SX MCLR 10K resistor to Vdd? Really?

Chicago Mike

So I'm shocked i can't find anything discussing this when I searched. I'm using an SX28, and have tied the MCLR to Vdd. (+5). This has worked fine, and I've been programming with the SX-key, which also ties directly to the MCLR on my circuit board. All is well with or without the SX-key plugged in.

So I read the datasheet on the SX and it says the SX MCLR should be tied to Vdd through a 10K resistor to run. So I did this, and with the SX key plugged in, the SX does nothing, with the SX key removed, the SX runs. Odd? So I need to be able to plug the SX-key and reprogram the device without removing a 10K reisistor. My question is, so how needed is the 10K resisitor, and is there a better way to do this that I'm missing?

Thanks,

PJMonty

Mike,

Can you be more specific about what happens when you plug the SX into the circuit with MCLR tied via a 10k resistor? You say the "SX does nothing." Do you mean it won't program and run? It debug? It won't let the SX generate a clock? Are you plugging it in after having programmed the chip an expecting it to run? I'm not sure what exactly is going on here.

Also, download the FAQ and search for MCLR. The pull-up resistor is mentioned a number of times in it.

Thanks,
PeterM

Post Edited (PJMonty) : 11/28/2008 11:44:07 PM GMT

PJAllen

http://forums.parallax.com/showthread.php?p=725046

Chicago Mike

Thanks!

here is more info.....

The SX does not start up, or run the program with the MCLR tied to Vdd via a 10K resistor and the SX-key plugged in. Unplug the SX-key, and the program starts up. Plug the key back in, and remove the 10K resistor and replace it with a wire, and the SX starts up as well.

Attached is a schematic of my circuit. 'Port4' is were the SX key gets plugged in. If the circled part of the schematic is replaced with a 10K resisistor and the SX key is plugged into Port4, the SX does not operate.

Zoot

If that schematic is correct, then your circuit is incorrect. The pins from the prog. port go should go right to Vdd, Gnd, OSC1, OSC2. MCLR should only be tied to Vdd through a 10k (and *should always be tied to Vdd through a 10k*). You have the connection from MCLR straight to the Vdd prog. pin and being tied to Vdd through the 10k. See the SX Data Sheet and SX-Key Manual for suggested setups.

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When the going gets weird, the weird turn pro. -- HST

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PJAllen

Yes, what Zoot said.

(see attached)

Chicago Mike

The schematic is trying to show how the circuit works, if I remove the circled wire, and replace it with the 10K, as the datasheet states, the circuit STOPS working when the key is plugged in.

Zoot

But removing the "circled" wire and replacing it with a 10k would be an *incorrect* circuit. Sorry, but either way, it's wrong. See PJ's correct schematic.

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
When the going gets weird, the weird turn pro. -- HST

1uffakind.com/robots/povBitMapBuilder.php
1uffakind.com/robots/resistorLadder.php

Chicago Mike

I see your point. I can't believe I missed that. Sorry.... works fine now. My mistake.

Guenther Daubach

Congratulations Mike

Nevertheless, maybe some words regarding the 10 kOhm resistor might be helpful:

For normal operation, it is necessary that the /MCLR pin "sees" high level, i.e. is connected to Vdd.

The SX datasheet specifies the maximum Voltage on any pin being 7 Volt. As long as you can be sure, that this limit will never be exceeded, you might directly tie any pin that requires high level to Vdd.

The SX inputs are protected against overrvoltage by diodes but these diodes can only stand a certain maximum of current. Therefore, it is a good idea to place a resistor between Vdd, and any input that shall be tied to Vdd as this resostor limits the current through the diode in case an overvoltage occurs. As inputs normally have high impedance, they don't sink remarkable current. IOW, the voltage drop across the resistor is almost zero, so the input is savely pulled to high, even with the resistor in place.

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Greetings from Germany,

G