1. ENEE 303 4th Discussion

2. Contents
Common Emitter Amplifier
BJT and MOSFET Current Mirrors

3. BJT: Hybrid-𝝅 Equivalent Circuit
From signal point of view, BJT behaves as voltage-controlled current source.
Accepts 𝑣 𝑏𝑒 between base and emitter
Provides current 𝑖 𝑐 at the collector
Input resistance is high but finite
Output resistance is high
𝐴 𝑣 =− 𝑅 𝐶 𝑔 𝑚
𝐴 𝑣 =− 𝑅 𝐶 ∥ 𝑟 𝑜 𝑔 𝑚
Small-signal models for the BJT: (a) neglecting the dependence of iC on vCE in saturation (the Early effect) and (b) including the Early effect
ENEE 303 Fall 2017

4. Analysis of Transistor Amplifier Circuits
Eliminate the signal source and determine the dc operating point of the transistor.
Calculate the values of the parameters of the small-signal model.
Eliminate the dc sources by replacing each dc voltage source by a short-circuit and each dc current source by an open- circuit.
Replace the transistor with one of its small-signal equivalent circuit models.
Analyze the circuit to determine the desired quantities.
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5. Characterizing Amplifiers
Input resistance 𝑅 𝑖 ≡ 𝑣 𝑖 𝑖 𝑖
Output resistance 𝑅 𝑜 ≡ 𝑣 𝑥 𝑖 𝑥
Open-circuit voltage gain 𝐴 𝑣𝑜 ≡ 𝑣 𝑜 𝑣 𝑖 𝑅 𝐿 =∞
Amplifier voltage gain 𝐴 𝑣 ≡ 𝑣 𝑜 𝑣 𝑖
𝐴 𝑣 = 𝐴 𝑣𝑜 𝑅 𝐿 𝑅 𝐿 + 𝑅 𝑜
Overall voltage gain 𝐺 𝑣 ≡ 𝑣 𝑜 𝑣 𝑠𝑖𝑔
𝐺 𝑣 = 𝑅 𝑖𝑛 𝑅 𝑖𝑛 + 𝑅 𝑠𝑖𝑔 𝐴 𝑣𝑜 𝑅 𝐿 𝑅 𝐿 + 𝑅 𝑜
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6. Example: CE Amplifier Calculate voltage gain given b = 100.
Determine bias point
Determine small signal model parameters
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7. The Common-Emitter (CE) Amplifier Configuration
moderate to low in value (typically kW); can be increased by lowering the bias current IC but this lowers the gain; for higher Rin need to modify this configuration
moderate to high in value (typically kW); can be reduced by lowering RC but this lowers the gain; for lower Ro emitter-follower configuration is needed
Avo = -gm (Rc||ro), no vpi
If RL is connected to the output, then
And the overall gain is,
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8. The CE Amplifier with Emitter Resistance
Re substantially increases the input resistance
Ro is unchanged,
The open-circuit voltage gain is,
Avo is reduced by 1 + gmRe
If RL is connected to the output, then
And the overall gain is,
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9. Summary: BJT Amplifiers
Emitter Follower (Common Collector)
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10. The Basic MOSFET Current Mirror
The drain current of Q1 is supplied by VDD through resistor R which in most cases is located outside of the IC chip.
If one considers transistor Q2, it is realized that it has VGS identical to Q1 .
The head of the circuit is transistor Q1, the drain of which is shorted to the gate, thereby forcing it to operate in saturation mode.
The output current is related to the reference current by the aspect ratio of the transistors.
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11. MOS Current-Steering Circuits
Once a constant current has been generated, it can be replicated to provide dc bias or load current for the various stages of the amplifier in an IC.
Current mirrors can be used to achieve this goal.
Q1 and R determine IREF
To keep the transistors in saturation,
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12. Example: MOSFET Current Mirror
For the circuit below, let 𝑉 𝐷𝐷 = 𝑉 𝑆𝑆 =1.5 V, 𝑉 𝑡𝑛 =0.6 V, 𝑉 𝑡𝑝 =−0.6 V, all channel lengths =1𝜇m, 𝑘 𝑛 ′ =200 𝜇A V 2 , 𝑘 𝑝 ′ =200 𝜇A V 2 , and 𝜆=0. For 𝐼 𝑅𝐸𝐹 =10 𝜇A, find the widths of all transistors to obtain 𝐼 2 =60 𝜇A, 𝐼 3 =20 𝜇A, and 𝐼 5 =80 𝜇A. It is further required that the voltage at the drain of 𝑄 2 be allowed to go down to within 0.2 V of the negative supply and that the voltage at the drain of 𝑄 5 be allowed to go up to within 0.2 V of the positive supply.
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13. Example: MOSFET Current Mirror
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14. Example: MOSFET Current Mirror
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15. Basic BJT Current Mirror Circuit
It works in a fashion very similar to the MOS mirror. However, with two important differences:
The non-zero bias current causes an error in current mirroring (magnitude of current conducted).
The current transfer ratio is determined by the relative areas of the emitter-based junctions of Q1 and Q2.
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16. Basic BJT Current Mirror Circuit
Assume 𝛽=∞
𝐼 𝑅𝐸𝐹 = 𝐼 𝐶1
𝐼 𝑜 = 𝐼 𝐶2
𝑖=0
𝐼 𝐶1 = 𝐼 𝑆1 𝑒 𝑣 𝐵𝐸 𝑉 𝑇
𝐼 𝐶2 = 𝐼 𝑆2 𝑒 𝑣 𝐵𝐸 𝑉 𝑇
Since 𝑉 𝐵𝐸 is the same for 𝑄 1 and 𝑄 2 ,
𝐼 𝑜 𝐼 𝑅𝐸𝐹 = 𝐼 𝑆2 𝐼 𝑆1 =𝑚= 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐸𝐵𝐽 𝑜𝑓 𝑄 2 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐸𝐵𝐽 𝑜𝑓 𝑄 1
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17. Basic BJT Current Mirror Circuit
Assume both have the same finite 𝛽.
𝐼 𝐶2 = 𝐼 𝑜
𝐼 𝑅𝐸𝐹 = 𝐼 𝐶1 + 𝐼 𝐵
𝐼 𝐵 = 𝐼 𝐵1 + 𝐼 𝐵2
𝐼 𝐶1 = 𝐼 𝑆1 𝑒 𝑣 𝐵𝐸 𝑉 𝑇
𝐼 𝐶2 = 𝐼 𝑆2 𝑒 𝑣 𝐵𝐸 𝑉 𝑇
𝐼 𝐵1
𝐼 𝐵2
𝐼 𝐵1 = 𝐼 𝐶1 𝛽
𝐼 𝐵2 = 𝐼 𝐶2 𝛽
𝐼 𝐸1 = 𝐼 𝐶 𝛽
𝐼 𝐸2 = 𝐼 𝐶 𝛽
Since 𝑉 𝐵𝐸 is the same for 𝑄 1 and 𝑄 2 ,
𝐼 𝑜 𝐼 𝑅𝐸𝐹 = 𝑚 1+ 𝑚+1 𝛽 , 𝑚= 𝐼 𝑆2 𝐼 𝑆1
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18. Basic BJT Current Mirror Circuit
If the Early effect is included,
𝐼 𝑜 𝐼 𝑅𝐸𝐹 = 𝑚 1+ 𝑚+1 𝛽 𝑉 𝑜 − 𝑉 𝐵𝐸 𝑉 𝐴2
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19. Example: BJT Current Mirror
Assume the availability of BJTs with scale currents 𝐼 𝑆 = 10 −15 A, 𝛽=100, and 𝑉 𝐴 =50 V, design the current-source circuit below to provide an output current 𝐼 𝑜 =0.5 mA at 𝑉 𝑜 =0.2 V. The power supply 𝑉 𝐶𝐶 =5 V. Give the values of 𝐼 𝑅𝐸𝐹 , 𝑅, and 𝑉 𝑂𝑚𝑖𝑛 . Also, find 𝐼 𝑜 at 𝑉 𝑜 =5 V. (Q1 & Q2 are the same)
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20. Example: BJT Current Mirror
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21. Example: BJT Current Mirror 2
For the circuit below, let 𝑉 𝐵𝐸 =0.7 V and 𝛽=∞. Find 𝐼, 𝑉 1 , 𝑉 2 , 𝑉 3 , 𝑉 4 , and 𝑉 5 for (a) 𝑅=10 kΩ and (b) 𝑅=100 kΩ.
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22. Example: BJT Current Mirror 2
ENEE 303 Fall 2017

23. Example: BJT Current Mirror 2
ENEE 303 Fall 2017