Understanding Transistor Voltage Amplification

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Transistor voltage amplification involves the relationship between base and collector currents, where the collector current is significantly larger than the base current. The output voltage at the collector is inverted compared to the input signal, meaning that an increase in collector current results in a larger voltage drop across the collector resistor, leading to a lower collector voltage. However, this does not negate amplification; instead, a small change in input voltage produces a much larger change in output voltage, typically in the opposite direction. The key to understanding this is recognizing that the AC gain of a BJT transistor in a common-emitter configuration is negative and greater than one. Overall, the transistor amplifies current, and the output voltage variation is a result of the collector current passing through the load resistor.
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Homework Statement


I've been reading about transistors now for a while and there is something that doesn't add up. It's about how the collector in a transistor amplifies the output voltage.

williamson-labs.com said:
As the voltage at the generator goes more positive; the base current increases; the collector current increases; the voltage drop across the collector resistor increases; and the voltage at the collector goes less positive or lower.

And then they say...

[QUOTE="williamson-labs.com]When operating with a collector resistor (RL): the output voltage from the collector is an amplified voltage.[/QUOTE]


What I was thinking is that when the collector current increases (amplification is usually around 100) the voltage is amplified, hence a high output at the collector. But high current at the collector would mean a big voltage drop in it's resistance and hence low output voltage in the collector.

So my question is basically...is it low current or high current in the collector that amplifies the voltage there?
 
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It doesn't make sense to say there is amplification "in the collector". The whole transistor is involved.
The transistor is basically a device that amplifies current. The collector current is many times the base current. The input signal varies the base current within some reasonable range, and the collector current varies a much larger amount in response. When the collector current passes through a resistor, you get a varying voltage across it. Input and output resistors can be chosen so the variation in the voltage across the output resistor is much larger than the variation in the input voltage.
 
Sigurdsson said:
What I was thinking is that when the collector current increases (amplification is usually around 100) the voltage is amplified, hence a high output at the collector. But high current at the collector would mean a big voltage drop in it's resistance and hence low output voltage in the collector.

So my question is basically...is it low current or high current in the collector that amplifies the voltage there?

Both of your quotations are correct. There are two things to keep in mind that might help you make sense of all of this:

(a) The output waveform (voltage as measured at the collector) is inverted, as compared to the voltage measured at the generator. So if the generator contains a sine wave component, you will see a sine wave at the collector too, except it will be 180o out of phase.

(b) There is still amplification. A small "wiggle" in voltage at the generator produces a comparatively large wiggle in the voltage at the collector. Granted, it is in the opposite direction, but it is much bigger wiggle.

Essentially, the amplification they are talking about is AC amplification. The DC voltage of the collector doesn't matter in AC amplification because you can always DC bias it accordingly (you need to do this anyway to minimize saturation).

In short, the AC gain (aka amplification) of BJT transistor in a common-emitter configuration is typically negative with magnitude greater than 1.
 
I think I got it...

http://www.play-hookey.com/digital/electronics/images/invert.gif

I was thinking with the output above the resistor in this circuit. If we have, let's say 5 volts, at v+ and there is a massive voltage drop at the RL resistor. Then there has to be large voltage at the other end.

[QUOTE="williams-labs.com]...and the voltage at the collector goes less positive or lower.[/QUOTE]

Like collinsmark pointed out the signal is inverted. So when they are saying in the above quote that the voltage at the collecter goes less positive, it's amplitude is increasing...while I thought decreasing.

*facepalm*


Correct me if I'm wrong, otherwise cheers guys!
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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