0
\$\begingroup\$

This is an image demonstrating the use of a pull up resistor.

enter image description here

I've just started studying electronics and have been learning about ohm's law. Ohm's law says that current within a circuit is equal to the potential drop across that circuit (or section of circuit) divided by the resistance of that circuit.

I understand that the pull up resistor prevents a short and limits current flow; however I am confused about how pressing the switch actually changes the current being delivered to the control chip. The circuit of the control chip (coming down from the top and exiting at the right) still has the same voltage drop and still has the same resistance regardless of whether the switch at the bottom is open or close. According to Ohm's Law this means that the current should be unaffected.

I'm sure there is an error in my logic, however I'm having difficulty figuring out where. I appreciate the help!

\$\endgroup\$
  • \$\begingroup\$ When you say the "control chip" are you talking about the 74HC04? \$\endgroup\$ – The Photon Apr 10 '15 at 4:12
  • \$\begingroup\$ @ThePhoton No. I likely misspoke when I said "control chip". For my purposes the wire exiting to the right connects to an Arduino digital input pin. But what I'm really curious about is why the state of the switch changes the current exiting at the right. \$\endgroup\$ – CorbinMc Apr 10 '15 at 4:16
  • \$\begingroup\$ Do you understand why closing the switch changes the input to the 74HC04? And do you understand why that changes the voltage at the output of the 74HC04? \$\endgroup\$ – The Photon Apr 10 '15 at 4:17
  • \$\begingroup\$ @ThePhoton No I suppose I do not. I thought that because the the digital pin on the arduino connects to ground that the potential difference from Vcc to that pin must be -5 V to maintain a 0 V difference around an entire loop of the circuit. \$\endgroup\$ – CorbinMc Apr 10 '15 at 4:20
  • \$\begingroup\$ The 74HC04 is able to create a voltage of either 0 V or 5 V at its output. \$\endgroup\$ – The Photon Apr 10 '15 at 4:32
3
\$\begingroup\$

I am confused about how pressing the switch actually changes the current being delivered to the [74HC04].

The 74HC04 has a high-impedance input. It will source or sink only miniscule currents, whether the input voltage is high or low (possibly it will sink more current at some in-between voltage, which is why we try to avoid driving it there).

So when the switch is open, there is essentially no current through the resistor. Therefore the voltage across it is 0, and the voltage at the input to the 74HC04 is VCC.

When switch is closed, it shorts the 74HC04 to ground. Since there's now 5 V across the resistor, some current (5 mA) will flow through it and everything is consistent. Still no current flows in or out of the 74HC04 input.

\$\endgroup\$
1
\$\begingroup\$

With a 74HC04, the input current is less than 1 uA. With the switch open, the chip current produces a voltage drop of 1 uA x 10k, or about .01 volts, and the voltage at the chip is essentially Vcc. With the switch closed, the voltage into the chip is 0 volts, since the input is now connected to ground. Regardless of the chip current, the input voltage variation is what causes the logic to function.

\$\endgroup\$
1
\$\begingroup\$

For all intents and purposes, HC CMOS inputs look pretty much like open circuits to DC, so your circuit simplifies to:

enter image description here

So, with the switch open there'll be no current through R1 and the junction of R1 and S1 will be at Vcc.

With the switch closed, however, all of the current through R1 will be shunted to ground, so the junction of R1 and S1 will be at ground.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.