??? 12/11/06 02:55 Modified: 12/11/06 03:02 Read: times |
#129262 - Why you should use a proper reset chip Responding to: ???'s previous message |
Ralph said:
I am newbie compared to all the engineers on this site. I'm no engineer, either. I studied physics and all I know today about electronics I learned by myself. Ralph said:
So I looked around found this in nuts and volts magazine in 1996 http://www.allen-systems.com/mb51.htm. Take care, I have had a look at the board of this project and noticed, that the lower of the two burden caps of oscillator could have a better (shorter) connection to the ground pin of AT89C2051. Ralph said:
I also notice on 80c31 part that ap notes leave off the 8.2k and use only 10uf cap will this even work most of the time? When you have a look at datasheet of 80C31 you will notice, that there's already a pull-down resistance of about 50...300kOhm at the reset line internally of micro, serving as charging resistor of RC-reset circuitry. Notice the huge manufacturing tolerance, though! The external resistor is only needed if you want to provide a clearly specified reset time, means if you cannot tolerate these huge tolerances. The reset time has to be designed to be at least longer than the built-in Pierce oscillator needs to start-up under worst conditions, adding the rise time of Vcc during power-up. 10µF plus 8.2kOhm form a time constant of about 80msec and provide a reset time of about the same value. Ralph said:
If it does will it work on 89c2051 part? This RC-reset circuitry stuff standardly published in the datasheets in the chapter "reset circuit" is the absolute minimum requirement needed to have the micro starting to run after power-up, at all. It does not at all mean, that this RC-circuit is working under all circumstances and eventualities or is recommended as optimal solution! The main disadvantages are, that you cannot expect a proper reset signal, when any supply voltage dips occur or when the power-up takes longer than the time constant of RC-circuit! More, since micros are supplied with internal flash code memory, improper reset schemes not providing a stable and reliable reset during power-up and power-down, are under suspicion to cause data loss and damage of code memory! So, I highly recommend to use a proper reset chip, providing a brown-out-detection, when using a flash micro like AT89C2051. To guarantee a valid reset signal even at very low supply voltages (when using such a reset chip), add a pull-up of about 4.7kOhm to the reset line. (The DS1833 does already contain this pull-up, internally.) How to design the reseting scheme? The threshold of brown-out-detection should be close to the nominal supply voltage of microcontroller and other involved chips. This, because critical timing specifications of a microcontroller circuit can severely change, when the supply voltage is relevantly reduced. Or by other words, a circuit which works at 5V+-5% can stop working properly below 4.75V, So, for a +5V application a reset chip showing a trip point of 4.62V +-0.12V (4.50V...4.74V) is ideal. As the trip point can be down to 4.50V for such a chip, your circuit should reliably work down to a supply voltage of at least 4.50V, guaranteed by your testing. (Of course, a reset chip showing a trip voltage of exactly 4.74V for a +5V application would be even more ideal. But tolerance free chips are not available so far...) One issue often overlooked is the minimum Vcc fall time the reset chip needs to reliably generate a reset signal on power-down. For the MAX1232 or ADM1232 this minimum Vcc fall time to fall from 4.75V to 4.25V is 10µsec. For the DS1232, on the other hand, this minimum slew rate is 300µsec! What does that mean? Assume you have an application drawing a certain supply current and the supply voltage is turned-on and -off by a switch in front of or behind the 5V regulator, then you need some minimum storage capacitance on the supply line to allow a sufficiently smooth power-down, means to guarantee the fullfilling of minimum Vcc fall time. This minimum storage capacitance can be calculated as follows: Assume the current consumption is 50mA, then you get: C = Q / U = I x t / U = 50mA x 300µsec / (4.75V - 4.25V) = 30µF So, you should have a 100µF cap on the 5V line somewhere on the board, or take a less demanding reset chip. By the way, designing a good reset chip is an art! So, don't be surprised, when the reset chip is more expensive than the micro... Kai |