FREQUENTLY ASKED QUESTIONS
February 2, 2016
Content Questions
How do you define output and input impedance? Do they change
with how you look at the circuit?
Output impedance is the impedance “seen” by a load connected across the
output terminals of a circuit. It’s determined by Thevenin equivalent impedance
found in the usual way: open circuit voltage across the terminals, divided
by short circuit current. The input impedance is the impedance “seen” by
something (nominally a supply) connected across the input terminals. It’s
determined operationally by the voltage across the terminals divided by the
current drawn.
What do Za and Zc represent physically in the 4-terminal network?
These are the input and input impedances of the 4-terminal network (with
no load attached). They represent the way the circuit-in-the-box acts: Za is
the equivalent impedance seen at the input terminals. You would find this in
the usual Thevenin equivalent way, by shorting terminals cd and by finding
the impedance between a and b. Zc is the equivalent impedance seen at the
output terminals, and you would find it by shorting terminals ab and finding
the impedance between c and d.
Why does reversing the network change the transfer function?
If you “reverse the network”, i.e. exchange output for input, you are looking
at a different network at the (new) output. To see just one example, think of
the RLC circuit we covered last class. The transfer function was determined
from the voltage divider equation, where the output voltage is taken across
R
. Now the reversed
the resistor, so vcd = H(jω)vab , H(jω) = R+j(ωL−
1
)
ωC
network would mean that the signal generator voltage would be applied across
the resistor, and the output voltage is not related to the input voltage to by
the same relation (reverse transfer function B(jω) would be the reciprocal
in this case).
How do you determine B(jω) from H(jω), if possible?
You can’t determine it just from the form of H(jω), without making specific assumptions about the circuit. B(jω) needs to be evaluated for each
particular circuit.
Why does vab = vT h = 0 and why does vcd = vT h 6= 0?
This applies to Thevenin equivalents of a 4-terminal network when we are
looking for the equivalent circuit at the input and the output. To find the
input vT h , you do the usual procedure for finding Thevenin equivalents: open
the circuit between the a and b terminals, and ask, what is the open circuit
voltage? The answer to that is vT h = vab (open) = 0, since there’s no real
power supply inside– so there’s no voltage across the terminals (the power
supplies we drew inside the box are abstract ones representing the operation
of the circuit when there is an external power supply). The input impedance
is can then be found in the usual way (and in general will depend on the load
impedance as well).
Now if we look at the output, with signal generator attached on the input:
the circuit does have a (sinusoidally-varying) voltage at the output when you
open c and d– that’s the non-zero Thevenin-equivalent output voltage.
Why is having a unity gain buffer useful?
If you don’t have a buffer, you will always get a voltage drop over a load, because the circuit acts as a voltage divider when including the source impedance.
You might not want that if you want the output voltage to follow the input voltage. A unity-gain amplifier “protects” the load from the source
impedance. A perfect such buffer will have infinite input resistance and
draw no current from the source. In general buffers “isolate” input circuits
from output circuits, so they don’t affect each others’ impedances (and they
are very common in circuit design... we’ll see examples).
For a buffer that creates vL > vs , does H > 1?
Basically, yes— you can have an amplifier with a gain greater than 1 that
gives you H > 1.
How do amplifiers amplify?
Aha, we will see. They use tricky solid state devices called transistors.
Do all amplifiers inherently buffer?
You can devise amplifier circuits that don’t buffer (i.e., you get voltage amplification but output sees effect of previous circuit’s impedance). Many
amplifiers in use do buffer though.
If we have only reactive elements (i.e., none resistive) in our load,
where does the power delivered to it go if it’s not dissipated?
If you have a purely reactive load (purely imaginary impedance), the phase
shift φ between current and voltage will be ±π/2 and average power dissipated (proportional to cos φ) will be zero. This corresponds to a situation
where energy is sloshing around between capacitors and/or inductors and
never gets dissipated. (In practice, of course, nothing really has zero resistance, so energy will always be dissipated, even if slowly. In superconductors,
energy dissipation is really tiny though, and energy can be stored for a very
long time!)