On Fri, 16 Jul 2010 13:51:39 -0700, Tim Wescott
tim_at_seemywebsite.com> wrote:
On 07/16/2010 11:40 AM, Jon Kirwan wrote:
On Fri, 16 Jul 2010 08:23:19 -0700, Tim Wescott
tim_at_seemywebsite.com> wrote:
On 07/16/2010 04:53 AM, Jon Kirwan wrote:
On Thu, 15 Jul 2010 20:09:35 -0700, Tim Wescott
tim_at_seemywebsite.com> wrote:
snip
If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak.
snip
Hi, Tim. That would ONLY seem correct to me in the extreme
case where f=t_on. (An impossible case.) Not in the case
where, let's say, t_on equals t_off so that D=50%, for
example, where it would seem to be pushed to 4X, not 2X.
In any case, very fundamental considerations would suggest
(and these completely ignore some other relationships that
are necessary, such as nearly fixed limitations on t_off due
to the allowable voltage across the L when transferring
energy to the cap):
I_peak = SQRT[ (2*P) / (f*L) ]
This is simple to observe, since it is nothing more than
figuring out the (1/2)*I^2*L energy at I_peak, times the
number of such pulses allowed in a second, which must match
the input power available, I'd think. (Power is energy per
unit time, after all.)
Well...
Note my constraint: if things are arranged (i.e. if frequency and
inductance is selected such that the inductor current goes to zero just
as the transistor switches on).
I made two points before your comment here, so I'm not sure
which of the two you are addressing.
The first was about 2X vs 4X and evolves from V = L*dI/dt and
depends exactly the point you make where the inductor current
goes to zero. Without intuition, here's the algebra:
V = L*dI/dt
Assuming inductor current goes to zero and dt=t_on:
V_in = L * I_peak / t_on
Now assuming that f=1/t_on (which we know isn't true):
V_in = L * I_peak * f
And therefore,
f = V_in/(L*I_peak)
But, with P=V_in*I_in (power):
I_peak = SQRT[ (2*P) / (f*L) ]
So, substituting for f from above:
I_peak = SQRT[ (2*P) / ([V_in/(L*I_peak)]*L) ]
I_peak = SQRT[ (2*P) / (V_in/I_peak) ]
I_peak^2 = (2*P) / (V_in/I_peak)
I_peak = (2*P) / V_in
But P=V_in*I_in, with 100% efficiency, so:
I_peak = (2*V_in*I_in) / V_in
Which works out conveniently to:
I_peak = 2*I_in
Of course, the problem in the above flow is that we assumed
f=1/t_on. Clearly, it isn't, as t_off is non-zero. If you
use the above and use f=1(2*t_on), assuming D=50%, then you
will get:
I_peak = 4*I_in
The second point I'd made above was about the basic idea of
multiplying power by time to get energy and then observing
what that _must_ mean regarding inductor current, once again
also assuming that it goes to zero each cycle. So this also
does not violate the assumption you called out, again. In
fact, I took it as a given.
That's all well and good, but it simply cannot be right. If the diode
conducts for the entire time that the transistor is off, then the output
side of the inductor is always either connected to the load (13.8V) or
ground (0V). The input side is constant, so the inductor current must
be a sawtooth. The average current is going to be the mean of the peak
and trough of this sawtooth. So if the average is 1A and the trough is
0A, then there's absolutely nothing left for the peak to be but 2A.
If your fancy math comes up with an answer that doesn't match, it means
that either your math or your assumptions are incorrect.
Perhaps you should simulate this on LTSpice and see what it says...
I need to do that, thanks.
When an inductor has a constant voltage across it, the current ramps at
a constant rate (assuming constant inductance with current, which isn't
always valid in power electronics -- close your eyes to that).
Agreed.
If the
current ramps up from zero to 2A, then 2A to zero, then _immediately_
goes from zero to 2A again, the average current is 1A.
I'm not sure I read this correctly. There is the highlighted
word _immediately_ that suggests 0-2A, but I know that _time_
is required to do that. So _immediately_ cannot mean zero
time. Which leaves me confused.
In any case, what I think needs to be focused upon is energy
and time, not average current.
However, without that intuition, the algebra above arrives at
9.65W (the 9.65V time 1A) available on input (and all this
assumes such a switcher even makes sense at all for a solar
panel, which I'm not convinced of because it wants to supply
a constant current and not a wildly swinging one)
That's what input caps are for -- to smooth out the current seen by the
panel, by letting the capacitor supply the AC portion.
I guess I failed to note the OP writing that part. I thought
about it, though. Just didn't see it. In any case, I wonder
about __proper__ design for solar panels here. Not the
simple stuff used in "photovore" robots, for example. But
real, meaningful design with a solar panel and something more
on the order of 10 watts and more varying downwards as the
sun moves.
divided by
the OP's 25kHz yields 386uJ as the energy storage required in
the inductor. From that, I get 3.93A for 50uH and 3.78A for
54uH. Neither of which are 2A and both of which are much
closer to the 4X factor, using D=50%.
Even then, it's probably higher still because there are other
considerations I can imagine. For one example, D will be
more like 30%, as the OP mentioned, and the t_off time will
likely be longer than the t_on, at steady state when the V
across the inductor during t_off will be 4V or so as opposed
to the t_on V of about 9.5V.
But very basic energy-in/energy-out considerations without
any duty cycle or other limitation considerations will
suggest that at 25kHz and 50uH (assuming those are even
tenable with each other, which is yet another question not
answered) the current needs to be higher than 2A here just to
meet the energy transfer per unit time requirement.
Or maybe I'm missing something. I am a modest hobbyist and
have had zero electronics training, after all.
Not so -- the inductor is not storing all of the energy that gets
transferred. Think about it -- for the 28us that the forward diode is
conducting you are still getting current flow from the input side -- the
inductor is only making up the difference between the 9.65V and the 13.8
(or whatever). Thus, you are overestimating the amount of energy that
must be stored in the inductor.
I'm thinking exclusively about the steady state case --
_after_ the output capacitor has reached it's design voltage.
There is allowable droop. But not in my wildest imagination
did I guess that the diode would be forward conducting every
cycle!! Only in the early startup time. Which I set aside
for thinking purposes.
L D
___
.-------UUU----o---->|-------.
| | |
| | |
Vin /+\ ||-+ /+\ Vout
( ) ||<- ( )
\-/ -||-+ \-/
| | |
| | |
=== === ===
GND GND GND
(created by AACircuit v1.28.6 beta 04/19/05
www.tech-chat.de)
Please explain how it is possible for the diode to _not_ conduct when
there is forward current in the coil and the transistor is off, as will
happen every single cycle at the moment that the transistor turns off.
It will forward conduct when the transistor is off, Tim.
Because of the inductor. That's never been at issue.
The point here is that the forward conduction is DUE to
energy stored in a magnetic field in the inductor. And
_that_ is the fact at issue in our discussion. This energy
_must_ be factored into any discussion.
Since the input side is at a lower voltage than the output,
at steady state, there is no other method by with power is
transferred, except by way of pulsed magnetic field energy
storage. As such, it's a very simple calculation to know how
much energy must be present in each pulse if there are
exactly 25000 of them per second. (25kHz.)
I'm not sure what is wrong about our dialog, yet. There is
something "cross purposes" about it. And I have to take the
blame here. You are the trained expert, not me.
Once the output, within ripple considerations, is
established, the ONLY way energy gets transferred is when
that diode is forward conducting. And if the output voltage
doesn't droop back down to 9V (which I can't accept it does
do), then this only happens when the inductor dumps. And, I
have to believe, the OP intends this to be just about all 10
watts at this steady state.
And if you're at all smart, you size the inductor so that it takes the
entire off-time of the transistor to "dump". Not just during startup --
all the time.
I'm not all that smart, Tim. I'm just a bystander looking
in.
Yes, I would be over-estimating in the case of "startup."
Obviously. But once the output voltage is established and if
the entire panel output of nearly 10W is supposed to continue
to supply energy to the output side at 100% efficiency (I
think we both take this as a given, for now) then the
inductor is fully involved in achieving that, I think.
Or?
I missed the OP's frequency spec. You know that if conduction is
continuous then the duty cycle has to be 30% just from the voltage
ratio.
Yes.
So you know that the FET is on for 12us, and that it reaches 2A
peak. To do this with that voltage in,
L = (9.65)(12us)/(2A) = 58uH.
If you want to do the energy balance equations go ahead, but take into
account that during forward conduction of the diode some of the energy
is going directly into the output. In fact, it'll be 70% of the input
power.
10W power-in needs to be transferred to 10W power-out, at
100% efficiency. Assuming that the output capacitor is at or
near it's rated output voltage of 12-14V, this seems to
suggest that the 10W must be divided out into (1/2)*L*I^2
pulses, since it cannot be directly flowing via a
reverse-biased freewheeling diode as the V_in is less than
10V and the V_out is decidedly higher.
As I had pointed out, this means almost 400uJ at 25kHz. Could
you help me by a more thorough description explaining by what
mechanism power might be transferred -- once steady state is
achieved? I'm not seeing it and that must be my fault.
So, you've got 9.65V on the input side, 4.15V across the inductor, and
you flow a charge of (1A)(28us) = 28uC, for a total energy of
(13.8V)(28uC) = 386uJ.
Yes, per pulse at 25kHz.
So how much of the energy is delivered by the
inductor, and how much is delivered by the input voltage source?
Okay. I think I'm following, a little.
Keep in mind that the only connection to the input source is the coil,
Okay, yes.
so any
current into the coil comes from the input source, which constrains the
input source current to that of the coil.
Okay. I think I follow the argument you are making here.
In my own words, what you are saying here is that the
inductor provides only "part" of the net voltage and that
both the inductor and the panel together provides the current
so that the effect is a "sharing" of net power. The result
of that is that only a portion of the total transfer of
energy per unit time needs to be present, stored temporarily
as energy in the magnetic field each pulse.
So I'm mistaken to imagine that approximately 10J of energy
needs to divided down by 25000 to get each pulse's energy
because some of the energy arrives directly by way of the
source voltage and current.
