Title: Current mirror
1Current mirror
- Current mirror is a programmable current source
- What is the load current (IC2)?
15V
IC2
IC1
2Twin - matched pair
- Q1 and Q2 are matched pair (TWIN)
- they have exactly the same performance because
they are fabricated on the same chip, at the same
temperature and have the same level of doping - Since Q1 and Q2 are matched, and have the same
VBE - IC2 IC1
- We can use the resistor in Q1 to control IC2
- IC2 IC1 (15V - 0.6V) / 15kW 1mA
3Current mirror with ratios other than 11
- Quiz What is the current ratio? Why?
4Current mirror with ratios other than 11
- Quiz What is the current ratio? Why?
5Temperature effects
- IC increase with the temperature
- if VBE remains constant,
- IC grows at about 9 / oC
- if IC remains constant,
- VBE falls at 2mV / oC
- a smaller VBE is enough to provide the same IC
because of the higher temperature
6Temperature effects
- The great thing about mirror is that IC2 is not
affected by temperature change ! - If we heat up Q1, will IC1 be changed?
- VBE can only vary from 0.5 to 0.8V
- IC remains more or less fixed at 1mA
7Negative feedback
- A deeper reason why IC1 is constant
- temperature gt IC1
-
- IC1 gt voltage drop across 15k resistor
- gt VBE1
- gt IC1
- This feedback mechanism keeps IC1 constant
8Temperature compensation for amplifier
- If VBE is fixed, what would happen if we heat up
the transistors in the following circuits?
9Auto-bias with feedback
- Left-hand circuit (without feedback resistor RE)
- extremely sensitive to temperature change
- IC at a rate of 9/oC, bias changed
- Right-hand circuit (with feedback resistor RE)
- temp , IC , VE , VBE
(as VB is fixed), - sqeezing VBE so that IC
- Another reason why IC is constant
-
10- The price to pay
- lower gain ( -RC/RE) in exchange for stability
- Can we have both stability (constant DC bias) and
high (ac) gain? - Yes !!
11Requirements of output driver
- High current output
- need large current to drive the mechanical voice
coil of the speakers driver - Voltage gain is not important
- because signal has already been amplified
- An emitter follower is perfect
- VIN VOUT
- small current in (IB), large current out (IE)
- Power out gt power in gt a power amplifier
12Output driver circuit
- A driver that produces large current over range
/-15V
13Output driver circuit
- Q1 and Q2 are both emitter followers
- What is the use of the current source in Q1?
- To replace RE
- Problem with RE
- If RE is too small gt small input impedance
- If RE is too large gt small IC gt small output
impedance - Active load (the current source)
- it has impedance gt large input impedance
- You choose large IC so that output impedance is
small
14Use a constant current source instead of resistor
- Q2 is a power transistor, needs large base
current - large resistor gives only small current
- ideal constant current source has infinitely
large impedance and can provide large current at
the same time - Clever !
- Current source is an example of active load
- Used extensively in integrated circuits because
it is easier to fabricate transistors than
resistors
15Output driver circuit
- DC offset voltage
- 0.6V (the VBE) for a simple follower
- Zero DC offset voltage in this circuit !
- Q1 (pnp) and Q2 (npn) are opposite type of
transistors - Note the 0.6V voltage drops of VBE in Q1 and Q2
cancels each other out - So that 0V DC input gives 0V DC output
16Current source and current sink
- Current source
- The ability of giving out current (flow out of
the circuit) - Current sink
- The ability of absorbing current (flow into the
circuit) - Consider Q2
- Range of VE /- 15V
- If VE gt 0, current flowing out to load (source)
- If VE lt 0, current flowing in from load (sink)
17Current source and current sink
- Max source current (when Q2 is fully on)
- If Q2 is fully ON (VCE0)
- max source current 15V/8W 2A
- Max sink current (when Q2 is fully off)
- IR ISINK IE
- ISINK is max when IE 0 (zero current from Q2)
- this happens when VE -15V
- Max ISINK 15 / 8 A
18Max output power of the amplifier
- Max output voltage swing
- /- 15V peak-to-peak
-
- Average power dissipated on a 8W load
19Root-mean-square (RMS) voltage
- Define RMS voltage as
- Advantage
- For ac voltage, the average power
- Hong Kongs power supply 240V
- It is a RMS voltage
20Quiescent (rest state, no signal) power
dissipation
- At rest state, 0V input gives 0V output
- power dissipated in resistor (0- -30)2/8
110W !! - power dissipated in transistor Q2
- quiescent collector current 30V/8 3.75A
- VCE 15V, IC 3.75A,
- power VCE IC 56W
- total quiescent power dissipation 110W56W
-
166W (HOT !!) - 14W amplifier gives out 166W heat (inefficient)
21Push-pull emitter follower (Class B amplifier)
- Problem with class A amplifier
- quiescent power dissipation is many times larger
than maximum output power - push-pull follower
22Class B amplifier
- Q1 (NPN) conducts on positive cycle
- Q2 (PNP) conducts on negative cycle
- 0V input, none of the transistor is conducting
- No current, no power dissipation at Q1 and Q2
- No resistor, no power dissipation
- Pros
- Efficient, no wasted power
- Cons
- Cross-over distortion
23Cross-over distortion
- input have to be greater than 0.6V or less than
-0.6V in order to make Q1 or Q2 conducting - Output 0V when input is between 0.6V and -0.6V
- Distortion
24Class AB amplifier
- Use diode to compensate for crossover distortion
- R can be replaced by current source
0.6V
25Amplifiers classification
- Class A transistors conduct all the time
(360o) - Class B transistors conduct only half of the
time (180o) - Class AB transistors conduct more than half of
the time, and less than full time - Class C transistors conduct less than half of
the time
26Differential Amplifier
- A differential amplifier has two input terminals
- only amplify the difference in signals
- common signal is NOT amplified
27Differential Amplifier
- The pair of input signals can be separated into a
common part and a differential part - e.g. (3V, 3.1V) inputs (0.05V, -0.05V)
(differential) - (3.05V, 3.05V) (common)
- Ideal differential amplifier
- Large differential gain
- Amplify the difference in signal
- Zero common-mode gain
- common signal produces zero output
28Long-tail pair
29DC Quiescent point
- 0V input (because base is tied to ground)
- VA -0.6V
- ITAIL
- ITAILis pretty constant because
- large negative voltage (-15V)
- variation of VA is small in practice
- A simple constant current source
- Can be improved by using a real current source
- By symmetry (same VBE), IC on both transistors is
about 1mA each
30Differential gain
- The key in understanding differential gain
- feed a DVIN/2 signal on one input, and a -DVIN/2
on the other, VA does not move
VA
DVIN/2
-DVIN/2
31Differential gain
- Because VA is fixed, only need to analyze the
part of the circuit above A - just a common emitter amplifier
32Differential gain
- Common emitter amplifier gain
- VOUT/VIN -RC/(REre)
- In out case, input -DVIN/2
- Differential gain
33Common mode gain
- Feed a common signal to both inputs, VA goes up
- model the 7.5k tail
- by two 15k in parallel
- I 0 because of symmetry
- split the circuit in two
- Common mode gain
- GCM -RC/ (reRE2(RTAIL))
Vin
34CMRR (Common-mode rejection ratio)
- CMRR differential gain / common mode gain
- Ideal differential amp has
- large differential gain
- small common mode gain
- Therefore CMRR should be as large as possible
- RTAIL should be as large as possible
35CMRR
- Replace RTAILby constant current source
- CMRR 100,0001 !!
- 100dB !
- dB 20log 105
- 205 100
36How to generate composite signal in the lab
- Composite signal common-mode differential
signal - use function generator 1 to generate the
common-mode signal (tied to true ground) - use function generator 2 to generate the
differential signal (floating ground, ground is
connected to the output of generator 1) - Make sure the ground of generator 2 is floating
(not connected to the main supply)
37Generator 2
Common signal
Generator 1
38Vdiff Vcommon
Floating ground !
Vcommon
Vcommon (from gen 1)