Lecture_23

Lecture_23 - V CE < 0, the transistor is likely biased...

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EE 310 Lecture 23 DC Analysis of Transistor Circuits Consider the two currents I B and I C : V BE (on) assumption: V BE I B Lecture 23 Page 1
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npn Example Find I B , I C , I E , and V CE : V BE (on) = 0.7 V, β = 200 EE 310 Lecture 23 Lecture 23 Page 2
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pnp Example Find I B , I C , I E , and R C such that V EC = 2.5 V: V EB (on) = 0.6 V, β = 100 EE 310 Lecture 23 Lecture 23 Page 3
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Graphical Analysis - the Load Line Find I BQ : Plot endpoints of load line: EE 310 Lecture 23 Lecture 23 Page 4
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Generalized Approach Assume that the transistor is biased in the forward-active region in which case V BE = V BE (on) , I B > 0, and I C = β I B . 1. Analyze the "linear" circuit with this assumption. 2. Evaluate the resulting state of the transistor. If the initial assumed parameter values and V CE > V CE (sat) are true, then the initial assumption is correct. However, if the calculation shows I B < 0, then the transistor is probably cut off, and if the calculation shows
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Unformatted text preview: V CE < 0, the transistor is likely biased in saturation. 3. If the initial assumption is proven incorrect, then a new assumption must be made and the new "linear" circuit must be analyzed. Step 3 must then be repeated. 4. Problem-Solving Approach (textbook page 311): EE 310 Lecture 23 Lecture 23 Page 5 Example of Device in Saturation V BE (on) = 0.7 V, β = 100. If in saturation, V CE sat = 0.2 V Recalculate assuming the device is saturated. Find the power dissipated in the transistor. Find I C and V CE assuming that the device is in the active region: EE 310 Lecture 23 Lecture 23 Page 6 Another Test for Saturation V BE (on) = 0.7 V, β = 100. If in saturation, V CE sat = 0.2 V Find the ratio I C / I B : I B I C EE 310 Lecture 23 Lecture 23 Page 7...
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This note was uploaded on 10/12/2010 for the course E E 310 at Penn State.

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Lecture_23 - V CE < 0, the transistor is likely biased...

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