E
Eeyore
- Jan 1, 1970
- 0
Winfield said:What IC was that? How is it used?
UC3843 presumably.
http://focus.ti.com/docs/prod/folders/print/uc3843.html
Graham
Winfield said:What IC was that? How is it used?
Winfield said:But, taking for example a boost converter (or an
equivalent flyback), operating in continuous mode
(which is much more efficient than discontinuous),
RMS current = 1.0 * load current if the duty cycle
is 50%.
Eeyore said:This is only true if the current flow is perfectly rectangular.
I would certainly NOT expect that to be true for a flyback converter.
now that's a subject you're acquainted with..Eeyore said:Winfield wrote:
Win !
Wash your mouth out.
Graham
actually, I've always used low valued R's in series with the + lineRobert said:Put a current sensor in series with the capacitor and then you will
know the answer (think about it, and you should then know the answer).
We don't know what kind of smps Simon has, do we?
But, taking for example a boost converter (or an
equivalent flyback), operating in continuous mode
(which is much more efficient than discontinuous),
RMS current = 1.0 * load current if the duty cycle
is 50%. It's lower for duty cycles higher than 50%,
and higher below 50%, e.g., rising to 1.5 Iout for
D = 25%, according to a quick notepad calculation.
But one thing, since power loss goes as I^2 R, a
low duty cycle, with very high charging currents
(e.g. 3x load current for D = 25%), will actually
cause more heating than the simple RMS calculation
implies, and means a greater safety margin should
be required.
I would but I don't have the time to get through your thick skull andEeyore said:Jamie wrote:
Yes !
I haven't seen anything from you that shows even the tiniest sign of knowledge
or experience of electronics to date.
Shorted electrolytic capacitor my arse !
Do please explain the shorting mechasnism won't you ?
Graham
With a load current of 1.5 - 1.7A, your 2 x 1000uF capacitors connected in
parallel will not have an adequate ripple current rating.
Jamie said:actually, I've always used low valued R's in series with the + line
before the caps if there is a feed back for regulation involved.
Phil said:"Eyesore"
** That very much depends on the actual temperature of the caps in
operation.
For sure that should be factored in as well.
Phil said:"Eeysore"
** Big FUCKNING SHAME how YOU failed to take ANY any account of the
matter.
It was one of the first questions I asked.
The said:I assume you are considering the simplest (not realistic) case
of a purely rectangular current pulse delivered to the diode.
In that case, wouldn't a D of 25% imply that the peak of the
current pulse would be 4x the load current?
[OOPS - off it went before I was finished!]The inductor and diode have 4x the output current,
but 1x of this goes into the load, so the capacitor
only has 3x while its being charged. However, the
correct multiplier is 1.73, not 1.5, and for the
The ripple current for a 1A7 output flyback will depend on duty cycle
and secondary inductance. If you're quoting measured secondary rms
current, however, then the capacitor rms ripple current should equal:
Icaprms = ( Isecrms^2 Ioutdc^2 )^0.5
However, the correct multiplier is 1.73
, not 1.5, and for the
[OOPS - off it went before I was finished!]
I take back my casual dismissal of the higher capacitor
dissipation for the actual case of discontinuous current.
The multiplier for 50% rises from 1 to about 1.25, and
for 25% to over 2x, which is certainly very significant.
It's too bad we don't know more about Simon's transformer
turns ratio, etc. One certainly doesn't want to fall
below 50% at full output, if it can be avoided.
Yes, of course, that's right. I stand corrected.
BTW, D is usually defined as the PWM switch-on time,
rather than the PWM power-delivery fraction as I've
used it here, so readers should reverse the D values.
What were your formulas?
The plots vs duty-cycle are
indeed useful, but I do have one comment: showing the
current rising from 0 to full for 0 to 100% duty cycle
is misleading. In reality the power supply might reach
full output at a 50% power-delivery fraction, etc., the
other 50% being used to charge the inductor/transformer.
Moreover, as we've used it here, not knowing the supply
design details, a more serious case occurs if we assume
that full power is reached for a lesser fraction than 50%,
such as 25%. That's when things get tough for the poor
capacitor, and surely would represent poor power-supply
design in this case. But that's exactly what happens
for a simple inductor boost regulator, e.g., 25% duty
for a 1:3 step-up ratio at maximum power.
The said:I'll do some more plots, normalized to constant power
delivered to the load.
I look forward to that.
Utter and complete nonsense.
You first sunshine.
What I stated is 100% correct.
I'm staggered quite frankly that so few people here know how to design something
as simple as a PSU competently.
I mean SERIOUSLY ! Not to know how to determine the required ripple current
rating of a capacitor ? Bizarre.