Electron flow with pulling resistors

I'm very confused on the idea of pulling resistors. I get that they set the default voltage to a value so that you don't get a floating high or low value, but I don't get the application in TTL.

So if we have 2 chips, Master and Slave (M and S for short). M & S communicate via TX and RX lines. M's TX like has a pull up, while the RX has a pull down respectivly. Assume Slave is reading TX and Master is reading RX. How does the flow of electrons change when RX and TX are switched on and off. assuming that this is a normal universe and electrons flow from ground to supply.

• If there is a voltage difference across the resistor, there is current flowing through it, which you can calculate. Generally you would look at the situation where the input with the pullup resistor was being actively driven low, however there is probably a small voltage mismatch in the high case too, especially as a TTL output won't drive to the supply rail. – Chris Stratton Nov 8 '15 at 4:01
• Electricity is not made of movement of electrons. I n electric shock, there is no electrons entering to your body causing pain. – GR Tech Nov 8 '15 at 6:36

they set the default voltage to a value so yhat you don't get a floating high or low value.

This is especially important in TTL which needs to have defined inputs to have defined output(s). If there is nothing driving an input, or what would typically drive that input is in a high impedance (tri-state) mode, a pulling resistor is necessary to provide a definite level on that input.

Check out these basic examples, based on the components you mentioned...

simulate this circuit – Schematic created using CircuitLab

The Tx line is "pulled HI" by R1. This example works better in a multi-master setup, but the idea is the same. The master transmitter will likely first check if the line is HI or LO. If it is LO, the slave is probably holding it LO, so the master will wait with its output in a tri-state mode (high impedance). The opposite would be true for R2 the pull-down resistor.

If neither the slave nor master are tryint to transmitting anything, then their outputs will be "turned off" in the high impedance tri-state mode. R1 will pull the master Tx HI; however, no current flows through R1 because there is no ground to sink the current. As such, the master Tx line is essentially the same voltage as Vcc.

Similarly, R2 pulls the master Rx line LO, and since there is no current source on either the master Rx line, there is no voltage drop across R2 so the master Rx line is essentially grounded.

Using this Drive / Tri-State operation, the chance of shorting master Tx to Slave Rx is greatly reduced, if not eliminated. Long story short, using the ideal laws of pin inputs, there should be essentially (virtually) no current flow at all through R1 and R2, hence, no flow of electrons. However,

• wait, so when a transistor opens its input to low, it's really grounding the connection, and when it's input is high, it's connected to the source? – user86234 Nov 8 '15 at 13:40
• @tuskiomi I'm a bit confused on your wording here, but like I said, when something is pulled HI or LO, they are essentially connected directly to source and ground. This is not the case for current controlled inputs (like a BJT base) since there will be current flow through the pulling resistor, hence a voltage drop across it. This only applies to gate-like inputs that have theoretically no current flow. – Kurt E. Clothier Nov 8 '15 at 21:53