SRC of lecture 4, we said: v_r(t) ~ Vr1 * sin(ws*t - phi_r) and i_r(t) = Ir1 * sin(ws*t - phi_r).
Also, we said that i_s(t) = Is1 * sin(ws*t - phi_s).
Since the resonant circuit is in series, one would think that i_s(t) = i_r(t). Then, Is1 = Ir1 and phi_s = phi_r. Is that true?
Regards,
Nitish
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3 comments:
Hi Nitish,
You have a good question. i_s(t) and i_r(t) are dependent on different portions of the circuit. i_s(t) is in reference to the input and related to input current i_g. i_r(t) is in reference to the output and related to output current I. Since this is the case, their respective peak magnitudes (I_s1 and I_r1) are not equal.
evaluated at T_s = (2/pi)*I_s1*cos(phi_s) (equation 19.4).
I = (2/pi)*I_r1 (equation 19.8)
As for the phase shifts, I think they will be equal at resonance. It is given that phi_r is the phase shift of i_r(t) with respect to v_s(t). phi_s is the phase of Z_in(s). When f = f_0, Z_in(s) = R_e resulting in v_s1(t) = v_r1(t). Because i_r1(t) is in phase with v_r1(t) and v_r1(t) now = v_s1(t), there is no phase shift for i_r1(t) with respect to v_s1(t). The result is phi_s = phi_r = 0, and H(s) = v_s1(t)/v_r1(t) = 1 --> V = V_g.
Something is missing from previous comment:
The line, "evaluated at T_s = (2/pi)*I_s1*cos(phi_s) (equation 19.4)", should read:
average of i_g evaluated at T_s =...
Hi Keith,
Thanks for trying to respond to the question. It turns out that i_s and i_r are the same current in a series resonant circuit. I emailed Prof. Maksimovic and his response was:
"Yes, these currents are obviously the the same in the SRC. The derivation is done in a more general manner so that the end result applies to converters that employ other types of resonant networks"
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