bleed resistor

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

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

| 1962 | I love to cook with wine. Sometimes I even put it in the food.

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

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

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

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)]

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

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

Tam

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

thanks

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

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

Graham

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.

--
Tony Williams.

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