1
\$\begingroup\$

I'm having serious problems figuring this out. Below is a part of an 8-bit computer schematic. The 62256-70 is a static ram chip. I understand these 8K2 resistors are there to provide a default value to the data bus.

Reading the specs, I see that LOW is considered < 0.8V and HIGH is considered > 2.2V.

What I'm trying to understand is: connecting a single resistor to a power rail will change its current. If I checked the voltage before and after the resistor, it would stay the same.

How then, is this dropping the voltage? Isn't the voltage going into D0 the same as the voltage at the other end of the resistor? Which is also equal to Vcc?

Thank you :)

Schematic

\$\endgroup\$
  • 1
    \$\begingroup\$ Wouldn't those be pull-UP resistors, preventing floating outputs or weak logic-1 outputs confusing a CPU reading D0..7? \$\endgroup\$ – RedGrittyBrick Jun 13 '13 at 15:01
  • \$\begingroup\$ RGB - I'd make that an answer if I were you. At a guess the device has open collector or weak outputs, although if the device is old enough it may well not have reliable pullups built into the device as most modern ones do. \$\endgroup\$ – John U Jun 14 '13 at 8:16
2
\$\begingroup\$

Looking at the resistor in isolation doesn't make sense. You need to consider the transistors on either end:

schematic

simulate this circuit – Schematic created using CircuitLab

Q1 represents the TTL output driving the data bus. Q2 represents the TTL input of a chip. Ignore the transistor part numbers, this is an example. R2, R3 and Q2 make up the input of a TTL chip.

When Q1 is switched off, a small current flows through R1, R3 and the base of Q2. This turns on the transistor and the input is considered "on". Because of the nature of transistor junctions, the base will tend to have a voltage not exceeding 0.8V. The other end of R3 can rise higher; the exact value of R3 is very dependant on the design of the TTL input.

When Q1 is switched on, the current through R1 is diverted entirely through Q1 and the bottom end of R1 has a voltage close to zero. No current flows in Q2 and it is "off".

\$\endgroup\$
0
\$\begingroup\$

Connecting a single resistor to a power rail will change its current.

If you connect a resistor between high and low voltage, creating a complete circuit, the applied voltage will induce current. Is this what you mean by 'change its current'?

If I checked the voltage before and after the resistor, it would stay the same.

This implies one of two things:

  • You don't have a complete circuit, or
  • If you do have a complete circuit, there's another resistance in series that's orders of magnitude higher than the resistor you're measuring across.

Simplisticly, this SRAM IC most likely has open-collector outputs. The data lines are connected to switches which are closed when a logic 0 is desired (pulling the pin to ground) or open (superhigh impedance) when a logic 1 is desired.

So, if the data line is open, the circuit isn't complete (there's no path through the open collector output), there isn't any current, and by Ohm's law, no voltage drop across the resistor.

How then, is this dropping the voltage?

If the SRAM changes the output to zero, you'll ideally have a (VCC - GND) volt drop across the resistor. (In reality, the switch won't be a dead short, and will have a few ohms of resistance). The purpose of the resistor isn't to drop the voltage per se, it's to limit the current to an acceptable amount so that when the open-collector output pulls the signal low, it doesn't fry due to excess power dissipation.

\$\endgroup\$

protected by clabacchio Jun 14 '13 at 11:50

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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