bleed resistor

[Paul Taylor]

hi all
I need to place a bleed resistor across a power supply reservoir capacitor
of 15000uf to discharge it on turn off is there any rule of thumb to
calculate the right value for shortest turn off and wattage required.

thanks

 

[Pooh Bear]

hi all
I need to place a bleed resistor across a power supply reservoir capacitor
of 15000uf to discharge it on turn off is there any rule of thumb to
calculate the right value for shortest turn off and wattage required.

Choose a time constant that you like.

The bleed resistor will dissipate under normal running conditions. I normally
choose one ( in a high power amplifier ) to be ~ 1W dissipation.

Graham

 

[Jim Thompson]

hi all
I need to place a bleed resistor across a power supply reservoir capacitor
of 15000uf to discharge it on turn off is there any rule of thumb to
calculate the right value for shortest turn off and wattage required.

thanks

For mathematical amusement, why don't you calculate it ?

...Jim Thompson

 

[Tam/WB2TT]

hi all
I need to place a bleed resistor across a power supply reservoir capacitor
of 15000uf to discharge it on turn off is there any rule of thumb to
calculate the right value for shortest turn off and wattage required.

thanks

Pick a wattage, and then calculate the resistor value. If you only have room
for a 1/8W resistor, the turnoff time will be longer than for a 5W resistor.
I would pick a wattage less than 1% of the PS rating. Also, you will want to
pull enough bleeder current to keep the supply in regulation when the
external load is removed.

Tam

 

[Winfield Hill]

Pooh Bear wrote...

Choose a time constant that you like.

The bleed resistor will dissipate under normal running conditions. I
normally choose one ( in a high power amplifier ) to be ~ 1W dissipation.

These two criteria are often at odds with each other. If the voltage
is high enough to be dangerous, and the instrument will sometimes need
working on, the time-constant is best to be no more than a few seconds.
Since that usually means excessive dissipation, my approach is to pick
the appropriately-timed resistor, with its wattage sized to handle the
discharge energy from the capacitors, and switch it with relay contacts.
When unpowered the relay contacts should be closed. When AC power is
applied the contacts will open an instant later, and the ac transformer/
diode-bridge/resistor simply handle the high current flow for twenty or
so ms. If the relay fails to open, the fuse should blow after a second
or two (make sure the resistor is small enough for this to happen).

 

[Ian Bell]

hi all
I need to place a bleed resistor across a power supply reservoir capacitor
of 15000uf to discharge it on turn off is there any rule of thumb to
calculate the right value for shortest turn off and wattage required.

thanks

The amount of energy stored in the capacitor is given by:

0.5CV^2 joules

A joule dissipated over a second is one watt so an 1/8th watt resistor
should not be expected to dissipate more than one joule every 8 seconds, a
quarter watt resistor in 4 seconds, half watt in two seconds and one watt
resistor in one second. Work out your joules and decide how quickly you
want to dissipate the energy and that will give you the resistor wattage.
Remember that this resistor will dissipate this power when the supply is on
as well so it would be better to over rate it e.g. use a 1watt type where
the dissipation is half a watt.

From the decided dissipation and the supply volts you can calculate the
resistor value W = V^2/R so R = V^2/W

HTH

Ian

 

[Jon]

There are several approaches. Theoretically the capacitor will never
discharge to zero volts.
Approach 1)
Decide what is a "safe" voltage (Vd)
Decide what is the maximum acceptable time (t) to discharge to VD.
Calculate R:
Let Vc = the normal charged voltage.
R = t*/{C*[ln(Vc) - ln(Vd)]}
Where ln(x) is the natural logarithm of
x.
Calculate the wattage according to Paul Taylor's advice.
~
Approach 2)
Decide what is a "safe" voltage (Vd).
Decide on the minimum value resistance that you can use, based on
dissipation.
Calculate the time required to discharge to Vd.
t = R*C*[ln(Vc)-ln(Vd)]

 

[Tam/WB2TT]

There are several approaches. Theoretically the capacitor will never
discharge to zero volts.
Approach 1)
Decide what is a "safe" voltage (Vd)
Decide what is the maximum acceptable time (t) to discharge to VD.
Calculate R:
Let Vc = the normal charged voltage.
R = t*/{C*[ln(Vc) - ln(Vd)]}
Where ln(x) is the natural logarithm of
x.
Calculate the wattage according to Paul Taylor's advice.
~
Approach 2)
Decide what is a "safe" voltage (Vd).
Decide on the minimum value resistance that you can use, based on
dissipation.
Calculate the time required to discharge to Vd.
t = R*C*[ln(Vc)-ln(Vd)]

With 15000 uF, I assumed it was not a high voltage power supply; but then,
why is he worried about it at all?

Tam

 

[Paul Taylor]

thanks for all your valuable information with this I have sort out the
problem.

thanks

 

[Robert Latest]

These two criteria are often at odds with each other. If the voltage
is high enough to be dangerous, and the instrument will sometimes need
working on, the time-constant is best to be no more than a few seconds.

Years ago I worked in a (German) laser company in the power
supply department. Once I asked the head engineer why he had
littered the PS enclosure with more than a dozen screws. He told
me that there was some safety regulation that required the SMPS
primary filter caps to have discharged to a safe voltage within
the time it takes to open the enclosure. Large bleed resistor --
many screws. The relay approach is more elegant but probably
entails more elaborate safety audits. Once some bureaucrat has
figured out the average time it takes to remove one screw the
math is easy.

robert

 

[Pooh Bear]

Years ago I worked in a (German) laser company in the power
supply department. Once I asked the head engineer why he had
littered the PS enclosure with more than a dozen screws. He told
me that there was some safety regulation that required the SMPS
primary filter caps to have discharged to a safe voltage within
the time it takes to open the enclosure. Large bleed resistor --
many screws. The relay approach is more elegant but probably
entails more elaborate safety audits. Once some bureaucrat has
figured out the average time it takes to remove one screw the
math is easy.

I'd like to know which IEC reg that was ! I suspect said engineer was taking
an imaginative approach.

Graham

 

[Tony Williams]

Years ago I worked in a (German) laser company in the power
supply department. Once I asked the head engineer why he had
littered the PS enclosure with more than a dozen screws. He told
me that there was some safety regulation that required the SMPS
primary filter caps to have discharged to a safe voltage within
the time it takes to open the enclosure. Large bleed resistor --
many screws.

Yes. I've had to work to similarly described specs.
Plus requiring that the user should have to use a
tool to gain access to the innards.... where a coin
was specifically not classed as a tool, barring the
use of such things as large-slotted Zeus fasteners
for access panels.

 

[[email protected]]

hi all
I need to place a bleed resistor across a power supply reservoir capacitor
of 15000uf to discharge it on turn off is there any rule of thumb to
calculate the right value for shortest turn off and wattage required.

thanks

I tried a clever trick when I was too young to be more sensible. Used a
2 pole 2 way mains switch, and in the off position it switched the
transf primary across the dc output. Thus the transformer discharged
the reservoir.

The problems with this are:
1. if mains supply is lost but panel switch is on, output stays charged
2. lack of adequate insulation reliability between switch contacts,
creating a failure path from mains side to output.

In principle this could be done with a relay to solve 1., but as
someone mentioned a R would be safer!

FWIW a constant current device would give better discharge time vs
power waste, and a miniature lightbulb, whose R drops at lower V, would
give even better result. These must be considerably underrun to ensure
they last more or less indefinitely. Since you can solder miniature
lamps straight into the pcb this becomes an option.


NT