Microelectronic Circuits, Kyung Hee Univ. Fall, 2015 6.1 Basic …tioh.weebly.com/uploads/2/5/4/3/25433593/14week__chapter... · 2018-09-10 · Microelectronic Circuits, Kyung Hee

Microelectronic Circuits, Kyung Hee Univ. Fall, 2015

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6.1 Basic Principles• When operated in the active region, realize

voltage controlled current source• 𝑖𝑖𝐶𝐶 = 𝐼𝐼𝑆𝑆𝑒𝑒 ⁄𝑣𝑣𝐵𝐵𝐵𝐵 𝑉𝑉𝑇𝑇

• Nonlinear control relationship• For obtaining almost linear amplification

• Interested in voltage amplifier• Trans-conductance amplifier

• Pass the output current through a resistor


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6.1.2 Obtaining a Voltage Amplifier

Figure 6.2: (c) Simple BJT amplifier with input vBE and output vCE. (d) The voltage transfer characteristic (VTC) of the amplifier in (a).

𝑣𝑣𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 − 𝑖𝑖𝐶𝐶𝑅𝑅𝐶𝐶


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6.1.3 Voltage Transfer Characteristic (VTC)

Figure 6.2: Biasing the BJT amplifier at a point Q located on the active-mode segment of the VTC.

𝑣𝑣𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 − 𝑅𝑅𝐶𝐶𝐼𝐼𝑆𝑆𝑒𝑒 ⁄𝑣𝑣𝐵𝐵𝐵𝐵 𝑉𝑉𝑇𝑇


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6.1.4 Biasing the TransistorFigure 6.6: BJT amplifier biased at apoint Q (bias point/dc operating point)with a small voltage signal vbesuperimposed on the dc bias voltageVBE.


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6.1.4 Small-Signal Voltage Gain• Slope of the tangent to the VTC at the bias point Q

• 𝐴𝐴𝑣𝑣 ≡ �𝑑𝑑𝑣𝑣𝐶𝐶𝐵𝐵𝑑𝑑𝑣𝑣𝐵𝐵𝐵𝐵 𝑣𝑣𝐵𝐵𝐵𝐵=𝑉𝑉𝐵𝐵𝐵𝐵

• 𝐴𝐴𝑣𝑣 = − 𝐼𝐼𝐶𝐶𝑉𝑉𝑇𝑇

𝑅𝑅𝐶𝐶 (because 𝑣𝑣𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 − 𝑅𝑅𝐶𝐶𝐼𝐼𝑆𝑆𝑒𝑒 ⁄𝑣𝑣𝐵𝐵𝐵𝐵 𝑉𝑉𝑇𝑇, 𝐼𝐼𝐶𝐶 = 𝐼𝐼𝑆𝑆𝑒𝑒 ⁄𝑉𝑉𝐵𝐵𝐵𝐵 𝑉𝑉𝑇𝑇)

• Gain: • Negative (180⁰ phase shift between the input and output)• Proportional to the 𝐼𝐼𝐶𝐶 and 𝑅𝑅𝐶𝐶

• 𝐴𝐴𝑣𝑣 = − 𝐼𝐼𝐶𝐶𝑅𝑅𝐶𝐶𝑉𝑉𝑇𝑇

= −𝑉𝑉𝑅𝑅𝐶𝐶𝑉𝑉𝑇𝑇

= −𝑉𝑉𝐶𝐶𝐶𝐶−𝑉𝑉𝐶𝐶𝐵𝐵𝑉𝑉𝑇𝑇

• For deciding on the location of Q, need to consider of gain and allowable signal swing at the output


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6.1.6 Determining the VTC by Graphics

𝐼𝐼𝐶𝐶

Figure 6.7 Graphical construction to determine the voltage-transfer characteristic of the amplifier in Fig. 6.4(a).

Figure 6.9 Two load lines and corresponding bias points. Bias point Q1 does not leave sufficient room for positive signal swing at the drain (too close to VDD). Bias point Q2 is too close to the boundary of the triode region and might not allow for sufficient negative signal swing.

𝑣𝑣𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 − 𝑖𝑖𝐶𝐶𝑅𝑅𝐶𝐶

𝑖𝑖𝐶𝐶 =𝑉𝑉𝐶𝐶𝐶𝐶𝑅𝑅𝐶𝐶

−1𝑅𝑅𝐶𝐶

𝑣𝑣𝐶𝐶𝐶𝐶𝑣𝑣𝐵𝐵𝐶𝐶𝑖𝑖𝐶𝐶

𝑣𝑣𝐶𝐶𝐶𝐶𝑉𝑉𝐶𝐶𝐶𝐶𝑉𝑉𝐶𝐶𝐶𝐶

Saturationregion

Cut-offregion

Rc

Gain & Allowable signal swingDetermined by 𝑣𝑣𝐵𝐵𝐶𝐶 and 𝑅𝑅𝐶𝐶


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6.2 Small-Signal Operation and Models

• 𝐼𝐼𝐶𝐶 = 𝐼𝐼𝑆𝑆𝑒𝑒 ⁄𝑉𝑉𝐵𝐵𝐵𝐵 𝑉𝑉𝑇𝑇

• 𝐼𝐼𝐶𝐶 = 𝐼𝐼𝐶𝐶/𝛼𝛼

• 𝐼𝐼𝐵𝐵 = 𝐼𝐼𝐶𝐶/β

• 𝑉𝑉𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 − 𝐼𝐼𝐶𝐶𝑅𝑅𝐶𝐶• 𝑉𝑉𝐶𝐶 > 𝑉𝑉𝐵𝐵 − 0.4 𝑉𝑉


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Collector Current / Transconductance• 𝑣𝑣𝐵𝐵𝐶𝐶 = 𝑉𝑉𝐵𝐵𝐶𝐶 + 𝑣𝑣𝑏𝑏𝑏𝑏• 𝑖𝑖𝐶𝐶 = 𝐼𝐼𝐶𝐶𝑒𝑒 ⁄𝑣𝑣𝑏𝑏𝑏𝑏 𝑉𝑉𝑇𝑇

• (𝑣𝑣𝑏𝑏𝑏𝑏 ≪ 𝑉𝑉𝑇𝑇)

• 𝑖𝑖𝐶𝐶 ≅ 𝐼𝐼𝐶𝐶(1 + 𝑣𝑣𝑏𝑏𝑏𝑏𝑉𝑉𝑇𝑇

) if 𝑣𝑣𝑏𝑏𝑏𝑏<10mV : small signal approximation

• 𝑖𝑖𝐶𝐶 = 𝐼𝐼𝐶𝐶 + 𝐼𝐼𝐶𝐶𝑉𝑉𝑇𝑇𝑣𝑣𝑏𝑏𝑏𝑏

• 𝑖𝑖𝑐𝑐 = 𝐼𝐼𝐶𝐶𝑉𝑉𝑇𝑇𝑣𝑣𝑏𝑏𝑏𝑏

• 𝑖𝑖𝑐𝑐 = 𝑔𝑔𝑚𝑚𝑣𝑣𝑏𝑏𝑏𝑏• 𝑔𝑔𝑚𝑚 = 𝐼𝐼𝐶𝐶

𝑉𝑉𝑇𝑇: trans-conductance

• Slope of 𝑖𝑖𝐶𝐶 − 𝑣𝑣𝑏𝑏𝑏𝑏 characteristic curve at 𝑖𝑖𝑐𝑐 = 𝐼𝐼𝐶𝐶

• Voltage controlled current source

• Output resistance: infinite• Finite output resistance due

to Early effect


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Base Current / Input Resistance• 𝑖𝑖𝐵𝐵 = 𝑖𝑖𝐶𝐶

𝛽𝛽= 𝐼𝐼𝐶𝐶

𝛽𝛽+ 1

𝛽𝛽𝐼𝐼𝐶𝐶𝑉𝑉𝑇𝑇𝑣𝑣𝑏𝑏𝑏𝑏

• 𝑖𝑖𝐵𝐵 = 𝐼𝐼𝐵𝐵 + 𝑖𝑖𝑏𝑏• 𝑖𝑖𝑏𝑏 = 1

𝛽𝛽𝐼𝐼𝐶𝐶𝑉𝑉𝑇𝑇𝑣𝑣𝑏𝑏𝑏𝑏

• 𝑖𝑖𝑏𝑏 = 𝑔𝑔𝑚𝑚𝛽𝛽𝑣𝑣𝑏𝑏𝑏𝑏

• 𝑔𝑔𝑚𝑚 = 𝐼𝐼𝐶𝐶𝑉𝑉𝑇𝑇

: trans-conductance

• Small-signal input resistance between base and emitter=looking into the base

• 𝑟𝑟𝜋𝜋 ≡𝑣𝑣𝑏𝑏𝑏𝑏𝑖𝑖𝑏𝑏

= 𝛽𝛽𝑔𝑔𝑚𝑚

= 𝛽𝛽�𝑉𝑉𝑇𝑇𝐼𝐼𝐶𝐶

= 𝑉𝑉𝑇𝑇𝐼𝐼𝐵𝐵


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Emitter Current and Input Resistance• 𝑖𝑖𝐶𝐶 = 𝑖𝑖𝐶𝐶

𝛼𝛼= 𝐼𝐼𝐶𝐶

𝛼𝛼+ 𝑖𝑖𝑐𝑐

𝛼𝛼= 𝐼𝐼𝐶𝐶 + 𝑖𝑖𝑏𝑏

• 𝑖𝑖𝑏𝑏 = 𝑖𝑖𝑐𝑐𝛼𝛼

= 1𝛼𝛼𝐼𝐼𝐶𝐶𝑉𝑉𝑇𝑇𝑣𝑣𝑏𝑏𝑏𝑏 = 𝐼𝐼𝐵𝐵

𝑉𝑉𝑇𝑇𝑣𝑣𝑏𝑏𝑏𝑏

• Small-signal input resistance between base and emitter=looking into the emitter

• 𝑟𝑟𝑏𝑏 ≡𝑣𝑣𝑏𝑏𝑏𝑏𝑖𝑖𝑏𝑏

= 𝑉𝑉𝑇𝑇𝐼𝐼𝐵𝐵

= 𝛼𝛼𝑔𝑔𝑚𝑚

≅ 1𝑔𝑔𝑚𝑚

• 𝑣𝑣𝑏𝑏𝑏𝑏 = 𝑖𝑖𝑏𝑏𝑟𝑟𝜋𝜋 = 𝑖𝑖𝑏𝑏𝑟𝑟𝑏𝑏• 𝑟𝑟𝜋𝜋 = ⁄𝑖𝑖𝑏𝑏 𝑖𝑖𝑏𝑏 𝑟𝑟𝑏𝑏 = (𝛽𝛽 + 1)𝑟𝑟𝑏𝑏


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Voltage Gain• 𝑣𝑣𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 − 𝑖𝑖𝐶𝐶𝑅𝑅𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 − 𝐼𝐼𝐶𝐶 + 𝑖𝑖𝑐𝑐 𝑅𝑅𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 − 𝐼𝐼𝐶𝐶𝑅𝑅𝐶𝐶 − 𝑖𝑖𝑐𝑐𝑅𝑅𝐶𝐶• = 𝑉𝑉𝐶𝐶𝐶𝐶−𝑖𝑖𝑐𝑐𝑅𝑅𝐶𝐶• 𝑣𝑣𝑐𝑐𝑏𝑏 = −𝑖𝑖𝑐𝑐𝑅𝑅𝐶𝐶 = −𝑔𝑔𝑚𝑚𝑣𝑣𝑏𝑏𝑏𝑏𝑅𝑅𝐶𝐶 = (−𝑔𝑔𝑚𝑚𝑅𝑅𝐶𝐶)𝑣𝑣𝑏𝑏𝑏𝑏• Voltage gain of this amplifier 𝐴𝐴𝑣𝑣• 𝐴𝐴𝑣𝑣 ≡

𝑣𝑣𝑐𝑐𝑏𝑏𝑣𝑣𝑏𝑏𝑏𝑏

= −𝑔𝑔𝑚𝑚𝑅𝑅𝐶𝐶 = − 𝐼𝐼𝐶𝐶𝑅𝑅𝐶𝐶𝑉𝑉𝑇𝑇

Separating Signal/DC• Replaced 𝑉𝑉𝐶𝐶𝐶𝐶 and 𝑉𝑉𝐵𝐵𝐶𝐶 with short circuits• Current increments (𝑖𝑖𝑐𝑐, 𝑖𝑖𝑏𝑏, and 𝑖𝑖𝑏𝑏) obtained

when a small signal 𝑣𝑣𝑏𝑏𝑏𝑏 is applied• Considered an equivalent small-signal circuit

model

Figure 6.23 The amplifier circuit of Fig. 6.20(a) with the dc sources (VBE and VCC) eliminated (short-circuited). Thus only the signal components are present. Note that this is a representation of the signal operation of the BJT and not an actual amplifier circuit.


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Hybrid-π Model• Voltage controlled current source• Input resistance looking into the base, 𝑟𝑟𝜋𝜋• 𝑖𝑖𝑐𝑐 = 𝑔𝑔𝑚𝑚𝑣𝑣𝑏𝑏𝑏𝑏 and 𝑖𝑖𝑏𝑏 = 𝑣𝑣𝑏𝑏𝑏𝑏

𝑟𝑟𝜋𝜋

• 𝑖𝑖𝑏𝑏 = 𝑣𝑣𝑏𝑏𝑏𝑏𝑟𝑟𝜋𝜋

+ 𝑔𝑔𝑚𝑚𝑣𝑣𝑏𝑏𝑏𝑏 = 𝑣𝑣𝑏𝑏𝑏𝑏𝑟𝑟𝜋𝜋

1 + 𝑔𝑔𝑚𝑚𝑟𝑟𝜋𝜋 = 𝑣𝑣𝑏𝑏𝑏𝑏𝑟𝑟𝜋𝜋

1 + 𝛽𝛽 = �𝑣𝑣𝑏𝑏𝑏𝑏𝑟𝑟𝜋𝜋1+𝛽𝛽

= 𝑣𝑣𝑏𝑏𝑏𝑏𝑟𝑟𝑏𝑏

• 𝑔𝑔𝑚𝑚𝑣𝑣𝑏𝑏𝑏𝑏 = 𝑔𝑔𝑚𝑚 𝑖𝑖𝑏𝑏𝑟𝑟𝜋𝜋 = 𝑔𝑔𝑚𝑚𝑟𝑟𝜋𝜋 𝑖𝑖𝑏𝑏 = 𝛽𝛽𝑖𝑖𝑏𝑏

Figure 6.24 Two slightly different versions of the hybrid-π model for the small-signal operation of the BJT. The equivalent circuit in (a) represents the BJT as a voltage-controlled current source (a transconductance amplifier), and that in (b) represents the BJT as a current-controlled current source (a current amplifier).


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T Model• Voltage controlled current source with control

voltage being 𝑣𝑣𝑏𝑏𝑏𝑏• Resistance between base and emitter, looking

into the base, 𝑟𝑟𝑏𝑏• 𝑖𝑖𝑏𝑏 = 𝑣𝑣𝑏𝑏𝑏𝑏

𝑟𝑟𝑏𝑏− 𝑔𝑔𝑚𝑚𝑣𝑣𝑏𝑏𝑏𝑏 = 𝑣𝑣𝑏𝑏𝑏𝑏

𝑟𝑟𝑏𝑏1 − 𝑔𝑔𝑚𝑚𝑟𝑟𝑏𝑏 =

𝑣𝑣𝑏𝑏𝑏𝑏𝑟𝑟𝑏𝑏

1 − 𝛼𝛼 = 𝑣𝑣𝑏𝑏𝑏𝑏𝑟𝑟𝑏𝑏

1 − 𝛽𝛽𝛽𝛽+1

= 𝑣𝑣𝑏𝑏𝑏𝑏(𝛽𝛽+1)𝑟𝑟𝑏𝑏

= 𝑣𝑣𝑏𝑏𝑏𝑏𝑟𝑟𝜋𝜋

• 𝑔𝑔𝑚𝑚𝑣𝑣𝑏𝑏𝑏𝑏 = 𝑔𝑔𝑚𝑚 𝑖𝑖𝑏𝑏𝑟𝑟𝑏𝑏 = 𝑔𝑔𝑚𝑚𝑟𝑟𝑏𝑏 𝑖𝑖𝑏𝑏 = 𝛼𝛼𝑖𝑖𝑏𝑏• Example 6.5~6.7

Figure 6.26 Two slightly different versions of what is known as the T model of the BJT. The circuit in (a) is a voltage-controlled current source representation and that in (b) is a current-controlled current source representation. These models explicitly show the emitter resistance re rather than the base resistance rπ featured in the hybrid-π model.


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Performing Small-Signal Analysis

Figure 6.32 Performing signal analysis directly on the circuit diagram with the BJT small-signal model implicitly employed:

(a) circuit for Example 6.5; (b) circuit for Example 6.7.


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Summary


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Documents

Microelectronic Circuits, Kyung Hee Univ. Fall, 2015 6.1 Basic …tioh.weebly.com/uploads/2/5/4/3/25433593/14week__chapter... · 2018-09-10 · Microelectronic Circuits, Kyung Hee