Advanced Circuits Techniques. The University of Iowa. Spring 2015.
Two-Port Network Test
Question 1 The op-amp in the circuit below is non-ideal and has the following parameters: 𝑅𝑖 = 25K,
𝑅𝑜 = 1K, and the gain is 104 . Further, 𝑅𝐼 = 1K, 𝑅𝐿 = 2K, 𝑅2 = 91K, and 𝑅1 = 10K.
𝑣1 = ℎ11 𝑖1 + ℎ12 𝑣2
𝑖2 = ℎ21 𝑖1 + ℎ22 𝑣2
(a) Feedback Amplifier
(b) h-parameters
Using two-port theory, determine the gain 𝐴𝑓 = 𝑣𝑜 ⁄𝑣𝑖 , and the input- and output resistances 𝑅𝑖𝑓 and 𝑅𝑜𝑓
of the feedback amplifier. Determine the input- and output resistances of the amplifier proper, namely
𝑅𝑖𝑥 and 𝑅𝑜𝑥 . Use h-parameters to model the feedback network. It is important that you show your work.
For example, show how you calculate the h-parameters, redraw the circuits with the amplifier and the
feedback network replaced with two-ports, etc. Show that ℎ11 = 9.01K, ℎ22 = 1⁄101K and ℎ12 =
1⁄10.1. You may assume no feed forward for the feedback network. (24 points)
Solution The figure below summarizes the calculations for the h-parameters for the feedback network.
Since there is no feed forward, ℎ21 = 0.
ℎ11 =
𝑣1
= 𝑅1 ‖𝑅2 = 9.01K
𝑖1
ℎ22 =
𝑖2
1
1
=
=
𝑣2
𝑅1 + 𝑅2 101K
Shown is the two-port equivalent of the
amplifier. Since there is no feed forward,
ℎ21 = 0, and the controlled current source
associated with this parameter was removed
from the circuit.
Next, turn off the feedback by setting ℎ12 = 0,
which is the same as shorting the voltage source.
The gain of the resulting amplifier is
1
ℎ12 =
𝑣1
𝑅1
1
=
=
𝑣2
𝑅1 + 𝑅2 10.1
Advanced Circuits Techniques. The University of Iowa. Spring 2015.
𝐴=
25K
1.961K
(104 )
= 4,729
1K + 25K + 9.01K
1.961K + 1K
The improvement factor is (1 + 𝛽𝐴) = (1 + ℎ12 𝐴) = 469. The closed-loop gain is
𝐴𝑓 =
𝐴
4,729
=
= 10.1
1 + 𝛽𝐴
469
The closed-loop input resistance is 𝑅𝑖𝑓 = (1 + 𝛽𝐴)(1K + 25K + 9.01K) = 16.4M.
The closed-loop output resistance is
𝑅𝑜𝑓 =
2K‖101K‖1K
= 1.41 Ω
469
Finally 𝑅𝑖𝑥 = 16.4M − 1K ≈ 16.4M and
1
1
1
=
−
⇒ 𝑅𝑜𝑥 ≈ 1.41 Ω
𝑅𝑜𝑥 𝑅𝑜𝑓 2K
2
Advanced Circuits Techniques. The University of Iowa. Spring 2015.
Question 2 Determine the 𝑦-parameters in terms of 𝑠 for the linear network shown below. (16 points)
𝑦11
𝑖
[ 1 ] = [𝑦
𝑖2
21
𝑦12 𝑣1
𝑦22 ] [𝑣2 ]
(b) 𝑦-parameter definition
(a) Linear circuit
Solution
(i) 𝑣2 = 0
(ii) 𝑣1 = 0
Set 𝑣2 = 0 and drive with 𝑣1 (see (i) above). This shorts 𝑅 and 𝐿 and places 𝐶2 in parallel with 𝐶1 . Then
𝑖1 𝑣1 (𝑠𝐶1 + 𝑠𝐶2 )
=
= 𝑠(𝐶1 + 𝐶2 )
𝑣1
𝑣1
𝑖2
𝑣1 𝑠𝐶2
=
=−
= −𝑠𝐶2
𝑣1
𝑣1
𝑦11 =
𝑦21
Next set 𝑣1 = 0 and drive with 𝑣2 (see (ii) above). This shorts 𝐶1 and places 𝐶2 , 𝑅, and 𝐿 in parallel.
Then
𝑖2 𝑣2 (𝑠𝐶2 + 1⁄𝑅 + 1⁄𝑠𝐿 )
=
= (𝑠𝐶2 + 1⁄𝑅 + 1⁄𝑠𝐿 )
𝑣2
𝑣2
𝑖1
𝑣2 𝑠𝐶2
=
=−
= −𝑠𝐶2
𝑣2
𝑣2
𝑦22 =
𝑦12
𝑌=[
𝑠(𝐶1 + 𝐶2 )
−𝑠𝐶2
]
(𝑠𝐶2 + 1⁄𝑅 + 1⁄𝑠𝐿 )
−𝑠𝐶2
3