Dang, Sorry I don't get this. I turn on one of the transistors and
nothing happens right away because of the inductor. But current
starts to flow with an L/R time constant... now what?
It might be 'easier' to see this way:
B
|
|
v
Vin Rsense L1 |\ |
(+) ---/\/\---UUUU-----+----| >|----+---> Vout
==== | |/ | |
^ | D1 |
| O |
| Q1 / ---
A / --- C1
O |
| |
| gnd
gnd
I've reversed D1 and the voltage polarity and replaced Q1
with a switch. But it is very basic switcher topology for
cases where Vout > Vin (if Vout < Vin by much, then D1 is
always active and the whole idea doesn't work "well.")
Q1 goes on. This places a near-constant voltage across L1.
Since V=L*dI/dt and you know V and L, you can see that dI/dt
is a constant value. In other words, 'I' ramps upward,
linearly, from some starting point often assumed to be zero.
At some later time, when 'I' reaches some desired value that
might be called Ipeak [which can be measured as a drop across
Rsense as (Rsense*Ipeak)] Q1 is turned off.
L1 cannot permit the sudden demise of 'Ipeak' (a change in I
requires, by definition, some time t to do that), so L1's
voltage polarity flips and sends the current via D1 to drop
it's energy onto C1. Joules is 1/2*L*I^2 (use Ipeak for I)
and on the cap is 1/2*C*V^2. So Ipeak energy is translated
to a change in V on C1, such that 1/2*C*(Vnew^2-Vold^2) is
equal to the energy on L1.
If you don't see why L1's polarity changes, it might be
easier to simply assign a direction to Ipeak, first. Imagine
that the direction arrow of Ipeak points from point A to
point B. Now, when Q1 opens and prevents further current
down through it to ground, L1 "needs" to maintain that arrow
direction somehow. Since point A is "nailed" to Vin, that
side remains at the same Vin value. However, point B is now
floating, in effect. L1 "wants" to maintain Ipeak and its
direction, until enough time t can take place to allow Ipeak
to return to zero. To do that, I must flow via D1. To do
that, the V at point B must rise up to some "more positive"
value such that D1 is forward biased and can permit this. If
C1 has some arbitrary Vc on it, then point B must be at least
one diode drop above that.
The math way to see it is, again, that V=L*dI/dt. Assume
dI/dt is positive when Q1 is closed at t=0, for convention's
sake, and that I=0 at t=0. Then Ipeak will be positive when
Q1 opens at t=t1. At that point, I will decline from Ipeak.
But this means that dI/dt is now negative, not positive.
Which means that the V across L1 must reverse to the opposite
of what it was beforehand. Before, the voltage moving from A
to B declined. So now the voltage moving from A to B must
increase. Which also confirms the point made, which is that
point B will rise to a value higher than Vin and sufficiently
higher than the voltage across C1 that D1 can forward conduct
and permit L1 to lose energy in such a way that C1 can accept
it (energy is neither created nor destroyed.)
It really doesn't matter where the other end of C1 is tied,
so long as it is a low impedance voltage rail. So I hope you
don't mind that I kept it at ground. I might as well have
placed it at Vin (which would mirror the circuit better, but
wouldn't change the time-domain considerations above.)
Jon
)
