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2 years ago via answers.hackaday.com

How does a transistor amplify?

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farso's Avatar
farso | 2 years ago
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A transistor amplfys by varying the Collector-Emitter resistance. A small amount of current entering the transistor at the base will change the resistance (and hence the current) at the base by a larger amount. This is usually around 100-200 times.

The transistor is a shortening of the name "transient resistor" device.

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satiagraha | 2 years ago Report

Yes, this is the way to think about it. Transistors are a resistor that you can vary by doing something to a third connection (the Gate for FETs or the Base for BJTs). So for very small differences in the third connection, you can get a very large change in the resistance between the first two connections.

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musashiaharon | 2 years ago
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I'm also trying to figure these buggers out, but I'll share what I've been able to figure out so far. Please correct me if I'm wrong!

For BJTs, I usually think of them as current-controlled current sources. That is, the current from the base to the emitter (I_be) controls the current from the collector to the emitter (I_ce):

I_ce = h_fe * I_be = beta * I_be

where h_fe = beta ~= 200, +/- 100. This relationship is pretty accurate until you run up against limitations like operating close to the power supply rails, or running the transistor close to its maximum current rating. To get around this, BJTs are often biased so that the "idle voltage" at the base is around Vcc/2. This gives the input signal enough room to move around; otherwise, it would be up against one of the supply rails and the signal would be clipped. Limiting current can be accomplished with resistors.

The difference between an NPN and a PNP is that in an NPN, the emitter is on the side closer to ground. In a PNP, the emitter is on the side closer to V+. This means that in an NPN, the I_be flows into the base and out the emitter, but in a PNP, I_be flows into the emitter and out the base. Voltage-wise, for an NPN, a big base voltage gives a big I_ce, and a small base voltage gives a small I_ce. For a PNP, a big base voltage gives a small I_ce and a small base voltage gives a big I_ce.

The reason why the bias, or Q-point (for quiescent operating point) is set a bit higher than Vcc/2 has to do with the construction of a transistor. Internally, the BJT has PN junctions just like a diode. And, just like a diode, there is a voltage drop whenever current flows through the junction, which is about 0.7V for a silicon junction. At base voltages less than 0.7V, the behavior of the transistor becomes very nonlinear and is not useful for amplifying signals. Therefore, the useful range of voltages is inside 0.7V until Vcc, and the center of that range is where the engineer tries to set the Q-point.

A typical audio amplifier has a few sections. The first section is the differential amplifier. The second is the output stage, which drives the speakers. The third is the feedback circuit.

Since I am still learning, I don't feel comfortable saying more beyond this. Right now, I'm still studying various websites about the class AB amplifier.

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hells's Avatar
hells | 2 years ago
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They can be used as current or voltage amplifiers. The idea is that you are not amplifying within the transistor but using a logic switch to turn on / off the flow from a higher voltage / amperage source. Take a look at the source link for a full description.

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mrgoogfan | 2 years ago
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don't forget about npn or pnp

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