active analogue circuits year 2 b. todd huffman. circuit theory reminders basics, kirchoff’s laws,...
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Active Analogue CircuitsYear 2
B. Todd Huffman
Circuit Theory Reminders
Basics, Kirchoff’s laws, Thevenin and Norton’s theorem, Capacitors, Inductors
AC theory, complex notation, LCR circuits• Passive Sign Convention
• What is “Passive Sign Convention”?
• Good Texts: • Electronics Course Manual for 2nd year lab.• “Art of Electronics” by Horowitz and Hill
October 2015 Todd Huffman
V0
I R1
R2
R3
-V0+IR1+IR2+IR3=0
0Vn+
+
+
+
–
–
–
–
–V0
+IR1
+IR2
+IR1
I1
I3
I2
I4
I1+I2–I3–I4=0
0In
Kirchoff’s laws
KCL
KVL
AC circuit theory
• Voltage represented by complex exponential
• Impedance relates current and voltage V=ZIin complex notation:
Resistance RInductance jLCapacitance 1/(jC)
and combinations thereof• Impedance has magnitude and phase
0V V cos t represented by real component of j t
0V V e
easily shown from dI
V Ldt
jZ Z e
Q=VC
• Current is given by
• So |Z| gives the ratio of magnitudes of V and I, and give the phase difference by which current lags voltage
• Notice that the time dependent part is a common factor– So ejwt can be removed and is “understood” to be present when
returning to the time domain.– WARNING!!! This is only true for circuits with Linear behaviour!
j tj t0 0
j
V V e VI e
Z Z e Z
Op-amps Gain is very large (A)
Inputs draw no current (ZIN=)
Feedback v+=v–
VOUT+
–
v+
v–
VIN
R1R2
Non-Inverting Amplifier Circuit
+VOUT
VIN
R1
Inverting Amplifier Circuit
R2
–v–
v+i
i
First Non-ideal model
+
-
A(w)dV
+
dV
Instead of infinite gain, the devicehas finite, and frequency dep. Gain.
V0-
+
A(w) behaves like an RC filter.
Magnitude |A(w)|
Phase fA(w)
With a gain factor of over a million; and a roll-off around w = 1 rad/s
A(w) ≈ 106/(1+jw)
Model of this non-ideal gain curve
• Vx = A0V1
• KCL• (V2 – Vx)/R + jwCV2 = 0
– Substitute expression for Vx above and some algebra
• V2(1 + jwCR) = A0V1
• V2/V1 = A0/(1 + jwCR) ≡ A(w)
A0
R
CV1
V2
Vx
Note:Also Draw filter on Blackboard
How does this effect our negative feedback circuits?
• KVL• VR1 – dV – Vin = 0
• VR2 + dV + V0 = 0
• KCL• VR1/R1 = VR2/R2
• And also the Gain relationship
• dVA(w) = V0
Solve on board
+VOUT
VIN
R1
Inverting Amplifier CircuitR2
–v–
v+i
i
𝑉 0
𝑉 𝑖𝑛
=−𝑅2
𝑅1+(𝑅1+𝑅2 ) (1+ 𝑗 𝜔 )
106
The Transistor!Go to diode part of lecture
Take a step back to demonstrate a key
technique – Small Signal response
Current flows easily!Almost no current flow
𝐼=𝐼 0(𝑒𝑞𝑉 𝐴𝐶
𝑘𝑇 −1)
Diode (& Transistor) is nonlinear
• Norton and Thevenin equivalents• Superposition• Ohm’s law does not exist for a non-linear
device.
• Kirchoff’s laws• Power = VxI (but not V2/I or I2R)
This means some tricks do not work!
Other principles are still OK.
Small signal analysis
Current flows easily!
R
++
VACV0
Simple Transistor Model
• It can be a “switch”– Flow is “on” one way– Flow is “off” the other way
• It can be an amplifier– The flow is proportional to
the amount you turn the valve.
– If you turn the valve fast enough you can communicate in Morse-Code-litres
Bipolar Junction Transistor curves
On black board!
BJT – How to approach this?!
• Technique just like for diode• Assume it is working as expected• Find an “operating point” using DC
parameters (check assumptions!)• Use some kind of “equivalent circuit”
which is linear• Solve linear circuit for “small signals”• Check consistency
vBE
Model works for npn and pnp(follow passive sign conv. on resistor)
CB
Cm
II
IkT
qg
+–
First Transistor Small Signal Model
gmvBEb/gm
base collector
emitter
Typical npn form shown
b is related to details of trans. Construction: = 100 b good start
How to use graphs?• Actually; start from Ebers-Moll equation
• A lot like Diode: if VBE ≈ 0.6 V or more IC starts to blow up. If IC changes by 10x, VBE still ~0.6 V
1. Assume VBE ≈ 0.6 V is true!
2. Assume VCE is ≥ 1 V (transistor is “active”)3. Assume b = 1004. Solve and rethink assumptions if inconsistency is
found.
10 e kT
qV
C
BE
II
Our First Transistor Circuit
VOUT
VIN
RB RC
+
+
How is it Biased?
What does it do?
The 741 op-amp’s actual diagram