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

2.) Construct the JFET circuit below.

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

I have used the following resistors:

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

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

So my calculated current becomes:

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

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

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

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

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

5.) Calculate [math]\displaystyle{ y_{fs} }[/math] for your JFET (20 pnts)

For common source configuration JFET:

[math]\displaystyle{ 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]\displaystyle{ y_{fs} }[/math] we need to know the functional dependence of [math]\displaystyle{ I_D(V_{GS}) }[/math]. Lets approximate this function by line using my measurements and plot above:

The line equation is:

[math]\displaystyle{ 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]\displaystyle{ V_{DS} }[/math] in saturation region from my first measurements of [math]\displaystyle{ I_D }[/math] as function of [math]\displaystyle{ V_{DS} }[/math] for four different values of [math]\displaystyle{ V_{GS} }[/math]. So we can take the partial derivative of [math]\displaystyle{ I_D }[/math] with respect to [math]\displaystyle{ V_{GS} }[/math] using the line equation above. Finally,

[math]\displaystyle{ 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]\displaystyle{ y_{fs} }[/math] depend on [math]\displaystyle{ I_D }[/math]? (10 pnts)

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

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

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

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