Oh, That Interview Question— a reprise

RE: Oh, That Interview Question— a reprise

Bruce Trump — Mon, 19 Nov 2012 14:16:07 GMT

Isuru-- TI's blog site is not public and all its blogs are written by TI employees (or perhaps occasionally by invited guests). You might consider whether your comments are better suited to our E2E forums, directed to a specific product area.

RE: Oh, That Interview Question— a reprise

Harry — Sun, 18 Nov 2012 02:17:10 GMT

can someone tell me as to how I can start writing a blog on TI.com?

RE: Oh, That Interview Question— a reprise

Fred McMurry — Thu, 15 Nov 2012 20:38:07 GMT

My point there is _no_ phase angle to an rms solution. And most people would have to think 1 V amplitude from your academic question. If you are talking about lagging 45 deg you are looking at the sinusoid and not your rms meter. I would bet most would think V = 1 sin wt at that point.

The math will work the same either way... So it is irrelevant to your question.

Robert's point was the transient is different, which is true and worth noting.

I got some things muddled up in my initial post as well. I think the main point is old men shouldn't blog:) (To be clear, I'm including myself)

RE: Oh, That Interview Question— a reprise

Bruce Trump — Thu, 15 Nov 2012 20:21:06 GMT

Gene-- I believe you may still be thinking of a time-domain solution with a starting transient. Mine is a steady-state solution. The input signal is 1VAC = 1V RMS. The voltage on the capacitor is -3dB relative to the input signal which is 0.707VAC = 0.707 V RMS. The phase of the voltage on the capacitor 45 degrees lagging relative to the input signal. I'm not sure what you mean by "rms phase."

RE: Oh, That Interview Question— a reprise

Fred McMurry — Thu, 15 Nov 2012 20:01:55 GMT

Didn't you write: The response at this point is -3dB (0.707V) at 45° lagging. The phase is different, of course.

What is the rms phase of your solution?

RE: Oh, That Interview Question— a reprise

Bruce Trump — Thu, 15 Nov 2012 19:52:58 GMT

No. Sorry, Gene, but VAC should always be assumed to be RMS. For example, 120VAC is the nominal line voltage in USA. That's 120 volts RMS. AC voltmeters read in RMS. Even though they may respond to average-rectified, averaged-peak, or averaged-peak-to-peak, they read in RMS.

RE: Oh, That Interview Question— a reprise

Fred McMurry — Thu, 15 Nov 2012 19:29:37 GMT

What?!!!

Vdc = Vrms _they_ have the same meaning.

Vac = A sin wt

Vrms = A / sqrt(2)

If Vac = 1 then Vrms = 0.707.

And the steady state solution is 0.707 Vac, which is 0.5 Vrms.

Also let me define period:

period = 1 / frequency

I meant any n period (where n is an integer < 10).

RE: Oh, That Interview Question— a reprise

Bruce Trump — Wed, 14 Nov 2012 17:04:18 GMT

I’ll jump in here with some clarification. There was nothing in the original question indicating a transient condition. This is a steady-state AC situation. I fear that the suggestion of otherwise may shake the confidence of some who might read this whole thread. You are all welcome to continue with the discussion of a transient case (waveform starting at t=0) if you wish but it’s probably best to avoid terminology involving “RMS” during the transient. The RMS value is continuously changing during the transient (it’s not yet a sinusoid) and you must define the period over which you are averaging. Going back to the original question--the steady-state solution is 0.707VAC = 0.707 VRMS (they have the same meaning). With that clarification—have fun, all!

RE: Oh, That Interview Question— a reprise

Fred McMurry — Wed, 14 Nov 2012 16:26:34 GMT

Oops, got ahead of myself, that should have read:

However, with the source at 1V over the first cycles the rms voltage across the _source_ will be less than [sqrt(2)/2]Vrms, further the rms voltage across the capacitor will be more than 0.5 Vrms and resistor will be less than 0.5 Vrms.

Note: The rms voltage across the resistor will be slightly _less_, over any n periods, than the steady state until it reaches steady state.

The rms voltage across the capacitor will be slightly _more_ than the steady state until it reaches steady state.

RE: Oh, That Interview Question— a reprise

Fred McMurry — Wed, 14 Nov 2012 16:16:38 GMT

What Robert Scott explains is incorrect.

The waveform of the source is 1VAC not 1Vrms.

At the specified 1V magnitude for the source, the voltage across the capacitor (and resistor) is 0.707V or 0.5 Vrms (steady state).

At the RS specified 1Vrms, the voltage across the the source is 0.707, and the voltage across the capacitor and resistor is 0.5V or 0.354Vrms (steady state).

It is correct the the first ten cycles or so will not exactly line up with the steady state.

However, with the source at 1V over the first cycles the rms voltage across the _source_ will be less than [sqrt(2)/2]Vrms, further the rms voltage across the capacitor and resistor will be less than 0.5 Vrms.

Note: The rms voltage across the capacitor will be _less_, over any n periods, than the steady state until it reaches steady state.

Robert, I am certain that with a 1 Vrms source no voltage in this circuit could exceed the peak of 0.707 [sqrt(2)/2] V generated.

Would you hire me on the spot if I told you the above?

RE: Oh, That Interview Question— a reprise

Robert Scott — Wed, 17 Oct 2012 15:53:53 GMT

Sorry, my last comment should have said "1/sqrt(2)" for the RMS cap voltages. A senior moment.

RE: Oh, That Interview Question— a reprise

Robert Scott — Wed, 17 Oct 2012 15:49:17 GMT

My terrible memory was driving me crazy so I figured out the "general" solution for the series RC circuit IF the voltage is applied at t=0 (but not before). ASSUMING a 1 Vrms, 1 Hz sinusoid input voltage starting at t=0, and C=1/(2*pi), the full solution for t>0 is Vcap(t) = [exp(-2*pi*t)]/sqrt(2) + sin(2*pi*t - pi/4). For the 1st cycle of applied input the AC reactance calculation yields an RMS cap voltage of sqrt(2) whereas the exact solution gives an RMS value over this cycle of 0.735. By the 9th cycle they both yield the same value: sqrt(2). There's surprisingly small differences even by the 2nd cycle! But the differing calculated cap voltage waveforms do look pretty different during that 1st cycle (ONLY!). If an interviewee gave me this answer, I think I'd hire him on the spot with no more questions asked.

RE: Oh, That Interview Question— a reprise

Jimmie Ford — Sun, 14 Oct 2012 19:11:34 GMT

This concept may be a good way to create a very efficient led lighting system; however the power companies may not like it. JF

RE: Oh, That Interview Question— a reprise

Kumar Ankur — Sun, 14 Oct 2012 16:30:19 GMT

Freak but interesting.

RE: Oh, That Interview Question— a reprise

Robert Scott — Sun, 14 Oct 2012 13:35:25 GMT

Perhaps a better question to ask, rather than the frequency, is "How long has the AC source been applied to the circuit?". If it's been applied "forever", then your answer is correct. However, it was only raised from zero VAC to 1 VAC very recently, the steady state solution given by an AC reactance calculation is not valid. There is both a general and a particular solution to the differential equation for the capacitor voltage. I don't recall the "particular" solution for this simple system (it's been a LONG time since I've been in a classroom), but I suspect that the initial "AC voltage", defined perhaps as the Fourier component at the applied frequency, will actually be different from your answer. Of course, your answer will approach perfection as time increases far past the RC time constant value (but is it a nanosecond or a million years?)!