I'm a newbie to transistors. Could anyone explain why my set up doesn't work ? The transistor is a 2N2222 NPN BJT, and the microphone is just a standard earbud. The loudspeaker is just one I ripped out from a toy.


  • 1
    \$\begingroup\$ Shouldn't the output signal be AC-coupled, centered on 0? \$\endgroup\$ – clabacchio Mar 6 '12 at 12:28
  • \$\begingroup\$ Hmm.. could you explain? Sorry, I'm new to this stuff. \$\endgroup\$ – user8210 Mar 6 '12 at 12:47
  • 5
    \$\begingroup\$ 1. Place a capacitor (100uF?) in series with speaker (stops speaker DC shunting collector voltage) 2. Put 33 ohm resistor emitter to ground. 3. Connect from 10 uF to 100 uF across emitter resistor. || Circuit will draw ABOUT 6 mA. Vcollector will be ABOUT 3V. Gain will be ABOUT 220. If Vc < 1V increase emitter resistor. If Vcollector > 3V decrease emitter resistor. \$\endgroup\$ – Russell McMahon Mar 6 '12 at 13:17
  • \$\begingroup\$ See added circuit and notes. \$\endgroup\$ – Russell McMahon Mar 7 '12 at 5:13
  • 1
    \$\begingroup\$ @user8210: interesting value of the capacitor: 1kµ. Isn't that the same as 1m? \$\endgroup\$ – Curd Jul 9 '16 at 7:28
  1. Place a capacitor (100uF?) in series with speaker.
    This stops speaker DC shunting collector voltage)

  2. Put 33 ohm resistor emitter to ground.
    Vb ~= 1k/(1k+10k) x 9V = 0.8V.
    Vbe ~=0.6V.
    So need to deal with 0.2V extra.
    Add Re so 6 mA flow so V1k in collector = 6V drop .
    Re = V/I = 0.2/0.006 = 33 ohms.
    Transistor now draws ABOUT 6mA so drop across Rc = 1k = V = IR = 6 mA x 1k = 6V.
    So Vc = 9V-6V = 3V ~~~

  3. Connect from 10 uF to 100 uF across emitter resistor.
    This bypasses AC signal around Re.
    Re is now just for DC bias.

Circuit will draw ABOUT 6 mA.
Vcollector will be ABOUT 3V.

Gain will be ABOUT 230.
Gain ~~= 38.4 x V across Rc = 38.4 x 6 =~ 230

  • NB - Nobody knows this deep magic and many people will rise up and say it is not true. It is :-)

Due to non idealities the above may not quite work, so ...

  • If Vcollector < 1V increase emitter resistor.

  • If Vcollector > 3V decrease emitter resistor.

With a gain of 230 or so you need about 20 mV pp signal in to saturate this amplifier. We set Vc = 3V so 3V negative is max swing available. So Vin going positive by 3V/230 = 13 mV will saturate collector voltage and drive collector to groujd. 10 mV max is about right.

If we had set Vc at mid rail we would have more swing but less gain. Gain goes up as Vc static rises. As above Gain = 38.4 x Vc_static.

Close enough to "max gain possible~~= 40 x Vbattery. (a bit less actually. )

As Olin has noted - the speaker load will prevent the potential gain being achieved. Adding an emitter follower as a power amplifier (not a voltage amplifier - it has a gain slightly less than 1) will allow the speaker to be driven at higher power AND the amplifier to achieve closer to its maximum gain.

An emitter follower is not an ideal power amp and the one transistor amplifier is far from an ideal audio amplifier BUT the 2 transistor circuit shows what can be achieved.

Here is an excellent laboratory experiment sheet from University of Colorado, Boulder. Physics 330 - Experiment 7 - Transistor Amplifiers
It not only goes through the design but provides a circuit that will suit your need with minor modifications, as shown below.

In your application the bigger Ce is the better - it sets the low frequency gain and even eg 470 uF would be good. (A 10V 470 uF Aluminum electrolytic capacitor is cheap and commonly available). Leave base resistors and emitter resistor as I noted previously.

I have left their component values on black and added "our" ones in red.
The speaker may be connected to Vout OR can replace Re' (right hand transistor emitter resistor). They, too, should have named their components.

enter image description here

  • \$\begingroup\$ Your gain is plausible without the speaker attached. With the speaker, the impedance at the collector goes way down, and so will the gain. Put another way using your terms, Rc is 1 kOhm only for DC purposes. It is much much less when considering the AC signals this circuit is intended to amplify. \$\endgroup\$ – Olin Lathrop Mar 6 '12 at 13:43
  • \$\begingroup\$ @OlinLathrop - Good point. I've decoupled the speaker in step one. AC collector load impedance is indeed massively shunted by speaker. Part of my brain goes 'Doh!', but part notes that for a given signal level change on the base the collector current will be affected equally DC wise and ... . :-) 3:30am. Work to do. May revisit. \$\endgroup\$ – Russell McMahon Mar 6 '12 at 14:37
  • \$\begingroup\$ Since you put a cap accross the emitter resistor (it would be good to see a schematic, it's a bit hard to make sure you catch everything with this handwaving), the collector current will go up with frquency. Eventually the emitter load impedance gets low enough to be comparable to what is in the transistor and the base impedance divided by Hfe. At that point gain becomes unpredictable. Collector voltage partially collapses because you have much lower impedance pulling down than the 1 kOhm pulling up. You are always stuck with this 1 kOhm trying to drive the speaker up. \$\endgroup\$ – Olin Lathrop Mar 6 '12 at 15:07
  • \$\begingroup\$ @OlinLathrop - yes - 1K does not make a good pullup for 8 ohm speaker :-). Base resistance is 26 ohm.mA. ie 26 ohms at 1 mA, 13 ohms at 2 mA etc, based on fundamental properties of Si junction. 26 ohm.mA = 0.026 Ohm.Amp. 1/0.026 = 38.4 which is where my gain figure of gain ~= 38.4 x V_load comes from BUT as you note, this is unloaded gain. I said "up to 100 uF emitter capacitor" and this has about 10 Ohms impedance at 300 Hz so emitter resistor will still not be adequately bypassed at lower freq end. At 3 kHz zcap ~+ 1 Ohm = ok. All good fun. Need 1k high Z speaker ;-) \$\endgroup\$ – Russell McMahon Mar 6 '12 at 16:21

Several things:

  1. Get a schematic editor already! This second grade crayon drawing isn't doing your credibility any favors. Even a neatly drawn schematic with pencil on white paper, then scanned would be fine.

  2. Use component designators. It gets really tiring to say the "top 1 kΩ resistor", "the bottom 1 kΩ resistor", etc. Didn't you notice that all real schematics have them? I am only going to discuss your circuit briefly because it is hard to talk about without component designators.

  3. The transistor is probably saturated, so nothing would come out of the speaker even if it were AC coupled.

  4. A 9V battery is not a good power source for driving a speaker. 9V batteries are very wimpy as such things go, and speakers need considerable bursts of current.

  5. Here is a circuit to try with roughly the same parts:

    This isn't a great circuit for a number of reasons, but it is something you should be able to rig up with what you have and get some results. Don't expect it to amplify from a microphone. It doesn't have enough gain for that. You can probably connect a similar speaker as the input, put your ear right up to the output speaker, and hear something when you gently scratch the input speaker with your finger.

  • \$\begingroup\$ Oh thanks! What does P1 and P2 stand for? \$\endgroup\$ – user8210 Mar 6 '12 at 13:12
  • \$\begingroup\$ Those are the component designators for the two input connection points. \$\endgroup\$ – Olin Lathrop Mar 6 '12 at 13:14
  • 1
    \$\begingroup\$ Just a note that this circuit actually drives an 8 ohm speaker relatively well but the standby C-E current is high (possibly hundreds of mA). For a small transistor this could cause thermal runaway / heat damage. My transistor nearly smoked, I ended up at something like the question asker's original circuit with collector resistor = 100 ohms and big cap in series with speaker. Not quite as loud but no smoke. \$\endgroup\$ – Georgie Jan 3 '15 at 9:07

There are a few problems with your circuit.

First of all, your speaker has low resistance (usually 4-8 Ohms), so in the current citcuit it basically shorts out the transistor. To avoid that, connect a capacitor in series with the speaker. The capacitor should be 100uF or more.

Second, the collector resistor is way too high resistance. As I said, the speaker has low resistance, so it needs a lot of current. You probably would get something like 0.5mW, which may not be audible.

Also, there usually is a resistor (and a capacitor in parallel to it) connected to the emitter. This helps set the gain and reduce distortion.

The voltage divider connected to the base is calculated to get the transistor in the active region so there is less distortion.

To read more, go to this website, it has the formulas needed.

  • \$\begingroup\$ What do you recommend my collector resistor be set to? \$\endgroup\$ – user8210 Mar 6 '12 at 23:49

Try connecting the speaker to the positive of your battery and the other end to the collector terminal of Transistor. Just remove all the resistance ,and place an inductor from the positive of battery to the base of Transistor.

  • 2
    \$\begingroup\$ Welcome to EE.SE. The question has an answer that was accepted several years ago. Your answer might be of interest to others but you would need to add a schematic. There's a button on the editor toolbar. "Transistor" isn't a proper noun so it gets a small "t". \$\endgroup\$ – Transistor Jul 9 '16 at 7:25
  • \$\begingroup\$ @transistor Unless it is used as a nickname, in which case the first letter should probably be capitalized... \$\endgroup\$ – dim Jul 9 '16 at 15:30
  • \$\begingroup\$ @dim: Ah, but that's not my proper name. ;^) \$\endgroup\$ – Transistor Jul 9 '16 at 15:34
  • \$\begingroup\$ That inductor will short the input to the battery (unless it's a really large inductor). It will also saturate the transistor. \$\endgroup\$ – Oskar Skog Mar 6 '17 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.