Difference between revisions of "Lab 17 RS"

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[https://wiki.iac.isu.edu/index.php/Electronics_RS Go Back to All Lab Reports]
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The JFET (Junction Field Effect Transistor n-channel)
 
The JFET (Junction Field Effect Transistor n-channel)
  
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1.) Complete the table below for the JFET.
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=1). Complete the table below for the JFET.=
  
  
 
{| border="2"  cellpadding="8" cellspacing="0"
 
{| border="2"  cellpadding="8" cellspacing="0"
|Parameter ||  Value
 
 
|-
 
|-
|<math> I_{DSS}</math> ||
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!scope="col" | Characteristic
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!scope="col" | Symbol
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!scope="col" | Min
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!scope="col" | Max
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!scope="col" | Unit
 
|-
 
|-
|<math>V_{GS(off)}</math> ||
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| Zero-Gate-Voltage Drain Current ||<math> I_{DSS}</math> || 2.0 || 20 || mAdc
 
|-
 
|-
|<math>P_{max}</math> ||
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| Gate-Source Cutoff Voltage || <math>V_{GS(off)}</math> || - || -8.0 || Vdc
 
|-
 
|-
|<math>R_G</math> || 3.3 M<math>\Omega</math>
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| Total Device Dissipation @ <math>T_A=25^oC</math>
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| <math>P_{max}</math>  
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| colspan="2" | 350
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| mW
 
|-
 
|-
|<math>R_D</math> ||  
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| Gait resistor
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| <math>R_G</math>
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| colspan="2" | 3.3
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| M<math>\Omega</math> 
 
|-
 
|-
|<math>y_{fs}</math> ||  
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| Drain resistor
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| <math>R_D</math>
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| colspan="2" | 1.0
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| k<math>\Omega</math> 
 
|-
 
|-
|<math>y_{is}</math> ||  
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| Forward Transfer Admittance  || <math>y_{fs}</math> || 2000 || 7500 || <math>\mu</math>mhos
 
|-
 
|-
|<math>y_{os}</math> ||  
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| Input Admittance || <math>y_{is}</math> || - || 800 || <math>\mu</math>mhos
 
|-
 
|-
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| Output Conductance || <math>y_{os}</math> || - || 200 || <math>\mu</math>mhos
 
|}
 
|}
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=2.) Construct the JFET circuit below.=
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[[File:TF_EIM_Lab17Circuit.png| 300 px]]
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=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)=
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I have used the following resistors:
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<math>R_G = (3.34 \pm 0.02)\ M\Omega</math>
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<math>R_D = (0.968 \pm 0.002)\ k\Omega</math>
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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>.
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So my calculated current becomes:
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<math>I_D = \frac{V_{R_D}}{R_D}</math>.
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[[File:Table01.png | 550 px]]
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[[File:Table02.png | 550 px]]
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[[File:Table03.png | 550 px]]
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[[File:Table04.png | 550 px]]
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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>
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[[File:L17 id vs vgs.png | 800 px]]
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=4.) Plot <math> I_D</math> -vs- <math>V_{GS}</math> (30 pnts)=
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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:
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[[File:Table21.png | 500 px]]
  
  
2.)Construct the JFET circuit below.
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And below is my plot of <math> I_D</math> -vs- <math>V_{GS}</math>:
  
[[File:TF_EIM_Lab17Circuit.png| 400 px]]
 
  
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)
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[[File:L17 id vs vgs 21.png | 800 px]]
  
4.)Plot<math> I_D</math> -vs- <math>V_{GS}</math> (30 pnts)
 
  
5.)Calculate <math>y_{fs}</math> for your JFET (20 pnts)
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=5.) Calculate <math>y_{fs}</math> for your JFET (20 pnts)=
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For common source configuration JFET:
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<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>
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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:
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[[File:L17 id vs vgs 22.png | 1000 px]]
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The line equation is:
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<math>I_D[mA] = (10.53 \pm 0.04)[mA] + (4.04 \pm 0.04)\cdot V_{GS}[V]</math>
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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,
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<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=
 
=Question=
  
#Does <math>y_{fs}</math> depend on<math> I_D</math>? (10 pnts)
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==Does <math>y_{fs}</math> depend on <math>I_D</math>? (10 pnts)==
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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
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<math>y_{fs} \equiv  \left ( \frac{\partial I_D}{\partial V_{GS}} \right )_{V_{DS}} = const</math>
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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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[[Forest_Electronic_Instrumentation_and_Measurement]]
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[https://wiki.iac.isu.edu/index.php/Electronics_RS Go Back to All Lab Reports] [[Forest_Electronic_Instrumentation_and_Measurement]]

Latest revision as of 16:41, 13 April 2011

Go Back to All Lab Reports


The JFET (Junction Field Effect Transistor n-channel)


File:JFET MPF102 DataSheet.pdf

JFET MPF102 Pinouts.png


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] 3.3 M[math]\Omega[/math]
Drain resistor [math]R_D[/math] 1.0 k[math]\Omega[/math]
Forward Transfer Admittance [math]y_{fs}[/math] 2000 7500 [math]\mu[/math]mhos
Input Admittance [math]y_{is}[/math] - 800 [math]\mu[/math]mhos
Output Conductance [math]y_{os}[/math] - 200 [math]\mu[/math]mhos

2.) Construct the JFET circuit below.

TF EIM Lab17Circuit.png

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].


Table01.png

Table02.png

Table03.png

Table04.png


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]


L17 id vs vgs.png

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:


Table21.png


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


L17 id vs vgs 21.png


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:

L17 id vs vgs 22.png

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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