Effect on Drain Voltage & Source Voltage

Discussion in 'Homework Help' started by rodn.m, May 4, 2006.

  1. rodn.m

    Thread Starter Member

    May 3, 2006
    I refer to the Fig 3 on the attachment (also below) in the first homework help forum.

    This asks:if the value of the source resistor in fig 3 was reduced, describe the effect this would have on the drain voltage and the source voltage of the circuit.

    the reply i received said that the biasing voltage Vgs would be effected. But i was under the impression that this voltage was zero, as it is negative w.r.t. the source voltage which is positive.

    so, i am asking you if the resistance of Rs is reduced, would the biasing voltage Vgs be reduced also?

    what about the drain voltage Vd, would this be reduced also?

    and the source voltage, Vs. the formula to get this is Vs=IdRs, so how can this be reduced by the resistance?

    So, basically, i am asking you to be more specific with your answer.

    watt's what!

    excuse me if i seem ignorant, but i am returning to studies after nearly 20 years on the tools, so i have a bit of trouble getting around electronic circuitry. i am electrician by trade.
  2. n9352527

    AAC Fanatic!

    Oct 14, 2005
    First, you need to realise that JFET is a different animal compared to a regular FET. JFET is a normally on device. This means that when the Vgs is zero, the channel between source and drain is wide open. When gate to source is reverse-biased (negative Vgs for N JFET), the channel is squeezed and the current flow is restricted. When Vgs equals to the pinch-off voltage, the channel is completely pinched-off and current ceases to flow.

    Looking at your self biasing JFET circuit, Vgs is equal to 0 - Vs or -Vs. Also, each biasing point on the transfer curve corresponds only to one value of Id and Vgs. We also know that Vgs is equal to -Vs and Vs is equal to Id*Rs. Ultimately, all of these mean that each of the biasing point on the transfer curve has an associated unique Rs for correct biasing. The value of Rs can be calculated from (-Vgs)/Id.

    This is an equilibrium point where the Vgs resulting from current (Id) passing through the Rs corresponds to the Id itself. If you somehow disturb this equilibrium, e.g. by changing the Rs value, then the circuit would always settle at a new equilibrium, a new biasing point along the transfer curve, that satisfies the relationship between Vgs, Id and Rs (Rs = (-Vgs)/Id).

    Now, changing the value of Rs. Lower Rs simply means that the biasing point moves up along the transfer curve towards maximum Id. Higher Rs means that the biasing point moves down along the transfer curve towards maximum (negative) Vgs.

    This is exactly the reverse of procedure to obtain Rs value for a particular biasing point from the transfer curve!

    The answers to your questions therefore, if Rs is lowered Id would end up being higher. Vs (or -Vgs) would end up being lower and Vd (Vdd - Id*Rd) would end up being lower.
  3. dbwgwee


    Mar 30, 2006
    quod vos can perago quis coepi per nonnullus succurro ex scientia
    (quoting someone...who undoubtedly quoted someone else :p)