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brunokasimin
Joined: 13 Jun 2008
Posts: 54
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 02 Jul 2008 23:36 boost regulator |
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in my attachment is a circuit showing a general boost regulator.Ub is battery voltage, Lg is inductance, Uq is voltage source.
information:
Ub=200V, Uq=400V
With a duty cycle(a) of 50%, fluctuation current is ∆ig =10 A
Lg=1mH and the circuit is running in steady state.
the question are:
1.what does steady state mean?
2.how to calculate pulse frequency, fp of step-up chopper?
3.how to calculate the average power taken by the batterie?
thx in advanced
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rikie_rizza
Joined: 05 Aug 2006
Posts: 328
Helped: 20
Location: Bikini Bottom, between a rock and a pineapple
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| 03 Jul 2008 3:37 boost regulator |
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Question 1: Steady state means a system has reach its stable condition.
Question 2 and 3: Don't know.
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trapoe
Joined: 02 Jun 2008
Posts: 56
Helped: 10
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| 05 Jul 2008 18:13 Re: boost regulator |
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Lets start from point 3 that's the easiest.
If all the components (L, S and diode) are ideal they don't dissipate -> power from Ub goes only in the load -> Ub • Ig = Uq • Iq. (reversing all voltage arrows).
1) a rough meaning of "steady state" is that nothing change. This is really true if you have only DC signals.
If you're working with with repetitive signal "steady state" means that you have always the same cycle. Or that you can find exactly the same situation after one or more periods
2) If we are in steady state Ig is a triangle. When S is closed Ig increase, when S is open Ig decrease and go back to the initial value (steady state -> same value after a period).
For an inductor Vg = Lg • dIg/dt but Ig is a triangle so dIg/dt is constant then
Vg = Lg • ΔIg/Δt -> ΔIg = (Vg • Δt) / Lg -> Δt = (Lg • ΔIg) /Vg
where Δt = δ • T = δ / fp
When S is closed we have Vg = 200 V, Lg = 1 mH, ΔIg = 10 A, now we can calculate Δt. Remembering δ = 50 % we have fp.
Hope this will be clear.
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