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EE 310 Electronic Circuit Design 1 Experiment 5 Active Biasing of the MOSFET Amplifie
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  • EE 310
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  • Pennsylvania State University

EE 310 Electronic Circuit Design 1 Experiment 5 Active Biasing of the MOSFET Amplifier Introduction: The purpose of this lab is extremely simple, build a current source and understand why Integrated Circuit (IC) designers use them. The type of current source that will be constructed in this...

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  • March 6, 2023
  • 10
  • 2022/2023
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  • ee 310 electronic circuit design 1 experiment 5 active biasing of the mosfet amplifie
  • experiment 5 write up pennsylvania state university ee 310

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Page 1


EE 310 Electronic Circuit Design 1

Experiment 5

Active Biasing of the MOSFET Amplifier

By: Michael J. Stachnik



Introduction:
The purpose of this lab is extremely simple, build a current source and understand why

Integrated Circuit (IC) designers use them. The type of current source that will be constructed in

this experiment is called a Current Mirror. A Current Mirror is a circuit designed to copy a

current through one active device by controlling the current in another active device of a circuit,

keeping the output current constant regardless of loading. A Current Mirror can be designed

from bipolar junction transistors (BJT) or field-effect transistors (FET), which are extremely

useful in biasing other transistors to carry out a number of analog circuit functions. Hopefully

this lab will reinforce the MOSFET analysis being taught in class as well as answer a few

questions on what exactly current sources are and how they can be constructed.

, Page 2


TASK A:


A1. Calculate RREF

iD = Kn * (VGS - VTN)2
100 μA = 201 μA * (VGS – 0.623 V)2
VGS = -0.082 V or 1.32 V

IREF = (V+ - VGS - V-) / RREF
100 μA = (10 V – 1.32 V – 0 V) / RREF
RREF = 86.7 kΩ



A2. Measure IREF

Using a 86.5 kΩ resistor and the steps supplied in Section 3. Experimental Effort parts 1
through 3, we were able to measure:
IREF = 98.56 μA


A3. Measure IO

By setting VDS2 = VDS1 = VGS and following the steps supplied in Section 3.
Experimental Effort parts 1 through 3, we were able to measure:

IO = 99.67 μA


A4. Derive the effect of Kn on IO / IREF

IO = [(Kn2 * IREF) * (1 + λn * VDS2)] / [Kn1 * (1 + λn * VGS)]
IO = (Kn2 / Kn1) * IREF
IO ≈ IREF



A5. Measure IO for two different VDS2 values

Set VDS2 IO
1 1.32 V 99.67 μA
2 4.0 V 102. 05 μA

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