Phase response of circuit

[The Jester]

How would you calculate the phase response for the following circuit?

[drachefly]

I'd use kirchoff's laws to get a differential equation, and solve it.

[The Jester]

I'd use kirchoff's laws to get a differential equation, and solve it.

Okay, not sure why you'd need a differential equation.

For the frequency response, I'm getting:

Af=R2jwL/(R1(R2+jwL)+R2jwL)

[Elessar]

Okay, not sure why you'd need a differential equation.

Because you're looking at a circuit with an inductor. Since you're working in frequency domain though, that's just mere nitpick.

For the frequency response, I'm getting:

Af=R2jwL/(R1(R2+jwL)+R2jwL)

If you have the frequency response equation, you should split it into imaginary and real components. From there, you can get the phase relation by arctan(b/a) (b being imaginary coefficients, and a being real).

I haven't checked your frequency response for correctness though.

[Braedley]

Your transfer function should come out to be

and acts as a really bad high pass filter. I'll see if I can find one of my old textbooks that has the phase shift equation in it.

[Braedley]

So instaed of trying to explain the math begind this that I have forgotten (and was told to rarely use anyways), I'll just tell you what the Bode plot of the phase shift frequency response should look like. If we set W=R1*R2/(R1*L1+R2*L2) (the corner frequency, measured in rads per sec), then at frequencies less than 0.1W, the phase shift has the straight line approx. of 90°. At frequencies greator than 10W, the approx. is 0°. Now connect those lines (from (0.1W,90°) to (10W,0°)) through (W,45°). This offers a very good approximation. To get a more accurate plot, use these fudge factors: @0.1W, -6°, @0.55W, +5°, @2W, -5°, and @10W,+6°

[Elessar]

So instaed of trying to explain the math begind this that I have forgotten (and was told to rarely use anyways), I'll just tell you what the Bode plot of the phase shift frequency response should look like. If we set W=R1*R2/(R1*L1+R2*L2) (the corner frequency, measured in rads per sec), then at frequencies less than 0.1W, the phase shift has the straight line approx. of 90°. At frequencies greator than 10W, the approx. is 0°. Now connect those lines (from (0.1W,90°) to (10W,0°)) through (W,45°). This offers a very good approximation. To get a more accurate plot, use these fudge factors: @0.1W, -6°, @0.55W, +5°, @2W, -5°, and @10W,+6°

I'm not sure if your algebra is correct in determining the coefficients of the corner frequency. However, the basic circuit model (along with your phase plot) is correct. The zero at the origin is what contributes the 90° phase shift for all frequencies, while the pole at some corner frequency W will contribute a negative 90° phase change (with the majority of the change at 0.1W and ending at 10W as you stated).

As for the transfer function. I do believe the initial nodal analysis (let Input will be A and Output will be B) is:
B = (B-A)/R1 + B/R2 + B/Ls

This reduces to H(s) = (1 + R1/R2 + R1/Ls - R1)^-1
I am too tired to verify the remainder of the algebra. My own chicken scratch suggets that the corner frequency should be (R1R2 + 1) / R1L(1 - R2)

Is this actually a high pass filter? The zero at the origin would suggest ever increasing magnitude response. However, my memory swears that this circuit is the basic inductive high pass filter.

[Elessar]

Ah answered my own question. Getting tired, I forgot about the pole reducing the effect of the zero at high frequencies. Depending on the location of the pole, this circuit could be either high pass or bandreject.

[Braedley]

All first ordered filters are either generic high pass or low pass. You can't even get into some of the slightly more complex filters like a low cut or a high cut, band pass or band notch, because they are all second order or greator.

And yes my algebra is correct for the corner frequency (unless I copied it wrong from the pic I posted). It has such an odd form because the inductor is in parallel with a resistor, which is one reason for this being such a crappy filter, because it also attenuates at frequency.

As for your math, you're equating voltages to currents. If we take A and B to be voltages, then you have a voltage on the left and a current on the right.

If we assume that the load resistance is very large (ie approaching infinity), which is a valid assumption, then we can easily use voltage devider rule (derived from KCL I belive, though it may be KVL) to find the relation between the input and output, as long as we're dealing with voltages (also usually reasonable to assume). As for my first assumption, as long as the load resistance or the input resistance to the next stage is much greator than R2//sL1 (ie RL>10(R2//sL1)) than the effect it has on the circuit is minimal. Otherwise we just throw the next resistance in parallel with R2 and L1.

[drachefly]

If we assume that the load resistance is very large (ie approaching infinity), which is a valid assumption

IF the load's input is buffered. If this is being used as a power supply, no.

Since it's not specified, we can't assume either one.

[drachefly]

BTW, solving differential equations is easy if you guess the form Ae^iwt

[Braedley]

It's a filter. There likely is another stage following it if it is in a power supply, because you don't want your output impedance to vary with frequency or load current (which this would). To add to that, all the circuit analysis is done assuming that there is not current draw on the output unless told otherwise. You wouldn't be able to solve the circuit if this weren't the case. Also, you may not totaly understand what very large is in this case. Large is considered to be about 10 times greator. Very large is about 100 times greator. The output resistance of this curcuit can be expected to be between a few ohms and a few 100 k ohms. The 100 k ohms is a little generous too. So 10 to 100 megaohms as a load resistor is not an unreasonable assumption. Put this through an op-amp and the problem's solved.