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Suppose the following audio amplifier configuration:

enter image description here

Suppose that Vc is correctly configured to be in the middle of the rail. So, at rest, Vc = 4.5V, Ve = 0 and Vb = 0,6V, right?

Suppose I inject a 1Vpp audio signal at the base.

As the signal rises from 0 to 0.5V, Vb increases from 0.6 to 1.1V. Because Vbe > 0.6V, the transistor will conduct. Ic will increase. As Ic increases the voltage drop on R1 increases, forcing Vc down. So, we see that Vc and Vb are 180 degrees out of phase.

What I do not understand is how will this occur on the negative cycle of the audio.

As the audio starts going from 0 to -0.5V, Vb will reduce from 0.6 to 0.1. Because Vbe will be less than 0.6 the transistor will cut off. No current will flow from collector to emitter.

In my mind, as the transistor cuts off, Vc will be 9V but this circuit is mentioned on the web as being an amplifier. It seem to me more like a half cycle rectifier...

What am I missing?

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  • \$\begingroup\$ You're not missing anything; the transistor won't conduct on the negative part of the cycle. \$\endgroup\$
    – Hearth
    May 11, 2018 at 14:06
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    \$\begingroup\$ To clarify: You should inject the audio signal via a capacitor, which allows the AC audio to be superimposed on a DC bias, so the transistor remains properly biased in operation. Even so, this is a very poor design for an amplifier, and I would advise against using it when you can make a better one using just a couple extra resistors. \$\endgroup\$
    – Hearth
    May 11, 2018 at 14:08
  • \$\begingroup\$ so how can this be an amplifier? \$\endgroup\$
    – Duck
    May 11, 2018 at 14:08
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    \$\begingroup\$ It amplifies positive signals just fine! Read my second comment for more information on how to improve it. \$\endgroup\$
    – Hearth
    May 11, 2018 at 14:08
  • \$\begingroup\$ @Felthry - ok, I know this circuit is not good as an amplifier. I am just trying to understand why it is referred as being an amplifier... \$\endgroup\$
    – Duck
    May 11, 2018 at 14:09

2 Answers 2

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If your input voltage goes below that necessary to bias the transistor on, it won't amplify it. This circuit will amplify decently well as long as your signal is positive only, which you can achieve by capacitively coupling your signal into the base node like so:

schematic

simulate this circuit – Schematic created using CircuitLab

Though personally I would highly recommend a design like this, which has improved linearity and is less likely to damage your transistor:

schematic

simulate this circuit

The use of the resistor between the emitter and ground allows you to control the gain of your amplifier, so that you don't have to deal with having far too much amplification which would result in signal clipping. It also improves linearity as well. Biasing the base with a resistor divider instead of with a resistor to Vcc also allows better control of the amplifier characteristics, improving linearity.

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For this to work reasonably well it has to be in the small-signal range of operation. You've shown the bias components (not a very good approach, but feasible).

To stay in the small-signal realm the input voltage has to be AC-coupled to retain the bias point and the input voltage has to be very small. If the possible linear range of output is 5Vp-p and the gain is 200 then the input voltage cannot be greater than 25mVp-p.

schematic

simulate this circuit – Schematic created using CircuitLab

Here is the output with 15mV input (30mV peak-to-peak), as you can see it's close to clipping.

enter image description here

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    \$\begingroup\$ This is the correct response, I think. +1 The OP was considering a huge signal and needs to understand the 10X change in collector current for a \$60\:\text{mV}\$ change at the base. So even that is too much since the collector current might swing from \$20\:\mu\text{A}\$ to \$2\:\text{mA}\$. Not to mention the highly distorted output for all that gain variation vs phase. A 1.5X variation might be okay. But this limits the peak to \$10\:\text{mV}\$ on the input. \$\endgroup\$
    – jonk
    May 11, 2018 at 16:53
  • \$\begingroup\$ @jonk That's a good rule of thumb to remember. I thought it might be more intuitive to put it in terms of the output (which obviously clips) but that 60mV=10x rule of thumb is worth committing to memory. \$\endgroup\$
    – Spehro Pefhany
    May 11, 2018 at 16:56
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    \$\begingroup\$ It is because it applies very broadly: in some other quite useful and realistic circuits, perhaps most especially the diff-amp pair. [I know you know it, but to just make sure it is clear, a simple derivation shows that for base-emitter voltage variation, \$v_\text{p}=V_\text{T}\cdot\operatorname{ln}\left(\frac{I_\text{C}}{I_{\text{C}_\text{Q}}}\right)\$.] \$\endgroup\$
    – jonk
    May 11, 2018 at 17:10

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