Chapter 4

Bipolar Junction Transistor

Base, Emitter, Collector

General BJT notes

• NPN: Not Pointing iN
• PNP: Pointing iN Please
• The arrow is for the emitter current.

• For an NPN the current flows into the collector. For a PNP the current flows out of the collector.

• Quiescent is a none referring to a state of being quiet, still, at rest, dormant, inactive. In an electronic amplifier or filter is said to be in a quiescent state when no signal is applied to its input

Large Signal vs Small Signal Analysis

	DC sources	AC sources	Capacitor	BJT
DC analysis	No change	0V or 0A	Open-Circuit	Active, Saturation or Cutoff
AC analysis	0V or 0A	No change	Short-Circuit	Small-signal model

Signal Notation

• Total = $lowercase_{UPPERCASE}$
• DC = $UPPERCASE_{UPPERCASE}$
• AC = $lowercase_{lowercase}$

Large-Signal DC models

BJT Regions Base-Emitter Collector-Base Region FB FB Saturation FB RB Active RB RB Cutoff RB FB Reverse Active

Region Constraints Region NPN PNP Saturation $I_{B}>0$ and $\beta I_{B}>I_{C}>0$ $I_{B}>0$ and $\beta I_{B}>I_{C}>0$ Active $I_{B}>0$ and $V_{CE}>0.2$ $I_{B}>0$ and $V_{CE}<-0.2$ Cutoff $V_{BE}<0.5$ and $V_{BC}<0.5$ $V_{BE}>-0.5$ and $V_{BC}>-0.5$

• Active: The normal mode of operation. This mode has the largest common-emitter current gain.
• Reverse Active: The collector and emitter roles are reversed. Most BJTs are not symmetrical, thus $\alpha$ and $\beta$ take on different values.
• Saturation: High current from the emitter to collector. Logical "on".
• Cutoff: Very little current flow. Logical "off".

• Instead of memorizing the region constraints for NPN & PNP BJTs, it is easier to remember the BJT regions as they work for both NPN & PNP.

Large-Signal DC Analysis

1. Assume an operating region for the BJT.
2. Solve the circuit to find $I_{C},I_{B},V_{CE}$
3. Check the values to see if they match the region constraints.

• Active region: The current source is on the collector side and points towards the emitter
• Saturation region: The voltage sources have the same orientation

Bias-Circuit Design

• The principle goal of bias-circuit design is to achieve nearly identical operating points for the BJTs, even though the BJT parameters may vary significantly from unit to unit. <ref>Electronics p245</ref>
• Fixed-base bias circuit is a very simplistic method that does not deal well with a wide variety of $\beta$ values. The base current does not adjust for changes in $\beta$. If you want a circuit that has a particular operating point on the collector load line, the base current must change with $\beta$.
• Using two voltage sources (one for the collector and one for the base) achieves an emitter current that is relatively independent of $\beta$. Usually only one source is available or we may wish to inject a signal into the base. In an ideal world you could simply place the voltage sources in series. Since the voltage sources are grounded, they would be put in parallel. The wire connecting the voltage source will see the potential difference (and a lot of current). The wire will burn up and bad things will happen to the voltage sources.
• Four-resistor bias circuit provides a constant base voltage independant of $\beta$. This design is great for discrete circuits, but is impractical in integrated circuits because of the excessive chip area they consume.

• How does this handle injecting an ac source? How is it not coupled?

Small-Signal Equivalent Circuits

BJT Small Signal equivalent (both PNP & NPN)

• $R_{\pi }={\frac {v_{be}(t)}{i_{b}(t)}}={\frac {V_{T}}{I_{BQ}}}={\frac {\beta V_{T}}{I_{CQ}}}$
• $i_{c}(t)=\beta i_{b}(t)\,$
• Tranconductance: $g_{m}={\frac {\beta }{r_{\pi }}}={\frac {I_{CQ}}{V_{T}}}$

Common-Emitter Amplifier

Common-emitter amplifier small-signal equivalent

• A four-resistor biasing network

• Coupling capacitor to separate the AC input/output from the DC biasing
• Bypass capacitor allows for a common-emitter for AC, yet provides a stable bias current for DC

• Voltage gain:

• $v_{in}=i_{b}r_{\pi }+(\beta +1)i_{b}R_{E1}\,$
• Why don't they use R1 & R2 when finding Vin?
• $v_{o}=-\beta i_{b}R_{L}^{'}\,$, the direction of the current necessitates the minus sign
• $A_{v}={\frac {v_{o}}{v_{in}}}=-{\frac {\beta R_{L}^{'}}{r_{\pi }+(\beta +1)R_{E1}}}$

• Input Impedance:

• $Z_{in}=R_{B}\parallel {\frac {v_{in}}{i_{b}}}=R_{B}\parallel (r_{\pi }+(\beta +1)R_{E1})$

• Output Impedance: $Z_{o}=R_{C}\,$

• The source is set to 0 to make the problem easier. You could leave the source on and use a voltage divider to determine the output impedance.

• Current Gain:

• $A_{i}={\frac {i_{o}}{i_{in}}}={\frac {v_{o}/z_{o}}{v_{in}/z_{in}}}={\frac {v_{o}z_{in}}{v_{in}z_{o}}}$

Common-Collector Amplifier (Emitter Follower)

Common-collector amplifier small-signal equivalent

• Voltage Gain

• $v_{o}=R_{L}^{'}(\beta +1)i_{b}$
• $v_{in}=i_{b}r_{\pi }+(\beta +1)i_{b}R_{L}^{'}$
• $A_{v}={\frac {(\beta +1)R_{L}^{'}}{r_{\pi }+(\beta +1)R_{L}^{'}}}$, thus the gain is less than one, but is positive

• Input Impedance

• $Z_{in}=R_{B}\parallel {\frac {v_{i}n}{i_{b}}}=R_{B}\parallel (r_{\pi }+(1+\beta )R_{L}^{'})$

• Output Impedance

• $Z_{o}=v_{x}/i_{x}\,$

• $v_{x}/R_{E}-i_{x}-(\beta +1)i_{b}=0\,$
• $R_{s}^{'}i_{b}+r_{\pi }i_{b}+v_{x}=0$

• $Z_{o}={\frac {v_{x}}{i_{x}}}={\frac {R_{E}(R_{s}^{'}+R_{\pi })}{R_{s}+R_{\pi }+R_{E}(\beta +1)}}$

BJT models?

?BJT Characteristics?

Type?	Input Resistance	Output Resistance	Voltage Gain	Current Gain	Frequency response
Common-emitter	$R_{B}\parallel (r_{\pi }+(\beta +1)R_{E1})$	$R_{C}\,$	$-{\frac {\beta R_{L}^{'}}{r_{\pi }+(\beta +1)R_{E1}}}$
Common-collector	$R_{B}\parallel (r_{\pi }+(\beta +1)R_{L}^{'})$	${\frac {R_{E}(R_{s}^{'}+R_{\pi })}{R_{s}+R_{\pi }+R_{E}(\beta +1)}}$	${\frac {(\beta +1)R_{L}^{'}}{r_{\pi }+(\beta +1)R_{L}^{'}}}$
Common-base

Questions

• Why do we always seem to use a common emitter configuration? Common-base and common-collector have different properties.
• For P3.17 make the glob of sauder assuming they're all on and check the currents running through each one. Make sure they go the right way, and add up?
• Why is the input/output impedance so important?
• P4.54c

References

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