??? 03/02/07 20:08 Modified: 03/02/07 21:10 Read: times |
#134146 - reset (and diode) Responding to: ???'s previous message |
hi,
Richard Erlacher said:
What does, "no, even diode cannot help due positive reset - sorry." mean?
Well, this is simplest schematic of RC-reset. (This example shows negative reset to understand easy). -----*------> Vcc | --- | | | | --- | (A) *------> /RST | --- --- | | === When power is off for long time then capacitor is empty, indeed. Now, system is turned on. Vcc goes to nominal value (assume it is 5V) and /reset grows up slowly (capacitor takes current itself) so it does provide negative reset to reset pin. It is good - more or less. Now system is turned off. Vcc falls down to zero but voltage on node A falls not so fast due high resistance of reset pin and current limit resistor from Vcc as well. All this does not allow capacitor to lost its energy fast. In some cases when Vcc is about, for example, 2V nevertheless /RST is still about 4 or more volts so no reset signal is provided in the whole. Internal parts of processor come unstable but reset is not generated yet. It is very bad. Now imagine that system is turned on once again immediately during this period -- but capacitor is not empty yet! In this case even power-on situation does not generate reset by this RC circuit. How processor starts in this case I cannot say because it will not take reset state! Such situation may occur, for example, when power losts for short time. I can agree that it may be not so dangerous in situation when power goes off (but it is the question of real device and application, sure). But as you can see such schematic may cause "no reset" state even at power-on process, what is not allowed in most cases, really! So what is about diode. If we place diode (low-drop one, catode tied to Vcc) over resistor then it helps us to keep voltage on node A not more than +0.4..0.7V of Vcc (average, depend on diode type) during power-off process. At least it helps capacitor to lost its energy more quickly. But is it good practice? My answer - no, it is not. Regards, Oleg P.S. Remember this note in instructions: "switch power off, wait for 10 seconds and then power on". Wait till capacitor... (= |