Thread: WHAT DO I DO WHAT DO I DO WHAT DO I DO WHAT DO I DO WHAT DO I DO Power Good

Here is the schematic for my power good circuit. I haven't checked the diagram all that well(drawn in paint). This is the circuit I needed to get my board going.

well, I'll try that circuit, and my mobo is an old P75, It doesn't have anything inbuilt not even the controller card, to play mp3s I have to have atleast 2 cards plugged in, yeah, just thought you might like to know....

What does this circuit do????
I found it, and it is supposed to be a power good circuit.... Is this what I want???


[ 07-16-2001: Message edited by: THE_SKINNY_MP3_KID ]

Originally posted by THE_SKINNY_MP3_KID:
<STRONG>How does a resistor and a capatitor give a pulse????
</STRONG>

It forms what is called an "RC circuit". The simplist of all timer. The circuit outputs a pulse as the capacitor charges through the resistor. It is much more like a very long single triangle wave than anything else, but seems to work for power good.

So,,that is what I will use??

hmm...
I think I will just experiment for a while....

RC circuits are simple. The capacitor is charged up slowly via the resistor when power is applied. The rate at which the capacitor charged is calculated using the RC time constant. The RC time constant is the time taken for the capacitor to reach 63% of full charge(strange I know) and it takes five time constants to charge fully. The formula for the RC time constant is:

RC time constant(seconds) = R(Resistance in ohms)*C(capacitance in Farads)

So for my powergood circuit:

RC time constant = 10 000 * 0.0001
RC time constant = 1 Second

So it takes 1 second for the 100uF capacitor to charge up to 63% of 5 Volts(3.15V). The LM311 is a comparitor. This means it compares the two voltages on the + and - inputs and multiplies them by around 15 000. connected to the - input is a resistor divider. These two resistors cause the 5V power rail to be split in half. So the - input is sitting at 2.5V. When the circuit it turned on, initially the + input is at 0V and the + input is at 2.5V. The comparitor "compares" the two voltages and multiplies the difference by 15 000.

(0V-2.5V)*15 000 = -37 500V

So the output of the comparitor is -37 500V. No just kidding. Because it is impossible for the ouput to go below ground(0V) the ouput sits at 0V.

Now the capacitor begins to charge and the voltage on the + input increases. Once the voltage passes the 2.5V threshold of the circuit(this will take approx. 0.5 seconds) the output of the comparitor will be:

(2.51-2.5)*15 000 = 150V

Now the output is limited by the supply rail of the circuit(5V) making the output 5V.

So the circuit produces a 0V output for approximatly 0.5 seconds and then stays at 5V while power remains connected.There is a very small transition period around the 2.5V mark but the rise time of the pulse is very fast.

Hope this help you understand how my powergood circuit works SkinnyBoy. (Sorry about the corresponding long winded explaination).

Okay, thanx, I still haven't gotten around to making a power good circuit, and am currently just tapping to wires together...

it works though...

lol