# Floating ttl input and ground short without pulldown

A pulldown resistor seems to always be used for configurations where when the switch is open the normal state is low. For example

The reasons always given are to avoid a short between voltage and ground when the circuit is closed, and also to avoid the input floating when the switch is open.

I understand the first point about the short but why would the input still be floating if we did in fact short it to ground and didn't use a pulldown resistor? Can someone explain? Wouldn't the noise / random charges have a path of least resistance to go into just the same as with a pulldown resistor?

• " but why would the input still be floating if we did in fact short it to ground and didn't use a pulldown resistor..." If you short the input to ground then it is, by definition, no longer floating. Commented Nov 20, 2022 at 13:46
• That's what I thought, intuitively, which is why all the stuff I've read online have confused me. Therefore am I correct in surmising that the only reason for the pulldown is to prevent a short between ground and source? Commented Nov 20, 2022 at 13:49
• Do you really mean TTL, as you mention MCUs and surely they are CMOS these days, not TTL? Commented Nov 20, 2022 at 14:55
• Either, I suppose. I'm just trying to make sense of pull-up abd pull-down resistors for starters Commented Nov 20, 2022 at 15:05

If you had a 1000 Ω pull-down resistor (for example), when the switch closes, the current from 5 volts to ground would be 5 mA. If you used a 1 Ω resistor, the current would be 5 amps. Either will work for the MCU but it would be ridiculous to pull 5 amps just to send a signal to an MCU. A short circuit can also be regarded as a pull-down resistor.

Wouldn't the noise / random charges have a path of least resistance

I'm picking you up on this sentence because it's a common mistake to assume that all current or charge flows through the path of least resistance. Current distributes across a range of parallel paths with the highest current density being the path of least resistance.

The best solution is a "dynamic pulldown resistor" that changes its resistance depending on the state of the switch - low when the switch is off, and high when the switch is on. For example, this is implemented in the complementary output stages by adding a second (reverse) switch implemented by another transistor.

# How to switch voltage

## Revealing the problem

To understand something (in this case a "pulldown" resistor), we must first reveal the problem it solves. Here the problem is how to switch (off) the voltage applied to a load. It depends on the load resistance. Let's see how.

SW ON: The input voltage is controlled by a switch in series that has some low resistance when on, so VL = Vin...

SW OFF: ... and some very high resistance when off, so VL = 0 V.

## The invisible voltage summer

High-resistance load: But besides it, there are always other stray leakage resistances (fortunately, very high) between the power source and the load (there are also stray capacitances but this is another matter).

SW ON: When the switch is on, it overrides them because Vin is directly connected to load while +5V and -5V sources through high leakage resistances.

SW OFF: When the switch is off, only these stray resistances remain active, and they form a parallel voltage summer with weighted inputs. Its input gains are significant (close to 1), and there is a voltage across the load.

Low-resistance load: A typical example is the ignition system of a traditional car where the ignition switch is connected in series on the power supply to a low resistance load.

SW ON: When the switch is on, Vin is directly connected to load.

SW OFF: When the switch is off, as above, the stray resistances form a parallel voltage summer with weighted inputs. But its input gains are negligibly small, and the stray voltage across the load is negligible.

In the car example above, when the ignition switch is off (open) there is no voltage across the load... no problem. Other examples of this are also any (traditional) electrical appliances (stoves, heaters, etc.)

# The remedy

Obviously, to solve the problem, we have to artificially decrease the equivalent resistance by shunting the load with low resistance when SW is off.

## Zero shunt resistance

SW1 ON, SW2 OFF: We can do it with another switch SW2 that is on when SW1 is off. Note, however, that this textbook circuit is practically unfeasible because in the short switching moment either...

SW1 ON, SW2 ON: ... both switches will be on...

SW1 OFF, SW2 OFF: ... or off at the same time.

SW1 OFF, SW2 ON

## Low shunt resistance

That is why switches are actually controllable resistors whose resistance quickly changes from one value to another (100 Ω to 1 MΩ or 1 MΩ to 100 Ω).

R1 LOW, R2 HIGH

R1 MID, R2 MID: At the time of switching, they overlap.

R1 HIGH, R2 LOW

## High shunt resistance

Finally we got the idea, "Why turn off R2? Cannot we just leave it on all the time?" For this purpose, we increase its resistance to an acceptable value.

SW ON: There is no conflict when SW is on.

SW OFF: The load is shunted with some resistance when SW is off.

• This answer could benefit from a schematic. :-) Commented Apr 27 at 1:55
• @Math Keeps Me Busy, "A picture is worth a thousand words"? I get the subtle hint :-) Commented Apr 27 at 2:41
• @Math Keeps Me Busy, Done! Commented Apr 27 at 18:47
• A long answer was not really what I had in mind. Just a schematic of a "dynamic pull down resistor that changes it's resistance depending on the state of the switch". Commented Apr 27 at 19:15
• @Math Keeps Me Busy, Good thing you reminded me, I had forgotten about that. In fact, such an element seems contradictory and rather complicated because it must increase its resistance when the switch is closed (there is an input voltage) and it must not do so when it is open and there is a stray voltage. But it's worth discussing... Commented Apr 27 at 20:33