# Resistor value for Enable/Disable

• likephysics
In summary, the resistor value for an IC Enable/disable pin is determined by calculating the range of allowable pullup/pulldown resistor values based on the output current of the input and the input logic high/low threshold voltage values. This resistor value should be low enough to meet the input logic voltage level while allowing the input bias current to pass through. Additionally, a 2x or 4x margin should be added for safety. The reason for using a resistor to tie off an input is for testability during In-Circuit-Test (ICT) in manufacturing. This is because without a resistor, it would be difficult to detect if the IC pin is connected or not. For unused pins, a smaller resistor value can be used for
likephysics
How is the resistor value for an IC Enable/disable pin decided.
I always use 10K. But not sure how to calculate it .
Do you connect Enable bar directly to ground or thru a resistor.

likephysics said:
How is the resistor value for an IC Enable/disable pin decided.
I always use 10K. But not sure how to calculate it .
Do you connect Enable bar directly to ground or thru a resistor.

You can calculate the range of allowable pullup/pulldown resistor values based on the output current of the input (bias current) and the input logic high/low threshold voltage values. So the resistor value must be low enough so that the input logic voltage level is met, while the input bias current is passing through the resistor. That gives you an upper bound on the reisistor value, and you'll choose something to give you 2x or 4x margin or so.

The reason to use a resistor to tie off an input (that is always at high or low) is for testability at the In-Circuit-Test (ICT) stage in manufacturing. That's the "bed of nails" test that checks continuity and basic component values on your PCB assembly.

berkeman said:
You can calculate the range of allowable pullup/pulldown resistor values based on the output current of the input (bias current) and the input logic high/low threshold voltage values. So the resistor value must be low enough so that the input logic voltage level is met, while the input bias current is passing through the resistor. That gives you an upper bound on the reisistor value, and you'll choose something to give you 2x or 4x margin or so.

I'm a bit confused. An example would help.

The reason to use a resistor to tie off an input (that is always at high or low) is for testability at the In-Circuit-Test (ICT) stage in manufacturing. That's the "bed of nails" test that checks continuity and basic component values on your PCB assembly.

How would the resistor help for ICT?
They can have a test point without resistor.

likephysics said:
I'm a bit confused. An example would help.

How would the resistor help for ICT?
They can have a test point without resistor.

I'll try to post an example later today if I can get back to the PF. For ICT, if an input is hard grounded, you cannot tell if the IC pin is connected or not. All the ICT test point on that net can detect is the hard ground. If you connect the input to ground with a pulldown resistor, you can measure the value of the resistor, and also detect that the IC pin is connected (because of the internal ESD protection diodes to Vcc and Ground).

Sorry for the delay -- crazy busy weekend.

So for a simple example, say you want to use a pulldown resistor to tie off an unused input to a 74LS00 quad NAND gate. From the old TI TTL databook (yes, the yellow one), you get:

Vil = 0.8V max

Iil = -0.4mA max at Vil = 0.4V

So for the second set of numbers, you could use a resistor R = 0.4V / 0.4mA = 1k Ohm, but it would be best to use a smaller resistor by 2x or so to have margin on Vil.

For a more modern example, we can check the 75175 line driver that you mentioned in your PM to me.

http://pdf1.alldatasheet.com/datasheet-pdf/view/5748/MOTOROLA/SN75175D.html

Vil = 0.8V max

Iil = -100uA at Vil = 0.4V

So the second set of numbers gives you R = 4k Ohms.

berkeman said:
Sorry for the delay -- crazy busy weekend.

So for a simple example, say you want to use a pulldown resistor to tie off an unused input to a 74LS00 quad NAND gate. From the old TI TTL databook (yes, the yellow one), you get:

Vil = 0.8V max

Iil = -0.4mA max at Vil = 0.4V

So for the second set of numbers, you could use a resistor R = 0.4V / 0.4mA = 1k Ohm, but it would be best to use a smaller resistor by 2x or so to have margin on Vil.

For a more modern example, we can check the 75175 line driver that you mentioned in your PM to me.

http://pdf1.alldatasheet.com/datasheet-pdf/view/5748/MOTOROLA/SN75175D.html

Vil = 0.8V max

Iil = -100uA at Vil = 0.4V

So the second set of numbers gives you R = 4k Ohms.
Thanks. This is for unused pins only, correct?
What about resistor values for Enable/disable?

likephysics said:
Thanks. This is for unused pins only, correct?
What about resistor values for Enable/disable?

The Enable input should have the same characteristics as the general inputs, I would think. Just double-check the datasheets to be sure.

## What is a resistor and its role in the Enable/Disable function?

A resistor is an electronic component that restricts the flow of electric current. In the Enable/Disable function, a resistor is used to regulate the voltage and current levels, which determines the on or off state of a device.

## What factors should be considered when choosing a resistor value for Enable/Disable?

The main factors to consider when choosing a resistor value for Enable/Disable are the voltage and current requirements of the device, as well as the desired on/off state and the sensitivity of the components involved. Other factors such as temperature and power dissipation may also need to be taken into account.

## What happens if the resistor value for Enable/Disable is too low?

If the resistor value is too low, it may allow too much current to flow through the device, potentially damaging it. This can also lead to excessive heat generation, which can affect the performance and lifespan of the device.

## What happens if the resistor value for Enable/Disable is too high?

If the resistor value is too high, it may restrict the flow of current too much, preventing the device from functioning properly. This can also result in a decrease in voltage, which can affect the performance of the device.

## How do I calculate the appropriate resistor value for Enable/Disable?

To calculate the appropriate resistor value for Enable/Disable, you will need to know the voltage and current requirements of the device, as well as the desired on/off state and the resistance of the other components in the circuit. You can then use Ohm's law (R=V/I) to determine the necessary resistance value.

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