Just to verify my reset circuit

My reset pin is active low, this would work right?

Vcc |------- | | | - R10k 1N4148 | + | | |-------To reset pin | C10uF | | GND

Reply to
Ant_Magma
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YES

Dan

Reply to
Dan H

what is the diode for?

Reply to
Figaro

The diode is to rapidly discharge the capacitor when the power is turned off. Without it the capacitor will remain charged for a long time and turning the power off and back on quickly will not cause a reset to start the microProc.

Dan

Reply to
Dan H

Hi, Ant. It depends. You have to look at the data sheet to see if a very slow-rise reset is OK for your chip. This will take dozens of milliseconds to go through the intermediate area between logic "0" and logic "1", which might be a problem for the reset circuitry internal to your chip. Look on the data sheet for maximum reset pulse rise and fall times.

If it is a problem, or if you're not sure, and you have a spare inverter gate and no time to research, it might be just best to reverse the R and C, and then use the inverter to square it up anyway. No points off here. The hysteresis of a 74HC14 can be invaluable for this. If I need an inverter on a board, I'll spec one of these so I can have this squaring up capability.

Good luck with your project. I hope things are going well.

Chris

Reply to
Chris

--
Yes, but why do you need such a large capacitor?
Reply to
John Fields

Hi, Ant. If this is the Reset for the DP83847, the data sheet specifies a 160us minimum pulse width at LLP pin 46, and doesn't specify a maximum. In other words, the data sheet doesn't say.

You'll notice on the IBM sample board that this signal comes from off the board (SYSRESET). That tells me the signal on the IBM board is almost certainly buffered to avoid noise problems.

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And if you look at the reset timing diagram on p. 46 of the data sheet

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your sequencing on hardware reset doesn't look like it's conducive to a lazy startup, although I don't see anything there that specifically prohibits it.

If you were working a real engineering project, you'd either be safe and square the reset up, or call the apps people at National and ask. They are almost always willing to give a hand for a design win on anything with any volume, especially something simple like this. They might even email you an answer same-day.

A complex chip can have complex reset circuitry, involving a sequence of internal events that occur when reset is unasserted. These may not happen in the right order with a slowly rising reset signal. I've made the mistake before with a processor board of using a simple RC to reset to save the trouble of a complex routing, and then not having things come up reliably. You really should check before you save a gate to be sure.

And my gut feeling with this chip is that, unless you're severely pressed, you should just scrounge a logic gate or add a TinyLogic gate, square it off, and be done with it.

Chris

Reply to
Chris

Reply to
crazy frog

Actually, pin 46 is a CMOS input, so it neither sources (blows) or sinks (sucks) current.

By the way, if you want a simple transistor circuit for your one minute delay, and you happen to have a 10,000 uF cap and a 10K pot handy, you might want to try this (view in fixed font or M$ Notepad):

| | SW1 | VCC-o | __--o-o-------o-------. | GND-o | | | | | | | | | .-. | | | .-->| |10K | | | | | | | | D - | '-' | | ^ | | | | | '----o | | | | | | | .-. | | | 1K| | | | | | | .---o | | '-' | | | | | |/ | | '-------o-| | | +| |> | | 10,000 --- | |/ | uF --- '-| TIP101 | | |>

| | | | === .--o | GND | | | D| C| | - C|RY1 | ^ C| | | | | ====== | GNDGND | (created by AACircuit v1.28.6 beta 04/19/05

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Simpler, and fewer components. This relies on the fact that the guaranteed minimum turn-on voltage for a 12V relay is almost always less than 9.1V Even if Vcc goes down to 10.5V, this circuit should still light up the relay. This circuit doesn't discharge the 10,000uF cap through a base junction on turn-off (eeeww), or rely on two diodes to limit Vbe to 1.4V (?ouch!) on turn-on. You don't have to worry about exceeding maximum wiper current if you turn the pot to min. You don't have a 100uF cap which performs no useful function at all. Nor do you have a second transistor which really doesn't do much except drop voltage. You will also be able to drive a wider range of relay coil currents reliably (can drive a 12 ohm coil, but over 200mA timing becomes beta dependent -- and be sure to use a heat sink!) This circuit has provision for quick reset on switching. However, just like your circuit, the time delay on this circuit will be power supply voltage dependent. Use 1N4002 diodes for D.

Call it a day. Mr. Fields may have been a little brusque, but you're embarrassing yourself. Big time.

Good luck Chris

Reply to
Chris

Thx guys for your replies.

Chris, the reset is not for the National IC, its obtained from Micrel KS8721's datasheet. I want to use it on my Bel power line module, since there is no documentation watsoever in the Bel's datasheet regarding reset circuitry.

crazy frog wrote:

Excuse me?

Reply to
Ant_Magma

OK, Ant. You know what to do. You can either spend a gate or find out if it's OK.

Good luck Chris

Reply to
Chris

OK -- Let's save the switch by discharging the 10,000uF cap through a diode and a series resistor.

| | SW1 | VCC-o | __--o-o-------o-------. | GND-o | | | | | | | | | .-. | | | .-->| |10K | | | | | | | | D - | '-' | | ^ | | | | | '----o | | .-. | | | 100 | | .-. | | | | 1K| | | | '-' | | .---o | | '-' | | | | | |/ | | '-------o-| | | +| |> | | 10,000 --- | |/ | uF --- '-| TIP101 | | |>

| | | | === .--o | GND | | | D| C| | - C|RY1 | ^ C| | | | | ====== | GNDGND | | (created by AACircuit v1.28.6 beta 04/19/05

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Chris

Reply to
Chris

It's just one of the USENET lunatics. My newsreader (Pan) has a "plonk author" selection in one of its menus, so I can ignore him forever with two mouse clicks. No, wait: three - I have to click the "never expires" button. ;-)

Cheers! Rich

Reply to
Rich Grise

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