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I have constructed this amplifier as shown in the circuit diagram:

schematic

simulate this circuit – Schematic created using CircuitLab

I have made a few inferences:

  1. Reducing the value of emitter bypass capacitor reduces distortion but also reduces the gain. How do I find the correct value for getting good performance at audio frequencies?
  2. Using this single stage I am able to get very less audio output. Its almost inaudible. If I use a second stage, I get highly distorted output. However it is audible to certain extent. What should I do to reduce the distortion without affecting the gain?
  3. Can you please provide the design steps involved in choosing the capacitor values?
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    \$\begingroup\$ Where did this terrible circuit come from ? Can you put it back there and leave it? \$\endgroup\$ – JIm Dearden May 19 '16 at 6:39
  • \$\begingroup\$ What is wrong with this circuit? \$\endgroup\$ – Sâu May 19 '16 at 6:43
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    \$\begingroup\$ The standard audio amplifier for loudspeakers has three stages. Differential input stage (long-tailed pair or similar) voltage amplifier stage (VAS) and output stage. Even for headphones, or line level, the 3-stage amp should work OK. \$\endgroup\$ – mkeith May 19 '16 at 7:40
  • \$\begingroup\$ can you please provide a good schematic? \$\endgroup\$ – Sâu May 19 '16 at 7:41
  • \$\begingroup\$ You have a 2.2k resistor from 9V to collector, and you are driving the 4 Ohm speaker at the collector. This means that the maximum output current during rising signals is 9V / (2.2K + 4) =~ 4mA. During low-going signals, you have 470 Ohms, so you are limited to maybe 20 mA. \$\endgroup\$ – mkeith May 19 '16 at 7:46
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What is wrong with this circuit?

Typically a circuit like this has a gain that is roughly numerically the collector impedance divided by emitter impedance. So if you had 2k2 in the collector and 220 ohms in the emitter you'd get a voltage gain of ten.

This type of circuit is usually pushed to have much higher gains by bypassing the emitter resistor with a capacitor. This makes the emitter impedance very low at audio frequencies but also significantly increases distortion.

Then if you really want things to go wrong you make collector impedance something really low by hanging a 4 ohm speaker on it. Collector impedance (AC) falls from 2k2 to about 4 ohms and you have virtually no gain and lots of distortion.

Reducing the value of emitter bypass capacitor reduces distortion but also reduces the gain. How do I find the correct value for getting good performance at audio frequencies?

Use a resistor in series with C1 to control the AC gain.

Using this single stage I am able to get very less audio output. Its almost inaudible. If I use a second stage, I get highly distorted output. However it is audible to certain extent. What should I do to reduce the distortion without affecting the gain?

See my observations above - don't hang a speaker on the collector - use an emitter follower or better still get an audio amplifier IC like the LM386.

Can you please provide the design steps involved in choosing the capacitor values?

I think this has been done.

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  • \$\begingroup\$ Thank you @Andy, that saves me writing the explaination to my comment (+1) \$\endgroup\$ – JIm Dearden May 19 '16 at 10:06
  • \$\begingroup\$ @JImDearden I was inspired by your comment! \$\endgroup\$ – Andy aka May 19 '16 at 10:25
  • \$\begingroup\$ You don't need to worry about the 4 ohm load affecting the collector impedance. At least, not at audio frequencies with a 1 uF coupling capacitor, LOL! \$\endgroup\$ – Brian Drummond May 19 '16 at 11:16
  • \$\begingroup\$ @Andyaka Can you please provide design procedure for an emitter follower. From my calculations, I've found that the output impedance of this circuit is roughly 2k. Also the Vce is 5 volts. This means the ac amplified voltage should be in the order of volts. How can I feed this signal into the base of an emitter follower? How should I bias the emitter follower? \$\endgroup\$ – Sâu May 19 '16 at 16:04
  • \$\begingroup\$ This is a simple q and a site. You should be able to find info on the web about using an emitter follower to buffer your circuit. \$\endgroup\$ – Andy aka May 19 '16 at 16:16

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