0
\$\begingroup\$

Using answer about microphone I have built a simple amplifier.

schematic

simulate this circuit – Schematic created using CircuitLab

I changed some of the parts due to my needs and resources.

  • R1: is now potentiometer and allows me to set the "0" of the sound to 2.5V
  • R3: should control amplification (but it doesn't seem to), so again a potentiometer
  • C1: unless the original answer meant electrolytic capacitor, there's no way to get 22uF capacitor. So I used just two 22nF caps.
  • C2: I have 10uF capacitor (salvaged from old sound card ironically), so I used that

So I have set up R1 to get 2.5 volts on silence. The image displays 80Hz sine wave rather than silence, I generated it here (works on Android).

image description

Now if I move the R3.2, instead of change to gain, I just get voltage offset, and an insignificant one. Picture describes full range (min - max - min) of that particular potentiometer:

image description

Note also that increasing resistance increases voltage on output.

I could use my 50k potentiometer instead but that would give me no precision.

So what is wrong with my circuit? How to control the amplification?

Note: I displayed the analog input using this processing script and this arduino program, both mine but I thought I might share.

\$\endgroup\$
  • \$\begingroup\$ AFAIK this circuit has a fixed gain. No gain control is available. It wholly depends on the voltage/current produced by the mic. Also, try moving your question to the start, as it got buried at the end. \$\endgroup\$ – Passerby Mar 16 '16 at 21:08
  • \$\begingroup\$ Also your values seem odd. The R3 is 36k, to R4 1k? That would be 0.14V center, not 2.5V. \$\endgroup\$ – Passerby Mar 16 '16 at 21:10
  • \$\begingroup\$ And Oli recommended C1 to be 1µF or greater. A 1uf ceramic would be fine. \$\endgroup\$ – Passerby Mar 16 '16 at 21:12
  • \$\begingroup\$ Oh, essentially to change the gain without the offset change, R3 and R4 need to be changed, proportionally, lower. \$\endgroup\$ – Passerby Mar 16 '16 at 21:23
1
\$\begingroup\$

Try this...

  • Remove R1 the way it is currently wired and all wires leading to it.
  • Connect a ~1k resistor from Vcc to the transistor collector.
  • Connect R1's end leads from the transistor collector to ground, respectively. A larger value for R1 might be a good idea, such as 10k, but it isn't critical.
  • Connect a 1µF ceramic cap from R1's wiper to the arduino.

Edit:

  • Also connect a 10k from Vcc to the arduino pin, and a 10k from that to ground. This will cause the output to center around 2.5v, which seems to be what you are after.

This way, the position of R1 forms a "tap" between full signal, and no signal. Then adjust R3's value to give the cleanest output with the loudest microphone signal expected. There is a technical term for adjusting these resistor values - biasing. Also this. Have you calculated what the values should be?

\$\endgroup\$
  • \$\begingroup\$ No, I didn't calculate anything. I just adjusted till it looked nice on the graph. \$\endgroup\$ – Tomáš Zato - Reinstate Monica Mar 16 '16 at 21:48
  • 1
    \$\begingroup\$ I think R4 needs to be larger, like was shown in your previous link. And C4 needs to be much larger like Passerby and Oli suggested. But you're very close. \$\endgroup\$ – rdtsc Mar 16 '16 at 21:59
  • \$\begingroup\$ Thanks a lot. I'll buy some tomorow and come back to accept the answer. \$\endgroup\$ – Tomáš Zato - Reinstate Monica Mar 16 '16 at 22:10
1
\$\begingroup\$

Your gain will be approximately $$a=-\dfrac{R1}{R2}$$ As you modulate the base voltage, the emitter voltage will increase by a roughly equal amount. This increases the current through R1. Almost all the current through R1 also flows through R2, causing a voltage drop.

After choosing the gain you want for your circuit, you'll need to adjust the bias using R3 to maintain your 2.5V zero point.

$$V_e = \dfrac{2.5V\cdot R2}{R1}$$ $$V_b = V_e + 0.7V$$ $$V_b = \dfrac{V_{cc}\cdot R4}{R3 + R4}$$

One more thing to keep in mind is the size of your 44nF capacitor. It will create a highpass filter with cutoff frequency $$f_c = \dfrac{1}{2\pi\cdot (R4||R3)\cdot c_1}$$ With the current values you have in there this comes out to be over 3kHz. You can hear much lower than that so I recommend choosing a larger capacitor.

\$\endgroup\$
  • \$\begingroup\$ What's the $(R4||R3)$? \$\endgroup\$ – Tomáš Zato - Reinstate Monica Mar 17 '16 at 15:25
  • \$\begingroup\$ The parallel combination of R3 and R4. R3*R4/(R3+R4) \$\endgroup\$ – user103923 Mar 17 '16 at 16:20

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.