OK the first picture is an idealized representation, the second is true in the case of the common emitter amplifier shown, because the common emitter is an inverting amplifier.
Normally the collector resistor in the diagram would be replaced by an LC 'tank' circuit and the transistor would operate with a narrow conduction angle (<180 degrees).
The output (close to a sine wave) would be taken from the tank circuit either from a tap on the coil, by making the inductor a transformer, or by a split capacitor 'transformer' to provide a match to the next section, usually a filter.
Normally the collector resistor in the diagram would be replaced by an LC 'tank' circuit and the transistor would operate with a narrow conduction angle (<180 degrees).
The output (close to a sine wave) would be taken from the tank circuit either from a tap on the coil, by making the inductor a transformer, or by a split capacitor 'transformer' to provide a match to the next section, usually a filter.
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Of course, silly me, it's widely known that a positive current flows when a negative voltage is applied.
No. This would result in frequency doubling. Class C can be used in audio where the requirement for fidelity is very low, such as in a loudhailer.
Yes. DF96 is wrong, his reply is misleading. My second answer was sarcastic (a joke), I apologise for that. In the (class C) common emitter amplifier the transistor conducts when the voltage on the base is sufficiently positive. Current flows into the collector and the voltage at that point drops.
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That simple single-transistor stage in fig 2 will invert the polarity of the signal. Regardless of class, the basic common emitter stage will always invert the polarity of the signal. Feed the output of the first stage through another similar stage and the circuit again inverts the waveform polarity, so two stages in sequence results in the original signal polarity. That is why in discussions of signal "phase" through an amp, yo hear mention of even number of stages versus odd number of stages.
The more positive the base is made, the more the transistor conducts. The more it conducts, the higher the current through the load resistor. And Ohm's Law tells us the more curent through the resistor, the more voltage across the resistor. The top end of that resistor is hopefully conected to the powr supply, so the voltage at that end stays the same. SO then increasing the voltage across the resistor must mean the bottom end of it - the collector of the transistor - will wind up at a lower voltage.
Forget polarity for a moment. Unless polarity was your only question. Look at fig 1. With a continuous sine wave into the class C amp, the resulting output is just those peaks. Now look at fig 2. With the continuous sine wave into the transistor, the output is just the peaks as well. Yes, they are now at opposite polarity, but the waveform is still only those peaks. The way it is drawn, those peaks are narrower looking than in fig 1, but the difference is the same as turning the horizontal sweep speed control on your scope. It only appears different.
The more positive the base is made, the more the transistor conducts. The more it conducts, the higher the current through the load resistor. And Ohm's Law tells us the more curent through the resistor, the more voltage across the resistor. The top end of that resistor is hopefully conected to the powr supply, so the voltage at that end stays the same. SO then increasing the voltage across the resistor must mean the bottom end of it - the collector of the transistor - will wind up at a lower voltage.
Forget polarity for a moment. Unless polarity was your only question. Look at fig 1. With a continuous sine wave into the class C amp, the resulting output is just those peaks. Now look at fig 2. With the continuous sine wave into the transistor, the output is just the peaks as well. Yes, they are now at opposite polarity, but the waveform is still only those peaks. The way it is drawn, those peaks are narrower looking than in fig 1, but the difference is the same as turning the horizontal sweep speed control on your scope. It only appears different.
You're quite correct but if you want to get nit picky, the old Phase Linear 400 and 700 had outputs that technically were operating class C where class C is defined as the device conducting less than 180 degrees. The drivers operated class AB and started the outputs conducting slightly less than 180 degrees AKA class C.
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all right.these makes sense about what happen to both of the output.now i'm clear about it.btw thanks everyone for the reply.
No, because each conducting period is of alternating polarity.
If they are exactly symetrical, no second harmonic will be created.
I thought your reply was a bit strange. Apology accepted. I believe my answer is correct. The two graphs show collector current flow and voltage. Confusion may arise because the current graph rises from zero, while the voltage graph drops down from the supply rail. If you think of them as showing magnitude rather than absolute value then it doesn't matter if they 'ought' to be inverted.counter culture said:
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