1
Thévenin example
1.1
Thévenin voltage
To calculate the Thévenin equivalent voltage, look at the voltage across the terminals A and B.
This is a simple voltage divider, giving
Rp x
=
Rp x + Rp (1 − x)
Rp x
= Vs
= Vs x
Rp
ET h = V s
1.2
Thévenin resistance
Draw picture with NO voltage source!
This gives RT h as the parallel resitance, i.e.
1
1
1
=
+
RT h
Rp x Rp (1 − x)
Rp (1 − x) + Rp x
Rp
=
=
Rp x · Rp (1 − x)
Rp xRp (1 − x)
1
=
,
Rp x (1 − x)
which gives
RT h = Rp x (1 − x)
Now the voltage across the load can be calculated, using voltage division as
RL
RL
= Vs x
R T h + RL
Rp x (1 − x) + RL
1
= Vs x
x (1 − x) Rp /RL + 1
V L = ET h
This means the ouput is non-linear w.r.t. x, due to the loading. The non-linearity is given by
1
N (x) = ET h − VL = Vs x − Vs x
x (1 − x) Rp /RL + 1
1
= Vs x 1 −
=
x (1 − x) Rp /RL + 1
Rp /RL x (1 − x)
=
= Vs x
Rp /RL x (1 − x) + 1
Rp /RL x2 − x3
= Vs
=
Rp /RL x (1 − x) + 1
If the impedance of the load is much higher than than the ouptut impedance (which is usually
the case), then this can be simplified to
N (x) ≈ Rp /RL x2 − x3
1
2
Norton example (piezoelectric crystal)
The Laplace equivalent of the impedances are
RL
1
sCN
1
sCc
The total impedance of the circuit is thus
1
1
=
+ sCN + sCc ,
Z
RL
which gives
Z=
RL
sRL (CN + Cc ) + 1
This gives the voltage across the output of the recorder
V L (s) = i (s) · Z = iN (s) ·
RL
sRL (CN + Cc ) + 1
Looking at the transfer function G (s) of the entire system, we see that
G (s) =
VL
RL
=
,
sRL (CN + Cc ) + 1
iN (s)
which is a first order system.
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