I am building a 2 to 4 decoder with 7408 And gates and need to put LED's on the output as indicators, active high. 2 questions

1. Would I put a current limiting resistor in series with the LED to ground or put the resistor and LED from the gate output to Vcc?

2. The LED draws 30mA at 2V, would I use the worst case Vo of the gate to calculate the required R value?

Thanks

 

If you read the 7408 data sheet (I realize that's asking a lot) it shows that the maximum low (sink) output current is 16mA at 0.4V output voltage and the maximum high (source) output current is only 0.4mA. Thus you would need to connect the diode to the +5V and the maximum diode current would be 16ma. The series resistor required is (5-0.4-2)/16mA=162 ohms (use 170-180 ohms for margin).

This connection lights the LED when the gate output is low which is opposite to what you want. To invert the signal you could use a 7400 Nand gate which has a low output when both inputs are high. If you need a high output for other purposes you could add a 7405 inverter at each output.

If you want to operate the LED with 30mA then you could use a 7406 buffer/ inverter at each AND gates output to drive the LED. The 7406 can sink 40mA maximum.

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Carl

 

Use a small power transistor with a low base current to be controlled by the ICs output. Connecting an LED with or without current limiting resistor will cause problems if the output signal is used to control other IC-inputs.

A BC547 will do the job nicely.

Boncuk

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Proper Planning Prevents Piss Poor Performance

 

To confuse matters, it turns out I need to have active high outputs.

I only have available a 7410 Nand (I need 3 inputs) I was going to place a 7404 inverter (all I have available) on the output to get active high output.

I would like the LED to light up when output is High. Circuits are a 2 to 4 and 3 to 8 decoder.

How will the inverter change anything in terms of driving the led's?

 

You can drive the LEDs as well as the 7404 inverter from the Nand output. Standard TTL outputs will sink a maximum of 16mA. The 7404 input will require 1.6mA, thus leaving about 14mA to drive the LEDs. Will that be bright enough for your requirement?

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Carl

 

So will I put the diode/resistor branch on the output of the inverter to Vcc? I want to LED to light when the output is high.

Also I only have 150 or 200 ohm resistors available, which side should I error on?

I think I was able to attach an image of the circuit?

Attached Images

301 Lab 3 2_to_4 Decoder Landscape 3.jpg (40.8 KB, 6 views)

 

No. If you connect the LEDs to the inverter outputs as you show, then they will light when the inverter's output is low. When the output is high there's no voltage across the diodes and thus no current flows.

If you want the diodes to light when the inverter output is high, then you need to connect them to the Nand output (inverter input). That way when the Nand gate is low (inverter output high) there will be voltage across the diode and it's resistor, and the diode will light.
If you have only those two resistor values available, use the 200 ohm. It will give about 13-14mA though the LED when on.

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Carl

 

Thanks a lot for your help, you've taught me a lot.