Ripple Voltage from Constant Power Load

[D from BC]

Here's something I'm stuck on...


120VAC----Bridge
60Hz ----Rectifier-----+------+ <<<What's the ripple voltage?
| | |
| 180uF 90W Constant
| | real power load
| | |
+----------+------+

A constant power load will draw whatever current to maintain 90Watts
of dissipation.
As the capacitor discharges, the current increases to maintain the
power dissipation.

I could spice for an answer but..
I haven't figured out yet how to model a constant power load in
LTSpice..


D from BC
British Columbia
Canada.

 

[legg]

Here's something I'm stuck on...


120VAC----Bridge
60Hz ----Rectifier-----+------+ <<<What's the ripple voltage?
| | |
| 180uF 90W Constant
| | real power load
| | |
+----------+------+

A constant power load will draw whatever current to maintain 90Watts
of dissipation.
As the capacitor discharges, the current increases to maintain the
power dissipation.

I could spice for an answer but..
I haven't figured out yet how to model a constant power load in
LTSpice..

For a capacitive rectifier, the droop in the capacitor will be the
same for constant power as it will be for equal average power.

The problem is determining the exact intersect between the rectified
sine and the capacitor droop to get the discharge period. That sounds
like integral calculus to me, but its probably just algebra and trig.

For rough estimation you can assume 2mSec charging in a 6mSec period,
that leaves ~4mSec to discharge.

C/2 x ( V1^2 - V2^2) = delta j delta j = P x t j = CV^2/2

1.8E-4/2 x ( 160^2 - V2^2) = 90 x 4E-3

root{ 160^2 - ( 180 x 4E-3 / 1.8E-4 ) = V2 = 147V

ppk ripple = 166 - 147 = 20V

Worst case ripple occurs at low line as the initial energy stored in
the caps reduces.

RL

 

[D from BC]

For a capacitive rectifier, the droop in the capacitor will be the
same for constant power as it will be for equal average power.

The problem is determining the exact intersect between the rectified
sine and the capacitor droop to get the discharge period. That sounds
like integral calculus to me, but its probably just algebra and trig.

For rough estimation you can assume 2mSec charging in a 6mSec period,
that leaves ~4mSec to discharge.

C/2 x ( V1^2 - V2^2) = delta j delta j = P x t j = CV^2/2

1.8E-4/2 x ( 160^2 - V2^2) = 90 x 4E-3

root{ 160^2 - ( 180 x 4E-3 / 1.8E-4 ) = V2 = 147V

ppk ripple = 166 - 147 = 20V

Worst case ripple occurs at low line as the initial energy stored in
the caps reduces.

RL

I figured out a model to confirm.
http://www.members.shaw.ca/chainsaw/SED/constantpwrload.jpg
470kb jpeg
Got Vripple = ~21.77Vpp
Those pesky diodes...

Good math.. Your rough estimate was accurate enough for my app.

Thanks


D from BC
British Columbia
Canada.

 

[Fred Bloggs]

Here's something I'm stuck on...


120VAC----Bridge
60Hz ----Rectifier-----+------+ <<<What's the ripple voltage?
| | |
| 180uF 90W Constant
| | real power load
| | |
+----------+------+

A constant power load will draw whatever current to maintain 90Watts
of dissipation.
As the capacitor discharges, the current increases to maintain the
power dissipation.

I could spice for an answer but..
I haven't figured out yet how to model a constant power load in
LTSpice..

If V is the DC capacitor voltage then the DC load current is I=P/V
making the incremental negative conductance dI/dV=-P/V^2 which is pretty
small. Taking the average V to be say 90% of Vpeak~170V, this gives
-4mA/V incremental or -250 ohms, which is less than 1Watt incremental,
and not worth considering in more detail.