Pullup resistors
How do we determine what the pullup should be? I mean 100 to 300 ohms or 10 K ohms doesn't exactly give me the data I need. (Does it vary with the circuit being driven for example?) I really want to trigger the circuit with a minimum of time so I want the pullup fast so that my transistors (Nice NPN's for switching) with about 4.7 K ohms lead in resistance trigger off pretty much as fast as any current comes on the pin to signal it.
I will be pushing a signal in the switch of the NPN transistor so that there is some current to pull a MOSFET fast. I need some current to move like NOW.
In any case since I just cooked a $15 chip and an X1 Board , I really want to know some answers. (I just burned up some money) I need to know what voltage appears on the pins of the chip when we set for example a PORTA.2 =1. or PORTA.2 = 0 Is it on PORTA.1 = 1 a positive or negative voltage. It was logic testing that cooked this equipment.
All advice will be appreciated.
Well I got a cross of a posting that occurred between my post and an answer.
The issue of Pullups seems to be a feature of the pin to cause it to draw some current to make it pull down a few thousandths of an ampere so that the circuit stays logic stable. I would think that if I am driving a transistor as I am the obvious value to use is about 10K because the transistor will have a 4.7 K or so leading out of the logic pin into the transistor Emitter. Or does this 4.7 to Vss draw down enough power so that the transistor needs no other resistor on the emitter?
Generally a 5 VDC to emitter of a switching transistor with logic state high would take a 4.7 K ohm in the way of the signal feeding into the emitter (P of the NPN). If you take a 5 VDC state with a 10 K ohm to Vss and the positive voltage on the pins as high logic state then a 4.7 to the emitter should keep enough current to the transistor to open the gate so I am guessing based upon the previous answer that a 10 K ohm is in order (???)
When we change from an input drain state as in the keyboard to an output control state, I would suspect that we add a current to bring the pin near switch on state but just below in order to logically control things, Is this so?
Thanks
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