and negative volts. I don't think in my humble electronic mind that the cap hurts but I dont' really see a use for it there.

And then a final question that I will probably get a slap for is... Is there a difference between connecting the speaker/cap wire to the collector or the emittor. And I think the answer has to do with: maybe if you put it on the emitter, you'll only get the posive part of the waves. I'm not sure though.

Seems like this whole experiment would have been much easier if I had a

1000=80 osciloscope.

I think you're actually responding to my post? I'm Walter; Jasen is someone else.

Think of a capacitor as being a way to block DC from flowing, while permitting AC to flow.

The DC voltage on the transistor's base is set by the resistive biasing network. It wants to be about 1/2 Vcc, roughly speaking. The DC voltage on the microphone wants to be zero (unless it's an electret mic, in which case it wants to be whatever the preferred bias voltage of the mic is; 1/2 Vcc might be fine). Since those are different DC voltages, you need some way to separate them, while allowing the AC signal to pass. A capacitor does that.

If you're using an electret mic and 1/2 Vcc is a good bias voltage for it, then you don't need the capacitor. But you do need to think about how much bias current the electret pulls (1mA would be a decent guess) and think about what that will do to your bias voltage. For instance, if you used 10k resistors for your bias network, and you had a 9V supply, you might think that would bias the transistor and the electret mic both at 4.5V; but you'd be wrong, because you can't get 1mA of current through a 10k resistor with a

9V supply, according to Ohm's Law. It would work to use 1k resistors, but then you have to realize that the mic is seeing a load of less than 500 ohms (that is, 2 1k resistors in parallel) and it may not be able to drive that. Sometimes it's easier to use a capacitor anyway, and bias the transistor separately from the microphone.

An example of this is the circuit at

Note that circuit provides voltage gain but not enough current gain to drive a speaker; but it is an example of biasing a common-emitter transistor stage driven by an electret mic.

There is a lot of difference, as I was trying to explain. If you connect to the collector, you can get voltage gain, but not a lot of current gain. If you connect to the emitter, you can get current gain, but no voltage gain. Which one you need depends on what you're trying to do. (To drive a speaker from a microphone you probably need both; so you probably need more than one stage of gain, that is, you need more than one transistor.)

It's not a scope you need, though it's a handy thing to have. It's a decent textbook, and a pencil and some paper.

Analog electronics gets frustrating quickly unless you understand at least a little bit of theory: if you just hook some stuff together without knowing why, the odds of it working are low. If you don't have some theory, the best you can do is build circuits that other people designed.

But the amount of theory you need is really not that big. It involves no math beyond basic algebra, and it can be described in a couple of chapters. Good books that will give you the knowledge you need to answer these questions include Horowitz and Hill's "Art of Electronics" 2ed., and the Handbook of the Amateur Radio Relay League (comes out annually; any edition after about 1980 is fine). In each case, although they are thick books, you only need the first 2 or 3 chapters to start getting somewhere.

if you need it becase the transsitor wants only positive voltage to operate correctly. it allows the transitor's base to be positive while the microphone is producing an AC signal.

in the simple one transistor amplifier circuit most of the output signal is on the transistor's collector.

there are other designs that use the emitter as the output terminal....

yeah...

as for wanting to use a dynamic microphone with a one transistor amplifier I get enough gain from a one transistor amp (configurted for maximum gain) to connect a 600 ohm dynamic mic to a soundcard input. it'd take a few more transistors to be able to drive a loudspeaker.

Bye. Jasen