I wanted to know, if an inductor is large, why would the current through it be constant? If an inductor is large, it means that it would be an open at moderate frequencies, and at DC the inductor is a short anyway, so how does any of that explain the constant current? Does it have something to do with the fact that the current cannot change instantaneously in an inductor? If that's the case, what does the size of the inductor have to do with that particular observation (current cannot change instantaneously)? I've attached an image of the problem. Thank you.

Razavi RF book problem

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    \$\begingroup\$ The value of L1 has to be large enough so that it can be assumed that the signal current is not flowing through it but flowing through the capacitor into RL. So the frequency of the signal and the values of the capacitor and RL determine how large the value of L1 should be. This is all mentioned in the question ! You concentrate on L1 having a large value, that is not the whole story. \$\endgroup\$ – Bimpelrekkie Mar 14 '17 at 10:37
  • \$\begingroup\$ Hey, thanks for your input, I understand the point about the signal current and the fact that L1 will be an open at those frequencies, but my question is about the DC current I_L1. You can see in the problem that I_L1 always flows through L1, whether M1 is on or off. What I don't understand is how is I_L1 constant due to the large L1. I tried doing the time domain analysis of the LCR circuit, but I still can't wrap my head around this concept! Maybe you're right, I'm not looking at it the right way. \$\endgroup\$ – eenn Mar 14 '17 at 11:42
  • \$\begingroup\$ I think I got it now, L1 is made so large that any current changes don't flow through it, but through RC, and the reason is because larger L's resist changes to current even more. I'm happy with that explanation! \$\endgroup\$ – eenn Mar 14 '17 at 12:16

The key is the instruction 'Assume L1 is large enough to act as an ac open circuit at the frequency of interest ...'

'The frequency of interest' means that for timescales related to changes in the signal voltage, L1 is so large that the current does not have time to change significantly. It will change, just not very much, not enough to change the bias conditions on the transistor.

When you are concerned with changes in current, the concept of 'DC' is not very useful, there are only lower and lower frequencies, slower and slower changes. Any voltage applied to an inductor will change the current by a significant amount given long enough.

When that circuit is switched on, the L1 current rises from zero in a time given by the applied voltage and its inductance. When M1 draws a varying current, if it does so fast enough, at frequencies of interest, then the L1 current changes so little during a cycle of modulation that essentially all of the current changes go through the load. If M1 draws a different current, and stays drawing that current, eventually the L1 current will change to match the new current. The larger L1 is, the longer this will take.

  • \$\begingroup\$ Thanks, though my question is about the DC current I_L1, as I mentioned in the above comment. \$\endgroup\$ – eenn Mar 14 '17 at 11:42
  • \$\begingroup\$ @eenn updated my asnwer a little. \$\endgroup\$ – Neil_UK Mar 14 '17 at 11:56
  • \$\begingroup\$ The interesting thing is that I_L1 could well be zero. The point is just that it's constant. (You must have some actual numbers somewhere else in the problem). You can consider it "DC" because you are looking for the peak current which, by definition, is happening instantaneously in time and therefore has no frequency component. \$\endgroup\$ – scld Mar 14 '17 at 11:58

When the current slows or stops, the magnetic field around the inductor will collapse and induce the current to continue to flow. An inductor opposes changes in current with the energy built up in its magnetic field.

A capacitor charges like a very fast charging battery. When the voltage increases the cap will absorb it as a charge. When the voltage drops the cap will supplement with the energy it has stored.

  • \$\begingroup\$ Thanks, which brings me to another point. Would the corollary be that a large capacitor would try to keep its voltage constant? I'm guessing that would be because the time constant would be too large for the initial voltage to change. \$\endgroup\$ – eenn Mar 14 '17 at 11:51
  • \$\begingroup\$ That is correct and also why a large inductor is an AC/high frequency open and a large capacitor is a DC/low frequency open \$\endgroup\$ – scld Mar 14 '17 at 11:57
  • \$\begingroup\$ @eenn see my update. \$\endgroup\$ – Misunderstood Mar 14 '17 at 12:02

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