- #1
L89
- 10
- 0
Hi, just logged on first time in ages. Got a message saying that my password was 563 days old lol.
Right now, I'm in first year of uni so very stupid![Smile :smile: :smile:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Anyways, we're studying about field-effect transistors. I understand how the gate controls the current via adjusting the size of the n-channel. What I'm quite confused right now is how it amplifies voltage. Maybe I already "know" the answer but I don't realize it.
In any case, I'll attempt to "explain" how it amplifies and anyone who can correct me please do so. I've included a basic circuit drawing containing a FET. So here goes...
When the voltage of the gate relative to the drain is negative, the n-channel tapers towards the drain terminal. Even if the n-channel no longer touches the drain terminal, a current still flows. However, as voltage G relative to D becomes even more negative, the resistance in the drain resistor increases proportionally. This means that the current stays constant, while the voltage of the drain resistor can be enlarged.
Okay, my problem is, I have this idea (wrong assumption?) that an amplifier uses a small input voltage and turns into a large output voltage. However, in my "explanation", all it did was to state that the voltage of the drain resistor can be enlarged while maintaining a constant current. It's like the analogy of a kid lifting a toy airplane above his head and saying "It can fly!".
So please can anyone get me out of this hole. It'll be appreciated that the explanation is done in accordance with my circuit attachment (or if the circuit is not for amplication, please tell me so and why).
Thanks in advance.
L89
Right now, I'm in first year of uni so very stupid
Anyways, we're studying about field-effect transistors. I understand how the gate controls the current via adjusting the size of the n-channel. What I'm quite confused right now is how it amplifies voltage. Maybe I already "know" the answer but I don't realize it.
In any case, I'll attempt to "explain" how it amplifies and anyone who can correct me please do so. I've included a basic circuit drawing containing a FET. So here goes...
When the voltage of the gate relative to the drain is negative, the n-channel tapers towards the drain terminal. Even if the n-channel no longer touches the drain terminal, a current still flows. However, as voltage G relative to D becomes even more negative, the resistance in the drain resistor increases proportionally. This means that the current stays constant, while the voltage of the drain resistor can be enlarged.
Okay, my problem is, I have this idea (wrong assumption?) that an amplifier uses a small input voltage and turns into a large output voltage. However, in my "explanation", all it did was to state that the voltage of the drain resistor can be enlarged while maintaining a constant current. It's like the analogy of a kid lifting a toy airplane above his head and saying "It can fly!".
So please can anyone get me out of this hole. It'll be appreciated that the explanation is done in accordance with my circuit attachment (or if the circuit is not for amplication, please tell me so and why).
Thanks in advance.
L89