Isn't it assumed (by definition of capacitor) that only the ripple current flows int he cap and the steady state current is taken by the load? Why does this method produce a different result like in HW1, first question (Q2.2 I think, it a buck-boost), part iii.
Thanks.
2 comments:
Interesting questions. You're post got me thinking, so I figured I would venture my thoughts about this. Sorry if I'm wrong.
I'm not familiar with the charge in/charge out method you've mentioned, but I'm guessing it's using the conservation of charge relating the total current in a loop to charge: I = Q/t, along with the relationship C = ΔQ/ΔV.
I actually found that you can get the same solution for the cap ripple voltage using these relationships:
Take the 1st interval of our buck-boost circuit, 0 <= t <= D*T_s:
I = Q/t, where t (time) = D*T_s
Solve for charge --> Q = I*t = I*D*T_s
Substitute for I from the circuit --> Q = (-V/R)*D*T_s
Now, plug into C = ΔQ/ΔV, taking care to note that ΔV = -2ΔV. The minus sign coming from the polarity of the capacitor current (and therefore voltage) during the 1st interval. The 2ΔV term accounts for the capacitor ripple which is one ΔV above and one ΔV below the DC voltage across the cap giving a total change of 2ΔV.
C = ΔQ/ΔV becomes C = ΔQ/(-2ΔV) -->
ΔV = ΔQ/(-2*C) = (-V/R)*D*T_s/(-2*C) = V*D*T_s/(2*R*C)
Using the charge relationship to solve for cap ripple voltage is what is done in slides 15-17 of lecture 4 as well.
A couple of things to mention based on the 2nd part of your post:
While it's true that capacitors effectively block DC currents, this doesn't prevent DC currents from charging capacitors and as a result, changing their voltage. For example, taking a look at the relationship between cap voltage and its current: dv_C(t)/dt = i_C(t)/C. If i_C(t) is a DC current, then i_C(t) is a constant. C is also a constant. Now you have dv_C(t)/dt = a constant. If you were to plot a waveform, this is a slope.
Regarding the current ripple through the cap. This isn't the only current going into the capacitor. As mentioned above, there is the DC portion of the current too. Current ripple, in the case of this buck-boost circuit, is the periodic deviation in the inductor current from the DC current being supplied to the inductor. Remember that the small-ripple approximation is based on the notion that the inductor is ideal, with inductance large enough to eliminate this deviation, making the inductor current = DC supply current.
In this buck-boost topology, the DC current through the capacitor and the load are periodically changing because they are essentially being connected and disconnected from the inductor. You can see this visually in the way the circuit can be drawn during the 1st interval compared to the 2nd.
Hope that's helpful.
Hey, thanks for your post but that is not really what I wanted.
On slide 42 of lecture 4 we go into estimating ripple voltage in 2 pole (LC) systems. Using the similar method for part iii of the buck boost question on HW1 I end up with
dV= D*Vg/(16f^2*LC), where we solve for C knowing dv= 0.1V giving me 26uF, but the HW solution says 50uF or something (I forgot).
So my question is why can't we use this method while ALWAYS finding ripple voltage on capacitor.
I understand your argurment that DC currents are used to charge up the cap. When I meant assume that only ripple current flows through the cap and the load takes the DC portion, I meant in steady state only.
Anyway thanks for the long post.
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