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Discuss Could somebody please tell me what the name of the below circuit is please in the UK Electrical Forum area at ElectriciansForums.net
Hi all, does anybody know the name of this circuit and how it operates and the purpose of each component please? any help is much appreciated.
Thanks in advance
Either that, or it's someone's homework !
TBH, I sort of missed out learning about MOSFETs, but I recognise this as the classic single transistor inverting class A amplifier circuit - but with a MOSFET instead of an NPN bipolar transistor.
C is just DC blocking.
In the absence of an input signal :
R1 & R2 bias the gate to a set voltage, we'll call it Vg.
To a rough approximation, the transistor will then turn on until the current through Rs causes the gate-source voltage to stabilise at a value that will cause the transistor to conduct that much current. With a silicon bipolar transistor, you'd use around 0.6V as a starting point for the base-emitter voltage, dunno what the equivalent for a MOSFET is.
The current through Rs also comes through Rd, causing Vout to take up some DC voltage in between Vdd and Vss.
If we now apply a rising voltage to the input, that will increase the voltage on the gate, the transistor will turn on a bit more, the current down the Rd-transistor-Rs path will increase until (approximately) the voltage across Rs increases to match the change in Vg. The increase in current also causes an increase in volt drop across Vd, so Vout reduces. The change in Vout for a given change in Vin is roughly proportional to the ratio of Vd to Vs - so if Vd is (say) 5 times the resistance of Vs, then the gain will be approximately 5, or more correctly, -5 as it's an inverting circuit.
These circuits are generally only used for small signals, otherwise they waste a lot of power. The transistor is always conducting (unless you feed in a very large signal, in which cause you'll get an 'orrible sound out if it's an audio amplifier), and there's always current flowing down the Rd-transistor-Rs path which generates a lot of heat - much more than the usable output power.
That's all dredging up stuff I learned "a few" decades ago and haven't used for "several" decades. So feel free to point out if it's all wrong.
I wish lolAs long as we're not doing your homework for you ;-)
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