L
Larry Brasfield
Guest
"lemonjuice" <exskimos@anonymous.to> wrote in message
news:1111498469.992802.231730@o13g2000cwo.googlegroups.com...
multiplied by its frequency like that. It may be simple
to you, but your formula appears to be taken from a
private language of your own. It mystifies me.
Perhaps we can simplify this by showing that Idc HAS TO
NOT change when the load current stays constant. I will
start with assumed conditions and points of agreement.
The average value of current taken out of the cap is
constant, is called Iload, and line frequency is constant.
The average value of current into the cap is called 'Idc'.
We are assuming an equilibrium where the peak and
valley voltages on the cap do not change from cycle
to cycle. In other words, repetitive waveforms.
Then we can say that, on a per cycle basis, the same
amount of charge is removed from the cap (going to
the regulator/load and now named 'Qcapout') as is
added to the cap (and now named 'Qcapin'.)
If we call the time separating charging cycles 'Tcy', then
Qcapin / Tcy == Qcapout / Tcy
Qcapin / Tcy is the same value as Idc.
Qcapout / Tcy is the same value as Iload.
Idc == Iload
Since Iload is constant, so is Idc.
Notice that you can change the C value all you like
without changing a bit of the above analysis. Note
also that it pertains whether the charging current is
a sinusoid fragment or not.
analysis I have done are what led me to differ with you.
If believing my own carefully done analysis until it is
shown to be wrong is 'stubborn', then count me in.
here is that the limits of integration that you need to
use to get a number out of it are changing under the
stated assumption that C is changing. I do not see
that incorporated into your result or thinking, yet.
When you do that, you will find that Idc and Irms
are no longer proportional. Another issue is that
under the assumption that capacitance is changing
(and that it matters), the currents in question are
not sinusoid fragments as your expressions claim.
Given these conditions, and I quote:
If you hold everything else constant, (including the
DC current taken from the cap by the regulator/load),
then you claim the following:
I'm changing the Capacitance which is not going to hold
everything constant.
Idc *is* constant as established above.
If we take as true your assertion (and I quote):
Idc is directly proportional to Irms
where Idc refers to that constant charge per cycle, then
Irms cannot change without violating your assertion of
proportionality.
I've posted a link in this thread to an analysis I've done.
(under subject "Re: 9V 1A schematic needed [link]")
Below is an extract from it that may get this cleared up.
You can lift and use the definite integral expressions
by plugging in your equivalent variable names.
<begin extract>
The model behind this analysis employs an ideal
transformer having an output resistance Ro and
open circuit peak output voltage Vp, rectifier
diodes that turn on at a fixed forward bias and
have no resistance [1], and an infinite output
filter capacitance [2]. An independent variable,
Vd, represents the difference between the filter
capacitor voltage limited by R and what it would
be if R was 0 Ohms. The transformer open circuit
output is assumed to be sinusoidal. The analysis
is for full wave rectification, but can be easily
adjusted for half wave.
[1. Diode resistance can be folded into the
transformer output resistance. The actual
P/N junction current/voltage characteristic
contributes negligible error because it only
affects the current flow when that current
is relatively close to zero anyway.]
[2. The infinite filter capacitor represents
the limiting (and worst) case for RMS/average
current ratio. The situation can improve for
lesser capacitors as the ripple becomes an
appreciable fraction of Vp.]
Get the rectifier turn-on level:
Vx := Vp - Vd
Substitute an angle, a, for the phase velocity:
a := 2 pi Fp t
(where Fp is line frequency and t is time)
The voltage charging the cap thru Ro:
v(a) = max(0, Vp cos(a) - Vx)
Derive (half of) the conduction angle
g = acos(Vx/Vp)
The squared charging voltage during conduction:
v(a)^2 = Vp^2 cos(a)^2 - 2 Vp Vx cos(a) + Vx^2
Mean charging voltage over a half cycle:
v_ = 1/pi Integral[from -g to +g] v(a) d(a)
= 1/pi (2 Vp sqrt(1 - (Vx/Vp)^2) - 2 Vx g)
Mean squared charging voltage over a half cycle:
vsq_ = 1/pi Integral[from -g to +g]
(Vp^2 cos(a)^2 - 2 Vp Vx cos(a) + Vx^2) d(a)
= 1/pi (Vp^2 g + Vp^2 sin(2 g) / 2
- 4 Vp Vx sqrt(1 - (Vx/Vp)^2) + 2 Vx^2 g)
Note that the integrals for finding means
are evaluated only within the conduction
angle and the interval over which the mean
is computed is a half cycle (pi radians).
These identities are were substituted:
sin(+acos(x)) = +sqrt(1-x^2)
cos(x)^2 = 1/2 + cos(2 x)/2
The mean and RMS output current will be:
I_ = v_ / Ro
irms = sqrt(vsq_) / Ro
<end extract>
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
news:1111498469.992802.231730@o13g2000cwo.googlegroups.com...
I have never seen the instantaneous value of a signal
multiplied by its frequency like that. It may be simple
to you, but your formula appears to be taken from a
private language of your own. It mystifies me.
Perhaps we can simplify this by showing that Idc HAS TO
NOT change when the load current stays constant. I will
start with assumed conditions and points of agreement.
The average value of current taken out of the cap is
constant, is called Iload, and line frequency is constant.
The average value of current into the cap is called 'Idc'.
We are assuming an equilibrium where the peak and
valley voltages on the cap do not change from cycle
to cycle. In other words, repetitive waveforms.
Then we can say that, on a per cycle basis, the same
amount of charge is removed from the cap (going to
the regulator/load and now named 'Qcapout') as is
added to the cap (and now named 'Qcapin'.)
If we call the time separating charging cycles 'Tcy', then
Qcapin / Tcy == Qcapout / Tcy
Qcapin / Tcy is the same value as Idc.
Qcapout / Tcy is the same value as Iload.
Idc == Iload
Since Iload is constant, so is Idc.
Notice that you can change the C value all you like
without changing a bit of the above analysis. Note
also that it pertains whether the charging current is
a sinusoid fragment or not.
My common sense about the circuit operation and
analysis I have done are what led me to differ with you.
If believing my own carefully done analysis until it is
shown to be wrong is 'stubborn', then count me in.
I have done that very integration. Part of the problem
here is that the limits of integration that you need to
use to get a number out of it are changing under the
stated assumption that C is changing. I do not see
that incorporated into your result or thinking, yet.
When you do that, you will find that Idc and Irms
are no longer proportional. Another issue is that
under the assumption that capacitance is changing
(and that it matters), the currents in question are
not sinusoid fragments as your expressions claim.
Yes, your 'constant' is my 'K where K is a constant'..
You did, if you followed your own logic.
Given these conditions, and I quote:
If you hold everything else constant, (including the
DC current taken from the cap by the regulator/load),
then you claim the following:
I'm changing the Capacitance which is not going to hold
everything constant.
Idc *is* constant as established above.
If we take as true your assertion (and I quote):
Idc is directly proportional to Irms
where Idc refers to that constant charge per cycle, then
Irms cannot change without violating your assertion of
proportionality.
I've posted a link in this thread to an analysis I've done.
(under subject "Re: 9V 1A schematic needed [link]")
Below is an extract from it that may get this cleared up.
You can lift and use the definite integral expressions
by plugging in your equivalent variable names.
<begin extract>
The model behind this analysis employs an ideal
transformer having an output resistance Ro and
open circuit peak output voltage Vp, rectifier
diodes that turn on at a fixed forward bias and
have no resistance [1], and an infinite output
filter capacitance [2]. An independent variable,
Vd, represents the difference between the filter
capacitor voltage limited by R and what it would
be if R was 0 Ohms. The transformer open circuit
output is assumed to be sinusoidal. The analysis
is for full wave rectification, but can be easily
adjusted for half wave.
[1. Diode resistance can be folded into the
transformer output resistance. The actual
P/N junction current/voltage characteristic
contributes negligible error because it only
affects the current flow when that current
is relatively close to zero anyway.]
[2. The infinite filter capacitor represents
the limiting (and worst) case for RMS/average
current ratio. The situation can improve for
lesser capacitors as the ripple becomes an
appreciable fraction of Vp.]
Get the rectifier turn-on level:
Vx := Vp - Vd
Substitute an angle, a, for the phase velocity:
a := 2 pi Fp t
(where Fp is line frequency and t is time)
The voltage charging the cap thru Ro:
v(a) = max(0, Vp cos(a) - Vx)
Derive (half of) the conduction angle
g = acos(Vx/Vp)
The squared charging voltage during conduction:
v(a)^2 = Vp^2 cos(a)^2 - 2 Vp Vx cos(a) + Vx^2
Mean charging voltage over a half cycle:
v_ = 1/pi Integral[from -g to +g] v(a) d(a)
= 1/pi (2 Vp sqrt(1 - (Vx/Vp)^2) - 2 Vx g)
Mean squared charging voltage over a half cycle:
vsq_ = 1/pi Integral[from -g to +g]
(Vp^2 cos(a)^2 - 2 Vp Vx cos(a) + Vx^2) d(a)
= 1/pi (Vp^2 g + Vp^2 sin(2 g) / 2
- 4 Vp Vx sqrt(1 - (Vx/Vp)^2) + 2 Vx^2 g)
Note that the integrals for finding means
are evaluated only within the conduction
angle and the interval over which the mean
is computed is a half cycle (pi radians).
These identities are were substituted:
sin(+acos(x)) = +sqrt(1-x^2)
cos(x)^2 = 1/2 + cos(2 x)/2
The mean and RMS output current will be:
I_ = v_ / Ro
irms = sqrt(vsq_) / Ro
<end extract>
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
The american bank we had in Germany used to nickle and