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	[math] I_{DSS}[/math]	2.0	20	mAdc
Gate-Source Cutoff Voltage	[math]V_{GS(off)}[/math]	-	-8.0	Vdc
Total Device Dissipation @ [math]T_A=25^oC[/math]	[math]P_{max}[/math]	350		mW
Gait resistor	[math]R_G[/math]	3.3M[math]\Omega[/math]
Drain resistor	[math]R_D[/math]	1.0		M[math]\Omega[/math]
	[math]y_{fs}[/math]
	[math]y_{is}[/math]
	[math]y_{os}[/math]

2.) Construct the JFET circuit below.

3.) Plot measurements of [math]I_D[/math] -vs- [math]V_{DS}[/math] by varying [math]V_{dd}[/math] for [math]\left | V_{GS}\right |[/math] = 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 [math]V_{DS}[/math] and [math]V_{R_D}[/math] and calculation of the current [math]I_D[/math]. Here I have used the meter to measure directly the voltage drop between the drain and source [math]V_{DS}[/math] and to measure the voltage drop on resistor [math]R_D[/math].

So my calculated current becomes:

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

And below I have plotted four curves [math]I_D[/math] as function of [math]V_{DS}[/math] for four different values of [math]V_{GS}[/math]

4.) Plot [math] I_D[/math] -vs- [math]V_{GS}[/math] (30 pnts)

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

And below is my plot of [math] I_D[/math] -vs- [math]V_{GS}[/math]:

5.) Calculate [math]y_{fs}[/math] 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 [math]y_{fs}[/math] we need to know the functional dependence of [math]I_D(V_{GS})[/math]. 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 [math]V_{DS}[/math] in saturation region from my first measurements of [math]I_D[/math] as function of [math]V_{DS}[/math] for four different values of [math]V_{GS}[/math]. So we can take the partial derivative of [math]I_D[/math] with respect to [math]V_{GS}[/math] 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 [math]y_{fs}[/math] depend on [math]I_D[/math]? (10 pnts)

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

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

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

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