Don't completely understand what you said but I think, mostly not.Am I right or not?
The way the circuit is drawn in the question looks like the source current is entering the gate, which is not possible.Don't completely understand what you said but I think, mostly not.
The gate-source voltage (Vgs) will be reverse biased by the voltage drop across Rs due to the current through it from the transistor drain-source current.
A JFET conducts it's maximum drain-source current (IDSS) when when Vgs = 0V and a negative Vgs will start to turn it off.
So from that you can solve the problem.
Also, the circuit drawn is confusing, because there should be a return path from the source current back to the drain to be a close or complete circuit.I think you are on the right tracks, actually. Yes, the drain current is the same as the source current and there is no gate current except for a very low level of leakage.
The arrow on the gate indicates a diode and that the gate is the anode or p-type. So it is not conducting, as you say, because it is either below the forward bias voltage (0.6V) or reversed because the source voltage is higher.
Yes, the bottom of Rs should be connected to circuit common (ground).Also, the circuit drawn is confusing, because there should be a return path from the source current back to the drain to be a close or complete circuit.
Yes, the bottom of Rs should be connected to circuit common (ground).
Also the drain supply should be a voltage, not a current source.
With an ideal current source the current will always be 7.5mA regardless of the value of Rx or anything else.
Thank you you made it very clear. I really appreciate your explanation. Thanks again,Here are two circuits where we might actually be able to use a current source. We need some way to divert some of the current away from the FET so a resistor would be a possibility. We only have to know the gate voltage required to cause 5ma of drain current to solve these, if solutions do in fact exist.
This makes it a little more interesting
You are welcome. Maybe we can try those circuits next just for the exercise.Thank you you made it very clear. I really appreciate your explanation. Thanks again,
I GOT IT. Thank you.First, you need to find the Vgs voltage that will ensure 5mA at the drain.
\[ I_D =I_{DSS}\left(1 - \frac{V_{GS}}{V_T}\right)^2 \]
As you have noticed we have a two solution:
Vgs = -5.449V and Vgs = -0.55V but only one has a physical meaning. Do you know which one? And why this one?
OK, below is my final attempt I KNOW it's wrong but not sure where is my mistake! Please help.First, you need to find the Vgs voltage that will ensure 5mA at the drain.
\[ I_D =I_{DSS}\left(1 - \frac{V_{GS}}{V_T}\right)^2 \]
As you have noticed we have a two solution:
Vgs = -5.449V and Vgs = -0.55V but only one has a physical meaning. Do you know which one? And why this one?
Hi,You are welcome. Maybe we can try those circuits next just for the exercise.
We know right off that the added Rds or added Rd has to pass 2.5ma so that helps right away.
Ok i will.Hi,
Can you kindly review my 3 different attempts I KNOW There is a mistake, but WHERE
Hello again,Hi,
Can you kindly review my 3 different attempts I KNOW There is a mistake, but WHERE