J
Jon Kirwan
Guest
On Thu, 20 Jan 2011 16:53:33 -0800 (PST), Bill Bowden
<bperryb@bowdenshobbycircuits.info> wrote:
mind I have had zero training, too. So I love it when you
press on, as it challenges me and I need that. Thanks. That
said, here's my shot at answering your question.
Imagine the case of discontinuous mode (tentatively, let's
think about it as applying when the load is very light.)
There are three periods of time in the fixed-frequency case
and I'll make my own choice about their order:
(1) switch-ON, inductor magnetic field rising from 0 to X
(2) switch-OFF, magnetic field collapsing back to 0
(3) switch-OFF, dead time from 0 ampere-turns till next
cycle starts again.
I'll call these SON, SOFF1, and SOFF2, respectively.
As the load is increased, more current is drawn. During
SOFF1, there is current flowing via the battery and inductor
to charge the cap. But during SOFF2 and the following SON,
the storage cap supplies the current. Which means the
voltage droops during this time.
When the next SOFF1 period starts after that droop, the
voltage across the inductor that is required in order to
freewheel the diode into conduction is lower. This lower
voltage implies, from dt=L*dI/V, that dt lengthens. (dI is
taken as I_peak in discontinuous mode and is entirely
determined by the duration of the SON period, given a fixed
inductor and supply voltage.) This means that more time is
required for the inductor to relax. Which means SOFF1
lengthens in duration, while SOFF2 shortens.
Eventually, as the load continues to increase and the droop
also increases, SOFF2 gets zero time and SOFF1 occupies the
entire remaining period. This is the moment of switching
between discontinuous and continuous mode. The load
increases and at some point you move from one into the other
domain.
Okay. So let's assume we have a basic feel for discontinuous
mode and move on to continuous mode as an extension. What
happens as the load continues to grow? ... ??
As this point is crossed, the droop goes still higher and dt
grows beyond the time you've allowed. Remember that we are
talking about a fixed period of time and fixed duty cycle,
for now. So when there is no longer enough time for the
inductor to relax to zero ampere-turns, the current doesn't
reach zero (obviously) and the following SON period will
start with a non-zero inductor current. Since the SON period
is fixed (by definition in this example), the dI=V*dt/L will
still be the same but will "build up" upon the baseline
current left by the prior cycle. So I_peak will not be
entirely determined now by the SON period of time and I_peak
will be higher than used in computations for discontinuous
mode.
Another way of writing this: I_peak is entirely computable,
regardless of load, by knowing the inductor, source voltage,
and SON time period in discontinuous mode. You can figure
that out without knowing the load, a priori. What changes
with varying load, in discontinuous mode, is the relative
apportioning of time between SOFF1 and SOFF2. However, in
continuous mode SOFF2 doesn't exist and SOFF1 time is
entirely fixed by your fixed frequency and duty cycle. (Let's
just call it SOFF in continuous mode.) Since SOFF time don't
change on load variations, obviously, something else has to
give. What gives is I_peak. In continuous mode, I_peak
rises with rising load. The SOFF period simply sets how much
of I_peak is allowed to relax.
Since current is flowing all of the time in continuous mode
and since that current is also on average higher and always
flowing during SOFF onto the cap, there is more charge
transfer to the cap during SOFF and therefore a higher load
is supportable.
In all cases, discontinuous and continuous mode alike, the
V_out*I_out must equal V_in*I_in (ignoring inefficiency for
simplicity's sake.)
In discontinuous mode, I_peak is known (determined.) Current
linearly rises from zero to I_peak during SON, falls linearly
from I_peak to zero during SOFF1, and does nothing at all
during SOFF1.
In continuous mode, I_peak isn't known (it depends upon the
load.) Current rises from some non-zero floor value to
I_peak during SON, falls linearly from there back down to the
non-zero floor during SOFF, where the difference between
I_peak and the floor value is the same as I_peak would have
been computed for discontinuous mode.
That's the qualitative angle. I've stayed clear of writing
equations and expressions in this post and instead focused on
'understanding' things with broad brush strokes. (I did
equations before and it left questions.)
.....
So let's take the discontinuous case and develop some
quantitative descriptions from the general understanding for
a situation where the load resistor, input voltage, duty
cycle, frequency, and inductor values are known but where we
want to know the output voltage and current that results.
Let's use your last tentative situation which I said ran in
discontinuous mode:
V_in = 9 V
f = 8300 Hz
L = 12 mH
R = 8200 Ohms
t_son= 50us
Add to the above, the following guesses:
V_sw = 0.1 V (BJT Vce drop, when on)
V_di = 0.5 V (freewheeling diode drop)
R_L = 0.4 Ohms (inductor resistance)
So,
duty = 50us*8300 Hz = 41.5%
Since in discontinuous mode, the inductor starts out by
entering SON with the ampere-turns being zero. So during
SON, we can compute:
I_peak = (V_in-Vsw)*t_son/L = (9-.1)*50us/12mH = 37mA
We want to know average values for V_out and I_out and I_in.
First off, let's write out the power equation:
P_out = P_in - (P_di + P_sw + P_L) = P_in
This illustrates that there are some losses. P_di is the
freewheeling diode loss, P_sw is the BJT switch loss, and P_L
is the inductor loss. Before continuing, let's get a bead on
them. A maximum figure would bound the problem. Let's see
where that takes us.
P_di = (1/2) * V_di * I_peak * t_soff1 * f = 5.4mW
P_sw = (1/2) * V_sw * I_peak * t_son * f = 0.77mW
P_L = (1/3) * R_L * I_peak^2 * t_son * f +
(1/3) * R_L * I_peak^2 * t_soff1 * f = 1.83mW
I used the worst possible value for t_soff1, which is 120.5us
- 50us, or 70.5us. In no case can it take longer than that,
if we stay in discontinuous mode. So, roughly speaking, 8mW
is the worst case loss. And the diode appears to be the main
culprit. The diode drop also factors in computing t_soff1.
So we can't entirely ignore it there, either.
Anyway, now that we have a worst case idea about losses,
let's use the figure 5mW for computations and label it
P_loss:
P_loss = 5mW
P_out = P_in - 5mW
V_out^2 / R + 5mW = V_in * I_in
So what is I_in, on average?
I_in = (1/2) * I_peak * (t_son + t_soff1) * f
That accounts for the rise during SON and the fall during
SOFF1. (It is zero during SOFF2.)
The problem is in figuring out t_soff1. It depends upon the
output voltage. But without controlling the duty cycle, you
don't know what V_out is, as it will depend upon t_soff1.
You can get a bead on it, though. From the above, we can
combine and re-arrange:
V_out = SQRT(R*(V_in*I_peak*(t_son+t_soff1)*f-10mW)/2)
and include,
t_soff1 = I_peak * L / (V_out - V_di - V_in)
Iterate. Start out by taking t_soff1 = 1/f - t_son and
compute V_out. Then plug that V_out into the t_soff1
equation and compute it. Then use that value back into the
V_out equation. Etc. Until it gets close. I get:
V_out t_soff1
0. ... 70.5 us
1. 36.4 V 16.5 us
2. 26.7 V 25.8 us
3. 28.6 V 23.2 us
4. 28.1 V 23.9 us
5. 28.2 V 23.7 us
6. 28.2 V 23.8 us
7. 28.2 V 23.75 us ... stop
It's actually a cubic equation that can be solved. But the
iterative approach works just fine.
So we get V_out = 28.2V with t_soff1 = 23.75us. Now we can
compute the average I_in as:
I_in = (1/2) * 37mA * (50us + 23.75us) * 8300 = 11.3mA
So,
P_in = 9 V * 11.3 mA = 101.7 mW
Clearly, using V_out = 28.2V, we get:
P_out = 28.2^2/8200 = 97mW
Which leaves us a difference of 4.7mW, which is very close to
the 5mW we used going into this process.
To reach the 24V you experienced:
P_out = 24^2/8200 = 70.2mW
That means that there must have been losses amounting to
about (101.7mW-70.2mW) or 31.5mW. Maybe 70% efficiency?
.....
Now for your continuous case, which I have to say I don't
completely understand, either. You will quickly see why.
V_in = 9 V
f = 8300 Hz
L = 12 mH
R = 22 Ohms
t_son= 50us
The dI value will be the same, namely 37mA. But it will ride
on top of some I_base value. But!!! You said you were
getting a low of about 250mA and a peak of 5A!! What??? How
can this be??
This equation is pretty solid:
V/L = dI/dt
It won't give an inch. This means:
dI = V/L * dt
We know V is almost 9V. We know L=12mH (or think we do.) And
we know that dt=50us because you set things that way. This
means there is no escaping:
dI = (V_in-Vsw)*t_son/L = (9-.1)*50us/12mH = 37mA
It just sits there. But you were seeing (5A - 250mA) or
4.75A!! How??? The equation says you cannot get there from
here.
Well, the assumption that L is 12mH has to be wrong. I
believe your 50us. And I cannot see that your 9V supply
suddenly jumps up to huge numbers on its own. The only other
way you can get 5A is if L significantly drops in value.
And it can do that. Through saturation.
It's going to be hard for me to go any further because of
that fact.
Perhaps someone else knows. But that case makes no sense to
me, for now, except deciding that you are experiencing a
saturated core.
In the remaining case:
V_in = 9 V
f = 8300 Hz
L = 12 mH
R = 1000 Ohms
t_son = 50us
t_soff= 70.5us
In this case, you reported approximately I_peak=45mA and a
base line of 10mA. So a difference of 35mA.
Note this is VERY CLOSE to the 37mA I've been estimating. So
very likely there is no core saturation here.
I'm going to stop for now and let things soak in a bit. If
you buy all the rest I've said, I'll be happy to take on this
middle case you presented. But I think you will see that it
comes out sensibly, too.
Until then, let me know what you think about the rest.
Jon
<bperryb@bowdenshobbycircuits.info> wrote:
I can try to clarify. Of course, one must always keep in
mind I have had zero training, too. So I love it when you
press on, as it challenges me and I need that. Thanks. That
said, here's my shot at answering your question.
Imagine the case of discontinuous mode (tentatively, let's
think about it as applying when the load is very light.)
There are three periods of time in the fixed-frequency case
and I'll make my own choice about their order:
(1) switch-ON, inductor magnetic field rising from 0 to X
(2) switch-OFF, magnetic field collapsing back to 0
(3) switch-OFF, dead time from 0 ampere-turns till next
cycle starts again.
I'll call these SON, SOFF1, and SOFF2, respectively.
As the load is increased, more current is drawn. During
SOFF1, there is current flowing via the battery and inductor
to charge the cap. But during SOFF2 and the following SON,
the storage cap supplies the current. Which means the
voltage droops during this time.
When the next SOFF1 period starts after that droop, the
voltage across the inductor that is required in order to
freewheel the diode into conduction is lower. This lower
voltage implies, from dt=L*dI/V, that dt lengthens. (dI is
taken as I_peak in discontinuous mode and is entirely
determined by the duration of the SON period, given a fixed
inductor and supply voltage.) This means that more time is
required for the inductor to relax. Which means SOFF1
lengthens in duration, while SOFF2 shortens.
Eventually, as the load continues to increase and the droop
also increases, SOFF2 gets zero time and SOFF1 occupies the
entire remaining period. This is the moment of switching
between discontinuous and continuous mode. The load
increases and at some point you move from one into the other
domain.
Okay. So let's assume we have a basic feel for discontinuous
mode and move on to continuous mode as an extension. What
happens as the load continues to grow? ... ??
As this point is crossed, the droop goes still higher and dt
grows beyond the time you've allowed. Remember that we are
talking about a fixed period of time and fixed duty cycle,
for now. So when there is no longer enough time for the
inductor to relax to zero ampere-turns, the current doesn't
reach zero (obviously) and the following SON period will
start with a non-zero inductor current. Since the SON period
is fixed (by definition in this example), the dI=V*dt/L will
still be the same but will "build up" upon the baseline
current left by the prior cycle. So I_peak will not be
entirely determined now by the SON period of time and I_peak
will be higher than used in computations for discontinuous
mode.
Another way of writing this: I_peak is entirely computable,
regardless of load, by knowing the inductor, source voltage,
and SON time period in discontinuous mode. You can figure
that out without knowing the load, a priori. What changes
with varying load, in discontinuous mode, is the relative
apportioning of time between SOFF1 and SOFF2. However, in
continuous mode SOFF2 doesn't exist and SOFF1 time is
entirely fixed by your fixed frequency and duty cycle. (Let's
just call it SOFF in continuous mode.) Since SOFF time don't
change on load variations, obviously, something else has to
give. What gives is I_peak. In continuous mode, I_peak
rises with rising load. The SOFF period simply sets how much
of I_peak is allowed to relax.
Since current is flowing all of the time in continuous mode
and since that current is also on average higher and always
flowing during SOFF onto the cap, there is more charge
transfer to the cap during SOFF and therefore a higher load
is supportable.
In all cases, discontinuous and continuous mode alike, the
V_out*I_out must equal V_in*I_in (ignoring inefficiency for
simplicity's sake.)
In discontinuous mode, I_peak is known (determined.) Current
linearly rises from zero to I_peak during SON, falls linearly
from I_peak to zero during SOFF1, and does nothing at all
during SOFF1.
In continuous mode, I_peak isn't known (it depends upon the
load.) Current rises from some non-zero floor value to
I_peak during SON, falls linearly from there back down to the
non-zero floor during SOFF, where the difference between
I_peak and the floor value is the same as I_peak would have
been computed for discontinuous mode.
That's the qualitative angle. I've stayed clear of writing
equations and expressions in this post and instead focused on
'understanding' things with broad brush strokes. (I did
equations before and it left questions.)
.....
So let's take the discontinuous case and develop some
quantitative descriptions from the general understanding for
a situation where the load resistor, input voltage, duty
cycle, frequency, and inductor values are known but where we
want to know the output voltage and current that results.
Let's use your last tentative situation which I said ran in
discontinuous mode:
V_in = 9 V
f = 8300 Hz
L = 12 mH
R = 8200 Ohms
t_son= 50us
Add to the above, the following guesses:
V_sw = 0.1 V (BJT Vce drop, when on)
V_di = 0.5 V (freewheeling diode drop)
R_L = 0.4 Ohms (inductor resistance)
So,
duty = 50us*8300 Hz = 41.5%
Since in discontinuous mode, the inductor starts out by
entering SON with the ampere-turns being zero. So during
SON, we can compute:
I_peak = (V_in-Vsw)*t_son/L = (9-.1)*50us/12mH = 37mA
We want to know average values for V_out and I_out and I_in.
First off, let's write out the power equation:
P_out = P_in - (P_di + P_sw + P_L) = P_in
This illustrates that there are some losses. P_di is the
freewheeling diode loss, P_sw is the BJT switch loss, and P_L
is the inductor loss. Before continuing, let's get a bead on
them. A maximum figure would bound the problem. Let's see
where that takes us.
P_di = (1/2) * V_di * I_peak * t_soff1 * f = 5.4mW
P_sw = (1/2) * V_sw * I_peak * t_son * f = 0.77mW
P_L = (1/3) * R_L * I_peak^2 * t_son * f +
(1/3) * R_L * I_peak^2 * t_soff1 * f = 1.83mW
I used the worst possible value for t_soff1, which is 120.5us
- 50us, or 70.5us. In no case can it take longer than that,
if we stay in discontinuous mode. So, roughly speaking, 8mW
is the worst case loss. And the diode appears to be the main
culprit. The diode drop also factors in computing t_soff1.
So we can't entirely ignore it there, either.
Anyway, now that we have a worst case idea about losses,
let's use the figure 5mW for computations and label it
P_loss:
P_loss = 5mW
P_out = P_in - 5mW
V_out^2 / R + 5mW = V_in * I_in
So what is I_in, on average?
I_in = (1/2) * I_peak * (t_son + t_soff1) * f
That accounts for the rise during SON and the fall during
SOFF1. (It is zero during SOFF2.)
The problem is in figuring out t_soff1. It depends upon the
output voltage. But without controlling the duty cycle, you
don't know what V_out is, as it will depend upon t_soff1.
You can get a bead on it, though. From the above, we can
combine and re-arrange:
V_out = SQRT(R*(V_in*I_peak*(t_son+t_soff1)*f-10mW)/2)
and include,
t_soff1 = I_peak * L / (V_out - V_di - V_in)
Iterate. Start out by taking t_soff1 = 1/f - t_son and
compute V_out. Then plug that V_out into the t_soff1
equation and compute it. Then use that value back into the
V_out equation. Etc. Until it gets close. I get:
V_out t_soff1
0. ... 70.5 us
1. 36.4 V 16.5 us
2. 26.7 V 25.8 us
3. 28.6 V 23.2 us
4. 28.1 V 23.9 us
5. 28.2 V 23.7 us
6. 28.2 V 23.8 us
7. 28.2 V 23.75 us ... stop
It's actually a cubic equation that can be solved. But the
iterative approach works just fine.
So we get V_out = 28.2V with t_soff1 = 23.75us. Now we can
compute the average I_in as:
I_in = (1/2) * 37mA * (50us + 23.75us) * 8300 = 11.3mA
So,
P_in = 9 V * 11.3 mA = 101.7 mW
Clearly, using V_out = 28.2V, we get:
P_out = 28.2^2/8200 = 97mW
Which leaves us a difference of 4.7mW, which is very close to
the 5mW we used going into this process.
To reach the 24V you experienced:
P_out = 24^2/8200 = 70.2mW
That means that there must have been losses amounting to
about (101.7mW-70.2mW) or 31.5mW. Maybe 70% efficiency?
.....
Now for your continuous case, which I have to say I don't
completely understand, either. You will quickly see why.
V_in = 9 V
f = 8300 Hz
L = 12 mH
R = 22 Ohms
t_son= 50us
The dI value will be the same, namely 37mA. But it will ride
on top of some I_base value. But!!! You said you were
getting a low of about 250mA and a peak of 5A!! What??? How
can this be??
This equation is pretty solid:
V/L = dI/dt
It won't give an inch. This means:
dI = V/L * dt
We know V is almost 9V. We know L=12mH (or think we do.) And
we know that dt=50us because you set things that way. This
means there is no escaping:
dI = (V_in-Vsw)*t_son/L = (9-.1)*50us/12mH = 37mA
It just sits there. But you were seeing (5A - 250mA) or
4.75A!! How??? The equation says you cannot get there from
here.
Well, the assumption that L is 12mH has to be wrong. I
believe your 50us. And I cannot see that your 9V supply
suddenly jumps up to huge numbers on its own. The only other
way you can get 5A is if L significantly drops in value.
And it can do that. Through saturation.
It's going to be hard for me to go any further because of
that fact.
Perhaps someone else knows. But that case makes no sense to
me, for now, except deciding that you are experiencing a
saturated core.
In the remaining case:
V_in = 9 V
f = 8300 Hz
L = 12 mH
R = 1000 Ohms
t_son = 50us
t_soff= 70.5us
In this case, you reported approximately I_peak=45mA and a
base line of 10mA. So a difference of 35mA.
Note this is VERY CLOSE to the 37mA I've been estimating. So
very likely there is no core saturation here.
I'm going to stop for now and let things soak in a bit. If
you buy all the rest I've said, I'll be happy to take on this
middle case you presented. But I think you will see that it
comes out sensibly, too.
Until then, let me know what you think about the rest.
Jon