TF EIM Chapt5 - Revision history

Foretony: /* vertical diffuse planar transistor */

2015-04-02T16:41:26Z

vertical diffuse planar transistor

Foretony: /* Bipolar transistor currents */

2015-04-02T16:38:00Z

Bipolar transistor currents

Foretony: /* The Bipolar transistor concept */

2015-04-02T16:35:38Z

The Bipolar transistor concept

Foretony: /* The Bipolar transistor concept */

2015-04-02T16:34:54Z

The Bipolar transistor concept

Foretony: /* The Bipolar transistor concept */

2015-04-02T16:34:33Z

The Bipolar transistor concept

Foretony: /* Temperature Dependence */

2015-04-02T16:19:41Z

Temperature Dependence

Foretony: /* Temperature Dependence */

2015-04-02T15:54:38Z

Temperature Dependence

Foretony: /* Temperature Dependence */

2015-04-02T15:46:12Z

Temperature Dependence

Foretony: /* Load Line */

2011-03-16T06:23:21Z

Load Line

Foretony: /* Bipolar Transistor Amplifier */

2011-03-16T06:23:04Z

Bipolar Transistor Amplifier

@@ Line 86: / Line 86: @@
 ==vertical diffuse planar transistor==
-Instead you start with a pure crystal of silicon that has been doped to to form the type os semi-conductor you wish to have as the COLLECTOR.
+Instead you start with a pure crystal of silicon that has been doped to to form the type of semi-conductor you wish to have as the COLLECTOR.
 For now let's assume p-type, this means you dope pure silicon with boron.  The collector is big (thick) because it has a lot of heat to dissipate.

← Older revision		Revision as of 16:38, 2 April 2015
Line 37:		Line 37:
	:<math>\Rightarrow I_E = I_B + I_C</math>		:<math>\Rightarrow I_E = I_B + I_C</math>

−	Here it is the case that the positive current is actually representing the flow of positively charged holes (in ~~non-semi-~~conductors the current ~~represent~~ the opposite direction of the flowing electrons)	+	Here it is the case that the positive current is actually representing the flow of positively charged holes (in conductors the current is in the opposite direction of the flowing electrons)

@@ Line 26: / Line 26: @@
 |}
-The bottom p-n junction, when forward biased, provides a large current with a small voltage bias.
+The bottom p-n-p junction, when forward biased, provides a large current with a small voltage bias.
 The top half of the p-n-p transistor is reversed biased, providing a "throttle" for the large current from the lower half.

@@ Line 20: / Line 20: @@
 | The p-n  junction.  The depletion region is represented as the shaded square. Notice the depletion region (and E-field in that region) is larger than the forward biased n-p juntion
 |-
-|[[File:TF_EIM_BipolarJunction.png| 200 px]][[File:TF_EIM_pnpBipTran.png| 200 px]] [[File:TF_EIM_BipolarJunctionDiodeRep.png| 200 px]] [[File:TF_EIM_BipolarJunctionCircuit.png| 200 px]] ||
+|[[File:TF_EIM_BipolarJunction.png| 200 px]][[File:TF_EIM_pnpBipTran.png| 200 px]] [[File:TF_EIM_BipolarJunctionDiodeRep.png| 200 px]] [[File:TF_EIM_BipolarJunctionCircuit.png| 200 px]]
 |-
 | The p-n-p  junction.

@@ Line 12: / Line 12: @@
 {| border="3"  cellpadding="20" cellspacing="0"
-|[[File:TF_EIM_ForwardBiased_bottomhalfBipolar.jpg| 200 px]] ||
+|[[File:TF_EIM_ForwardBiased_bottomhalfBipolar.jpg| 200 px]]
 |-
-| The n-p junction.  The depletion region is represented as the shaded square || The potential
+| The n-p junction.  The depletion region is represented as the shaded square
 |-
-|[[File:TF_EIM_ReverseBiased_tophalfBipolar.jpg| 200 px]] ||
+|[[File:TF_EIM_ReverseBiased_tophalfBipolar.jpg| 200 px]]
 |-
-| The n-p  junction.  The depletion region is represented as the shaded square. Notice the depletion region (and E-field in that region) is larger than the forward biased n-p juntion || the potential
+| The p-n  junction.  The depletion region is represented as the shaded square. Notice the depletion region (and E-field in that region) is larger than the forward biased n-p juntion
 |-
 |[[File:TF_EIM_BipolarJunction.png| 200 px]][[File:TF_EIM_pnpBipTran.png| 200 px]] [[File:TF_EIM_BipolarJunctionDiodeRep.png| 200 px]] [[File:TF_EIM_BipolarJunctionCircuit.png| 200 px]] ||
 |-
-| The n-p  junction.  The depletion region is represented as the shaded square. Notice the depletion region (and E-field in that region) is larger than the forward biased n-p juntion || the potential
+| The p-n-p  junction.
 |-
 |}

@@ Line 224: / Line 224: @@
 :<math>I_C = \alpha I_E + I_{CBO}</math>
-The collector current increases and as a result the transistor temperature increases
+As<math> I_{CBO}</math> increases with temperature, the base voltage <math>V_B</math> becomes more positive (the increase <math>I_{CBO}</math> increases <math>V_{BE}</math> and thus you need less <math>V_B</math> to overcome the the base-emitter junction).
 The base current becomes
@@ Line 230: / Line 234: @@
 :<math>I_B = (1-\alpha)I_E - I_{CBO}</math>
-So the base current becomes less but it is usually kept constant by the power source.
+The base current becomes less but it is usually kept constant by the power source.
 You have "thermal runaway" and the transistor may burn out.
-[[File:TF_EIM_LoadLineTempControlledCircuit.png| 200 px]]
+To compensate for this you add a resistor (<math>R_E</math>) between the base and emitter as shown below which holds <math>V_{EB}</math> constant. (Now <math>V_B</math> and <math>V_E</math> are both held constant)   As <math>V_B</math> goes more positive due to the reverse current I_{CBO} the resistor <math>R_1</math> will make <math>V_E</math> go more positive as well.
-As<math> I_{CBO}</math> increases with temperature, the base voltage <math>V_B</math> becomes more positive (the increase <math>I_{CBO}</math> increases <math>V_{BE}</math> and thus you need less <math>V_B</math> to overcome the the base-emitter junction).
+[[File:TF_EIM_LoadLineTempControlledCircuit.png| 200 px]]
 =Bipolar Transistor Amplifier=

@@ Line 246: / Line 246: @@
 As <math>V_B</math> goes more positive, <math>I_C</math> and <math> I_E</math> increase.  But since <math>V_E = I_E R_E</math>, <math>V_E</math> will increase.
-To prevent this you add a resistor (R_1) between the base and emitter as shown below.   As V_B goes more positive due to the reverse current I_{CBO} the resistor R_1 will make V_E go more positive as well.
+To prevent this you add a resistor (<math>R_1</math>) between the base and emitter as shown below.   As <math>V_B</math> goes more positive due to the reverse current I_{CBO} the resistor <math>R_1</math> will make <math>V_E</math> go more positive as well.
 =Bipolar Transistor Amplifier=

← Older revision		Revision as of 15:46, 2 April 2015
Line 245:		Line 245:

	As <math>V_B</math> goes more positive, <math>I_C</math> and <math> I_E</math> increase. But since <math>V_E = I_E R_E</math>, <math>V_E</math> will increase.		As <math>V_B</math> goes more positive, <math>I_C</math> and <math> I_E</math> increase. But since <math>V_E = I_E R_E</math>, <math>V_E</math> will increase.
		+
		+	To prevent this you add a resistor (R_1) between the base and emitter as shown below. As V_B goes more positive due to the reverse current I_{CBO} the resistor R_1 will make V_E go more positive as well.

	=Bipolar Transistor Amplifier=		=Bipolar Transistor Amplifier=

@@ Line 170: / Line 170: @@
 | npn||low || high|| low||high ||high || high
 |}
 ==Load Line==

← Older revision		Revision as of 06:23, 16 March 2011
Line 410:		Line 410:
	:<math>V_B = \frac{R_1}{R_1+R_2} V_{CC}</math>		:<math>V_B = \frac{R_1}{R_1+R_2} V_{CC}</math>

−	~~=Bipolar Transistor Amplifier=~~

−
−	~~[[File:TF_EIM_Lab14a.png]]~~
−
−
−	~~Consider the above circuit.~~
−
−	~~Observations~~
−
−	~~1.) Notice the input goes through a high pass filter with a break point of<math> \frac{1}{R_1 C_1}</math>.~~
−
−	~~2.) The output also goes through a high pass filter with a break point of<math> \frac{1}{R_l C_2}</math>.~~
−
−	~~3.) Kirchoff voltage rule~~
−
−	~~:<math>V_{CC} -I_C R_C - V_{CE} - I_E R_E = 0</math>~~
−	~~:<math>V_{CC} - V_{CE} -I_C (R_C + R_E) = 0</math> <math>I_C \approx I_E</math>~~
−
−	~~:<math>\Rightarrow I_C =\frac{ V_{CC} - V_{CE}}{R_C + R_E }</math>~~
−
−	~~4.) Kirchoff law~~
−
−	~~:<math>V_{CC} -(I_B + I_D) R_2 - V_{EB} - I_E R_E = 0</math>~~
−
−	~~5.) INput resistance~~
−
−	~~<math>\frac{1}{R_{in}} = \frac{1}{X_B} + \frac{1}{R_1} + \frac{1}{R_2}</math>~~
−
−	~~6.)~~
−
−	~~:<math>V_B = \frac{R_1}{R_1+R_2} V_{CC}</math>~~

	[[Forest_Electronic_Instrumentation_and_Measurement]]		[[Forest_Electronic_Instrumentation_and_Measurement]]