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Can an 74AUP1G06 Inverter drive an LED?

lax

Sep 12, 2021
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I'm trying to turn on an LED when a signal is in LOW (0V) state. This is on a very small PCB. SO I used a very small single bit inverter.
On paper, this should work, but the LED is not turning on. If I bypass the inverter and simply drive the LED, it works (obviously on when signal is high instead of low). The Blue LED displaying power is working correctly.

Should this work, or am I supposed to use a transistor to turn on the LED?
 

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Alec_t

Jul 7, 2015
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It should work IF the LED current you need is less than the current the inverter can provide without exceeding its rated maximum and without its output dropping below the Vfwd of the LED. Have you checked the inverter's datasheet?
 

lax

Sep 12, 2021
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It should work IF the LED current you need is less than the current the inverter can provide without exceeding its rated maximum and without its output dropping below the Vfwd of the LED. Have you checked the inverter's datasheet?
Could you help me decipher?

From: TI 74auo1g06 Datasheet

Absolute Maximum Ratings
Supply voltage, VCC –0.5 4.6 V
Input voltage(2), VI –0.5 4.6 V
Voltage range applied to any output in the high-impedance or power-off state(2), VO –0.5 4.6 V
Output voltage range in the high or low state(2), VO –0.5 VCC + 0.5 V
Input clamp current, IIK VI < 0 –50 mA
Output clamp current, IOK VO < 0 –50 mA
Continuous output current, IO ±20 mA
Continuous current through VCC or GND ±50 mA

And the LED

Kinbright Super Bright Orange

Forward Voltage IF = 20mA VF [2] Super Bright Orange 2.05 2.5 V
Reverse Current (VR = 5V) IR Super Bright Orange - 10 μA

My understanding was that the output is 20mA and @ 20mA the forward voltage was more than adequate.
I'm not sure if I understand the specs correctly
 

lax

Sep 12, 2021
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Nevermind .. Checked on Digi-Key website:

Texas Instruments SN74AUP1G06DRYR

Current - Output High, Low-, 4mA
Mystery solved.

I guess I have to now find an inverter of the same size with 20mA or more output.
Open to suggestions on part ...
 

Sunnysky

Jul 15, 2016
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Any '74LVC will work.
Reason: 3.6V max technology has low Ron ~ 25 ohms CMOS to 2.05V Orange from 3.3V , R=1.25V/20mA = 62 ohms then subtract 25 ohm driver, = 38 ohms (with tolerances)

'74AUP is ~ 100 ohms open drain
 
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Alec_t

Jul 7, 2015
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It is rarely a good idea to run components at their maximum ratings. That LED should give adequate brightness at much less than 20mA current.
In your posted circuit Vcc = 3.3V, R23 = 560R. From the D9 datasheet, Vfwd = 2.05V minimum @ 20mA.
So the maximum D9 current is ~(3.3V-2.05V)/560R = 2.2mA, ignoring the internal resistance of the inverter. That's well within the rated maxima of the LED and inverter, so both would be happy.
 

lax

Sep 12, 2021
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It is rarely a good idea to run components at their maximum ratings. That LED should give adequate brightness at much less than 20mA current.
In your posted circuit Vcc = 3.3V, R23 = 560R. From the D9 datasheet, Vfwd = 2.05V minimum @ 20mA.
So the maximum D9 current is ~(3.3V-2.05V)/560R = 2.2mA, ignoring the internal resistance of the inverter. That's well within the rated maxima of the LED and inverter, so both would be happy.

I am confused (sorry). Are you saying that the circuit should have worked? It definitively didn't, so either bad PCB or something else.
I've created a new design with the LVC version of the inverter and submitted to manufacturing (not approved yet). If there is a design flaw as opposed to a component flaw, how can I tell?
 

Harald Kapp

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Why not use a single transistor? When you use a logic level MOSFET you don't even need a base resistor (which is a must for a bipolar transistor).
If there is a design flaw as opposed to a component flaw, how can I tell?
Make a breaboard circuit and test it first before sending a PCB to manufacturing.
 

lax

Sep 12, 2021
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Build a prototype (even on a breadboard) and get that working before submitting for manufacture.
The part I'm using is EXTREMELY small and can't (at least I can't) be used on a breadboard.
 

lax

Sep 12, 2021
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I have tried the new part and it still doesn't work. Looking at the TI Datasheet, I have a confusing schematic. The LED *seems* backwards to me. I am literally trying to use the same design. What am I missing?? I guess I'm also confused by "MCU" and "VPU"
 

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Delta Prime

Jul 29, 2020
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Looking at the TI Datasheet, I have a confusing schematic. The LED *seems* backwards to me.
As it should when you first look at it!
Logic gates. it is called an inverter.
When there is a high or when the input is connected to VCC.
VCC is power positive side let's say 3 volts.
So input 3 volts output 0 volts or ground.
Output is at ground. Therefore current canal travel through the resistor through the LED and the inverter output which connects to ground.
It closes the circuit allowing the LED to light.
 

Delta Prime

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So when the input to the inverter is low or connected to ground the output of the inverter will be at VCC or 3 volts the LED will not light up.
On an inverter whatever the input is the output is directly opposite Input high Output low.
Input low Output high.

1684786735292.png
An inverter is also called a "NOT" gate.
 
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lax

Sep 12, 2021
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I swear, I understand inverter as far as logical signal.

I guess I'm confused as far electrical properties. From my understanding of LEDs and their symbols, the current (amperage) only goes in one direction and should go to *physical* ground so the circuit can function. A logical 0 is not a physical ground for the current to be able to circulate.
 

lax

Sep 12, 2021
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1684793408580.png1684793773664.png

TI design vs my design. TI's LED diode seems backwards ...

I'm simply trying to turn ON an LED on logical 0.
 

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lax

Sep 12, 2021
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As it should when you first look at it!
Logic gates. it is called an inverter.
When there is a high or when the input is connected to VCC.
VCC is power positive side let's say 3 volts.
So input 3 volts output 0 volts or ground.
Output is at ground. Therefore current canal travel through the resistor through the LED and the inverter output which connects to ground.
It closes the circuit allowing the LED to light.
From your explanation, when the input is high (1), the circuit will conduct. This DOES NOT do an inversion in my mind.
I want to turn on the LED when the input is 0 (Invert the 0 to a 1). So this part is not doing this????
 

lax

Sep 12, 2021
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So I guess this device is NOT the right device for my need. Any suggestion on a logical inverter as opposed to a current inverter? I want to turn on an LED when the signal is at 0 at the input.
 

lax

Sep 12, 2021
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1684796652340.png
I guess I should build it like this? Can this drive an LED directly or do I need an extra resistor?
 
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