Lab 17 RS

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The JFET (Junction Field Effect Transistor n-channel)

File:JFET MPF102 DataSheet.pdf

1). Complete the table below for the JFET.

Characteristic	Symbol	Min	Max	Unit
Zero-Gate-Voltage Drain Current	$I_{DSS}$	2.0	20	mAdc
Gate-Source Cutoff Voltage	$V_{GS(off)}$	-	-8.0	Vdc
Total Device Dissipation @ $T_A=25^oC$	$P_{max}$	350		mW
Gait resistor	$R_G$	3.3		M$\Omega$
Drain resistor	$R_D$	1.0		k$\Omega$
Forward Transfer Admittance	$y_{fs}$	2000	7500	$\mu$mhos
Input Admittance	$y_{is}$	-	800	$\mu$mhos
Output Conductance	$y_{os}$	-	200	$\mu$mhos

2.) Construct the JFET circuit below.

3.) Plot measurements of $I_D$ -vs- $V_{DS}$ by varying $V_{dd}$ for $\left | V_{GS}\right |$ = 0, 0.5, 1.0, 1.5 V. (40 pnts)

I have used the following resistors:

[math]R_G = (3.34 \pm 0.02)\ M\Omega[/math]
[math]R_D = (0.968 \pm 0.002)\ k\Omega[/math]

Below is the table with my measurements of voltages $V_{DS}$ and $V_{R_D}$ and calculation of the current $I_D$. Here I have used the meter to measure directly the voltage drop between the drain and source $V_{DS}$ and to measure the voltage drop on resistor $R_D$.

So my calculated current becomes:

[math]I_D = \frac{V_{R_D}}{R_D}[/math].

And below I have plotted four curves $I_D$ as function of $V_{DS}$ for four different values of $V_{GS}$

4.) Plot $I_D$ -vs- $V_{GS}$ (30 pnts)

For every measured $V_{GS}$ values I have picked up the current $I_D$ values in the middle of saturation region of each line as follow:

And below is my plot of $I_D$ -vs- $V_{GS}$:

5.) Calculate $y_{fs}$ for your JFET (20 pnts)

For common source configuration JFET:

[math]y_{fs} \equiv  \left ( \frac{\partial I_{out}}{\partial V_{in}} \right )_{V_{out}} = \left ( \frac{\partial I_D}{\partial V_{GS}} \right )_{V_{DS}}[/math]

So to calculate $y_{fs}$ we need to know the functional dependence of $I_D(V_{GS})$. Lets approximate this function by line using my measurements and plot above:

The line equation is:

[math]I_D[mA] = (10.53 \pm 0.04)[mA] + (4.04 \pm 0.04)\cdot V_{GS}[V][/math]

Also note that this line equation was obtained using about the same voltage $V_{DS}$ in saturation region from my first measurements of $I_D$ as function of $V_{DS}$ for four different values of $V_{GS}$. So we can take the partial derivative of $I_D$ with respect to $V_{GS}$ using the line equation above. Finally,

[math]y_{fs} \equiv  \left ( \frac{\partial I_D}{\partial V_{GS}} \right )_{V_{DS}} = (4.04 \pm 0.04)\ \frac{mA}{V} =  (4.04 \pm 0.04)\ mS[/math]

Question

Does $y_{fs}$ depend on $I_D$? (10 pnts)

No. As we can see from calculation above $y_{fs}$ is constant and does not depend from $I_D$. That is true if we are working in saturation region where the functional dependence of $I_D$ with respect to $V_{GS}$ is line so

$y_{fs} \equiv \left ( \frac{\partial I_D}{\partial V_{GS}} \right )_{V_{DS}} = const$

If we are in active region of $I_D$ as function of $V_{DS}$ the functional form of $I_D$ with respect to $V_{GS}$ is not the line anymore and $y_{fs}$ will depend on $I_D$.

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