# E12 Assignment 07

Description:

Problems:
1. Practical Op Amp Differentiator

For the circuit shown (which uses an ideal op-amp)

a) Find the transfer function H(s)=Eout(s)/Ein(s) in terms of Rf, Ri, and C.

b) Find an expression for the unit step response in the time domain.

c) Sketch the step response (there are no component values, so you can only show the shape of the response).  Give expressions for the initial value (t=0+), final value (t=∞) and time constant in terms of circuit parameters.

Solution:

a)

b)

c)  Initial value=-Rf/Ri, final value=0, τ=CRi.

2. Op Amp with Positive Feedback

This problem explores how the circuit behaves if an op-amp has finite gain.  Therefore do not assume that the two input terminals of the op-amp are at the same voltage though you may still assume no current into the inputs.  The last part of the problem explore difficulties that ensue if the circuit does not have negative feedback.

A real op-amp not only has negative feedback, it is also not infinitely fast.  We represent this finite speed by giving it a transfer function of the form

where A0 is a large positive quantity (typically 105 or more) and τ0 is a positive number (typically on the order of 0.1 seconds).  The output is defined to be

where E+ and E- are the voltages at the inverting and non-inverting terminals.

a) For the circuit shown, find an expression for the transfer function H(s)=Eout(s)/Ein(s).  It should be expressed in terms of a ratio of polynomials in positive powers of s.

b) What is the time constant of the circuit (i.e., H(s)) if  Zi=Zf, A0=106 and τ0=0.1?  Remember, you can get the time constant of a system by looking at the denominator of the transfer function.  Note that, with feedback, the circuit is much faster (smaller time constant) than the op-amp alone.

c) Find H(s) if the circuit is built as shown below (with the op-amp terminals switched).  What fundamental problem does this create?  This is a circuit with positive feedback.   Hint: you can re-use your previous work if you just make A0 negative.

Solution:

a)

b) τ ≈ 2e-7  (i.e., 200 nSec - much faster than the op amp alone). The speed goes up dramatically, but the gain of the amplifer drops from 106 to approximately 2. When using negative feedback in this circuit we trade speed (the circuit with feedback is much faster) for gain (the circuit with feedback has much lower gain).

c) We can use the same equation as before but change the sign in front of A0.  The value of τ becomes negative, so the homogenous solution grows exponentially with time (this is a behavior that is not seen in circuits with only capacitors, inductors and resistors).  In this case we have positive feedback instead of negative feedback.

3. Impedance Translation with Transformer

In class (gears on a bicycle) we saw how the effective load changes when gears are being used.  By varying the turns ratio (N1/N2) we can match the apparent load in a circuit (R'load) to the source load (Rs) to maximize power transfer.   See image below - the circuit on the right is equivalent to the one on the left (i.e., currents and voltages are equal).  Show that

Recall that

where e1 and i1 are the voltage across and current through the left coils (and R'load), and likewise for e2 and i2 and the left coils (and Rload).  You can start by writing equations for Rload in terms of e2 and i2, and for R'load in terms of e1 and i1.

For the transformer (as for gears and levers) power in is equal to power out.  A lever trades force for velocity (the input can be low force with high velocity, with an output at higher force but lower velocity); a transformer trades current for voltage.

Solution:

4. Efficiency Improvement with Transformer

If the 8 ohm load is connected directly to the source (shown below) - you may assume Vsource is 1 volt.

1. Find the power dissipated in Rs (call this PRs),
3. Find tthe total power, Ptot, and
4. Find tthe efficiency, η, which is the power dissipated in the load divided by the total power.  This is a measure of how well power is transferred from the source to the load.
5. Determine the turns ratio (N1/N2) needed for a transformer to make the 8 Ohm load appear as 600 Ohms.
6. Repeat a-d with the 600 Ohm apparent load. Note that although total power supplied by the source decreases by half the efficiency increases enough so that the power dissipated in the load increases by a factor of about 20.
Solution:

5. E12 Dual/Analog 2a

Write equations for the currents summed at the nodes labeled 1 and 2.  Write the currents in terms of ia, e1 and e2.

Solution:

6. E12 Dual Analog 2b

a) For the circuit shown, draw over the mechanical system to find the force-current analog.

b) Draw free body diagrams and write equations of motion.

c) Rewrite the equations of motion substituting in the analogous force-current quantities.

d) Verify that the equations from part c are the same as the equations obtained by summing currents at the nodes labeled 1 and 2.

Solution:

a) Draw over electrical system (background), replacing mechanical elements with their force-current analogs.

Redraw.

b)

c)

d) They are the same.

7. E12 Dual/Analog 2c

a) Find the dual of the circuit shown by drawing over the circuit and substituting the appropriate dual elements.

b) Write the node equations for the original circuit (shown above)circuit and substitute dual quantities.

c) Verify that the equations from part b are appropriate loop equations for part a.

Solution:

a) Get the dual by drawing over the original system and substituting dual elements.

b)

c)