Solving DC Operating Point of Common Source Amp

My lab partner and I are really stumped. We are given a circuit that has a FET with a $V_{DD} = -V_{SS} = 15\mbox{ }V$. We are told $A_V = 20$ and $I_{DS} = 1\mbox{ }mA$. The DC circuit is straightforward, containing an $R_D$ and $R_S$ between $V_{DD}$ and $V_{SS}$, respectively, and the FET drain and source terminals. There is a resistor $R_G = 1.5\mbox{ }M\Omega$ on the gate terminal of the transistor, connected to ground on the other end.

We are asked what the DC current through this resistor ($R_G$?) is. We are then asked what $V_{OV}$ is. It seems to imply that we only need values that I have outlined so far, but maybe it would require values from a data sheet too, I'm not sure.

Then it says, using $V_{GS}$, which I would know how to calculate if I knew $V_{OV}$, to calculate $R_S$. It also says we don't know $V_{DS}$ nor $R_D$ yet.

Any help would be appreciated. I have tried reading the notes and textbook, but I can't find what I need. I would also appreciate any sort of good sources/guides to this, if you don't want to just give me the answer. I too, would like to understand so I feel less like a cheater, haha.

• Where is V_OV? Can you draw a circuit? – endolith Mar 31 '11 at 14:38
• $V_{ov}$ is usually the symbol for overdrive voltage and is given by $V_{ov}=V_{GS}-V_t$ where $V_{GS}$ is gate-source voltage and $V_t$ is threshold voltage. – Adam P Mar 31 '11 at 18:33
• $V_t$ is the parameter that needs to be retrieved from the datasheet. – Kevin Vermeer Mar 31 '11 at 19:20
• What text are you using? – Adam P Apr 1 '11 at 0:28

I will try to help without giving too much away. I just took an analog IC design course last semester, so I'm all psyched up about this stuff. I hope that this does not come off as too academic...

In a common source FET amplifier, the input (gate) resistance is really really high, so the gate current is usually assumed to be zero. So the current through $R_G$ should be zero.

Also, you should know that the drain current in a FET operating in the saturation region is given by

$$I_D = \frac {k'} 2 \frac W L (V_{GS} - V_t)^2$$

Where $k'=\mu_n C_{ox}$ and is related to process parameters. $\mu_n$ is electron mobility (in NMOS) and $C_{ox}$ is gate oxide capacitance. W and L are the width and length of the channel, respectively. So you have an expression relating drain current to gate-source voltage. And $V_{ov}$ as well, since $V_{ov}=V_{GS}-V_t$.

You already know the drain current, $I_D=1m\text{A}$.

You also should know that in a common source amplifier, the open circuit voltage gain $A_v$ is related to the drain and source resistances, $R_D$ and $R_S$.

And you know that $V_{GS}$ should be equal to the voltage across $R_S$. (And you know the current through $R_S$, which is $I_D$. Correction: $V_{GS}$ is not the drop across $R_S$! But you should be able to determine the relationship from the information you have.

Without having a circuit drawing in front of me, and without having thought about it too much (please check my work!), I think that should give you enough info to make these calculations. You may have to leave in a "transconductance" parameter in the expressions for things if you do not know $k'$ or $\frac W L$.

Analysis and Design of Analog Integrated Circuits by Paul R. Gray and Robert G. Meyer et. al is my go-to reference for transistor circuits. I have the 5th edition. It's been around forever. You would probably be okay with a 4th edition. I would highly recommend this if you are interested in linear transistor amplifiers (op amps).

• This is assuming that the FET is a MOSFET or, if it is a JFET, that the gate-source junction is not forward biased. – Adam P Mar 31 '11 at 18:40
• I should also note that the drain current equation written above would also need to be adjusted if the device were a JFET. – Adam P Mar 31 '11 at 19:28

You're not saying what kind of FET you're using, a JFET or a MOSFET. In the latter case the gate is isolated from the channel and there will flow no current through the resistor. If it's a JFET it's different. Depending on the type the gate is either the N- or the P-part of a diode formed by the gate-channel junction. This diode is reversely polarized, but there will be a leakage current which may generate a small voltage over a large enough resistor.

If I understand correctly, there is a resistor connected between the gate of an FET and ground. In order to figure out the exact (very small) current through this gate resistor, you'd need a datasheet, and to get a datasheet, you'd need a part number. Did they give you one?

For practical purposes, you don't need one, since this is an FET, and the gate resistance will be extremely high, making the current extremely low, essentially 0. (But since this is a homework assignment, it might not be very practical.)

• No, $V_{SS}$ is -15V. If the source were also at 0V, then $V_{GS}$ would be 0V, and you'd need to look at the datasheet for leakage current. However, $V_{GS}$ is somewhere between 0V and 15V, so the current is not available directly from the datasheet. – Kevin Vermeer Mar 31 '11 at 17:23
• @reemrevnivek The DC current through RG would have to pass through the gate, so you'd need to look at a datasheet to find out the gate resistance, no? – endolith Mar 31 '11 at 20:53