5.3 symmetry

Post on 05-Apr-2018

226 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

TRANSCRIPT

8/2/2019 5.3 Symmetry

http://slidepdf.com/reader/full/53-symmetry 1/2

8/2/2019 5.3 Symmetry

http://slidepdf.com/reader/full/53-symmetry 2/2

Circuits Symmetry 2 M H Miller

Another well-known application of symmetry is the circuit configuration commonly called a ‘bridge’drawn below (ignore the dashed connection between nodes A and B for the moment). We suppose that

the two devices marked ‘1’ are identical, as are the two devices marked‘2’ (not necessarily identical to each other). Invoking ‘symmetry’ is a

sufficient basis to assert that the voltage difference from A to B is zero.After all would nature be so capricious as to behave differently along

identical paths? Moreover is not the current Ix = I/2; why would animpartial observer (as we expect Nature to be) choose to split current

between identical paths differently? Moreover if a connection (dottedline) were to be made between nodes A and B would not as much current

flow from A -> B as in the opposite direction. It would; the current in

such a connection must be zero. (The bridge is said to be ‘balanced’.)And of course is not the current in branch 2 (either side) Ix.

Symmetry finds wide application in integrated circuit technology. For a final illustration consider the

circuit drawn below. This is a simplified circuit equivalent for a ‘differential amplifier’, a ubiquitous

component in many integrated circuits. Two identical devices represented by the current-controlled

sources (actually transistors) produce a voltage difference

va – vb = ßR(ia – ib)in response to signal currents ia and ib. Each current can be written as a

sum of an even (‘common mode’) part and an odd (‘differential mode’part as follows:

ia = (ia + ib)/2 + (ia – ib)/2ib = (ia + ib)/2 - (ia – ib)/2

Note that these equations simply are identities.

Now apply Superposition, i.e., consider the circuit response separately tothe common mode input ia = ib = (ia + ib)/2, and the differential mode input ia = -ib = (ia – ib)/2. The

output voltage difference va – vb is zero; this is a conclusion which follows from the symmetry of the

circuit and the equality of the two input signals.

On the other hand the differential mode output is ßR (ia - ib), because of the anti-symmetry of the input

signals. It happens that spurious noise induced in the circuit, for example by electrical radiation, tends

to be the same for both inputs; the inputs ordinarily are very close together. This common mode signalthen is suppressed by the circuit symmetry. On the other hand by using differential mode for a desired

input signal, i.e., making ia = - ib, the desired input is amplified.

top related