I have been learning about the 7489 memory chip. I had two of these that turned out to be broken (see here for previous question on this). I got more and I have finally gotten everything to work.

From this answer, I learned the basic idea of how to set up the circuit. So now I have the following (schematics more or less copied from here). The following works well!


simulate this circuit – Schematic created using CircuitLab

I am still new to digital electronics and I am just trying to learn. My question is: How can one determine an appropriate pull-up resistor?

  1. For example, in the answer to the earlier question, the resistors R1, R2, R3, and R4, were suggested to be 2.2 kOhms. I chose 1kOhm. Does it matter? Is all that matters what the LEDs need or is there something relating to the 7489 that helps determine the size of the pull-up resistors? If it is only the LEDs (forward voltage 2V and current 0.20 mA) I would guess that I could actually use 220 Ohm. Is that right?

  2. Also, for the inputs (address and data input) I did choose 4.7 kOhm. What here determines the size? There are no LEDs so I assume it has to do with the IC, is that correct? What should I be looking for in the datasheet?

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    \$\begingroup\$ This question has already been answered in some detail. Please look at the "Related" column to the right and click on "How do I calculate the required value for a pull-up resistor?" \$\endgroup\$ – Sparky256 May 31 '16 at 23:37
  • \$\begingroup\$ @Sparky256: I see that for the question on how big a pull-up resistor in general should be. I will study these other answers. Thanks. So just to double check, about the resistors R1 - R4, they are only there for the LEDs and have nothing at all to do with the IC? \$\endgroup\$ – Thomas Jun 1 '16 at 0:54
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    \$\begingroup\$ That is correct. Most of today's LEDs have a Vdrop of about 3.0 volts, and the IC data outputs are open collector. The equation for LED current is 5V-3V=2V/R. This IC can sink 12mA max per output pin. At 150 ohms the current is 13.3mA, but you want some safety margin so a 330 ohm resistor will give you a bright 6mA of current. With all LEDs ON the IC only has to sink 24mA, or 120mW of power, well within safe limits. \$\endgroup\$ – Sparky256 Jun 1 '16 at 2:29

R1 - R4 are current-limiting resistors for the LEDs - they are required regardless of what is controlling the LEDs. If the LEDs drop 2 volts, assume there is 3 volts across the resistors (there's be a bit less, because the 7489 can't pull its outputs quite to ground), then for 20 mA, you'd want 150 Ohms. LEDs aren't very fussy about current, as long as it is less than their Absolute Maximum rating. Lower current gives less light, but for indicator light use I find tha 10 mA is more than enough, so I'd use 300 Ohms or more.

4.7K is a common value for general pull-up resistors. For high-speed signals, a lower value may be required to ensure that the signal changes fast enough when the signal should go high.

  • \$\begingroup\$ Thanks for the answer. So if the outputs are used as inputs for some other TTL chip, then no resistor (R1 - R4) is needed? (I just want to make sure that I understand before I do any damage.) \$\endgroup\$ – Thomas Jun 1 '16 at 0:53
  • \$\begingroup\$ Since the outputs are open collector (the output is an NPN transistor with the emitter grounded, and the collector connected only to the output pin), you should use a pull-up resistor of 4K7 or so when connected to a TTL input. \$\endgroup\$ – Peter Bennett Jun 1 '16 at 1:57

A resistor obeys Ohm's law, and choosing a resistor value is a matter of the permissible voltage and current, and power (voltage drop x current) according to application. For your switched inputs, a pullup resistor ought not to burn up, nor prevent a logic LOW level suitable for the 7489 inputs (0.8 volts max) when the switch is closed. Switch resistance of 0.01 ohms, power dissipation of 0.1 watt per resistor, and the datasheet limit of current from the 7489 input pin in low state (1.6 mA) all might be important; in practice, the power dissipation dominates, here.

Any value of input pullup resistor greater than 250 ohms will suit for LOW input normal function.

For input HIGH, switch leakage current, switch holdoff voltage, and logic HIGH preferred value (about 3V) don't limit your resistor choice at all. What MIGHT limit it, is the drive impedance with respect to input capacitance of the input pin; if there are transmission lines connected, a 220 ohm pullup and 330 ohm pulldown (to ground) gives good drive (circa 130 ohms to 3V), suitable for the load of the cable. Low impedance also protects against accidental signal pickup (stray capacitance, static discharge).

For LED drive, you do the same thing: compare the output voltage and current drive (tested up to 16 mA into no more than 0.45V) of the 7489 output pin with the current and voltage drop of the resistor and LED in both the HIGH and the LOW condition, and find resistor values that (1) do turn the LED sufficiently ON when output is low, and (2) don't turn the LED ON when output is high, and (3) don't burn up resistors or logic.

220 ohms works, will allow ~11 mA per LED.

Optional considerations: if logic is also driven from the outputs as well as LEDs, it, too, will require a current allowance. A shunt resistor in parallel with the LED will ensure lights-off, and may improve the high-side logic margin (because of the up-to-2V offset from the LED).


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