Newbie Question - Pullup Resistors

I'm in a verbose frame of mind, just remember that you asked;)

Electricity comes in two forms, + electricity and - electricity.  Ben Franklin got the sign wrong, but that wasn't proven until about two centuries later.  Let's say we have a voltage source of one volt.  No such thing really exists, it implies having access to infinite power.  There would be a plus terminal and a minus terminal.  Potential is always measured across two points.  We assign 0V to the negative terminal of the voltage source as a notional convenience.

If we only have resistors for loads, all potentials must be between zero and one.  We could put a voltage ladder across the power supply.  This would be two conductors in series, let's make them both 2S(Ohm/2) each 'cause I am lazy.  The series combination of the two loads creates a single load of 1S(1Ohm) with a tap in the center as the power supply sees it. To meet its promises, the supply sources an ampere of current.  Now there are three potentials, 0V, 0.5V and 1.0V.  With reactors and switches, I could get supervoltages, but I'm not goin' there now.  If I want to raise the middle potential I need to make the load  in the upper leg of the pair more conductive, or make the load in the lower leg more resistive, or both.

The basic vacuum tube is pretty much analogous to an n-FET.  So, a typical circuit would have the lower half of the voltage divider replaced by the conduction channel (they didn't call them than then) of the tube.  In signal work there was a resistor (typically) between the top end of the conduction channel and (top-of-supply) T.o.S.  Using a smaller resistor caused greater dissipation while typically making things faster by making more current available to charge the conduction channel + load capacitance.  Instead of a resistor we can use a (constant-current-source) CCS or active load, in a sense making load resistance zero or even negative.  This is pseudo-complementary or cascode.  We can also use magnetics to get push-pull, but it is tough.  An output stage than can slew both the top load and bottom load, in opposite directions, of course, is more efficient than one with only one device active in the output pair.

Logic chips sometimes complicate things still further by having a 'tri-state' mode in which both output switches are off.

Transistors come along.  Electrons, in Si, are lighter and faster than holes.  So n devices predominate.  Many output stages use a resistor pull up and an NPN pull down.  This is why the term 'pull-up resistor' is more common than 'pull-down resistor.'  But the development of semis did allow for complementary symmetry, an n device and a p device in the same totem-pole, often their control terminals could be tied together.

To repeat, a more conductive pull-up will make you run faster and hotter.

I'm not familiar with the ZX Spectrum itself, but the usual reason for pull-up resistors on a bus is to force the signals high if nobody is actively driving the bus.  Digital signals have two defined values: "1" (greater than a "high" threshold voltage) and "0" (less than a "low" threshold voltage).  Depending on the transistor technology, it can be a bad thing to have a signal between the threshold voltages for an extended period of time.  Pull-up or pull-down resistors are a way to ensure that a signal goes to a well-defined "0" or "1" if it's not being driven by a pull-up or pull-down transistor.  A transistor is stronger than a typical pull-up resistor, so it wins it it's turned on.

At the risk of pure pedantry, there is "positive logic" and "negative logic."

uPs are all positive logic, not all jellybean circuits.

Before CMOS, positive logic trues were less energetic than falses, your compy may use falling-edge interrupts and is using the pu's to avoid strays.  More conductive ones would give you a little more protection,  but raise your power dissipation.

Hi Nicholas,

Just to add to what has been posted, pull up resisters allow you to stabilize the EM field along the circuit paths.

As others have pointed out, consistent current switching stabilizes the overall standing wave down the circuit path.

Also, some input circuits are very susceptible to static charge flow if you do not properly ground and terminate the data path.

Essentially, very bad things happen when static charge exceeds the breakdown voltage for some devices.

When using a parallel bus path, the pull up resistors balance the line into a tuned circuit and keeps the digital data in sync as it moves down the parallel wires.

DAB

The logic families have their own characteristics, bipolar families often had fairly high-Z positive logic true and fairly low Z positive logic false.  CMOS has pretty symmetrical loading, but the high Z-in can expose the reactive aspects of the bus.  Obviously, differential signaling, for all of its complexity can have its merits in situations.  Don't even go beyond the HF zone without good reason if you are a beginner is my advice. 

Just to add some food for thought. I like to attack these understanding problems by asking, "What would happen if those resistors were not there?" Try thinking about this problem, and sometimes you will lead yourself to the answer.

Good thinking, MW!  IIRC Vache Volperian has a suite of fast analytic techniques that operates by substituting zero and infinite Rs for the Rs and comparing these extreme situations.  What if this resistor was infinite? is a good question, sometimes what if it was zero? is a good one as well.