AND Gate + Inverter Question??? HELP

Thread Starter

lisandro

Joined Jul 26, 2007
5
I'm testing a Quad 2-INPUT AND Gate (SN54/74LS08) to which I added a Hex Inverter (NTE7406) to obtain a logic sequence needed to run an H-bridge.

I'm having problems attaching an output from the AND Gate to an input on the Inverter in order to get the inverted signal from the inverter output thru a resistance and finally into the diode. When measured, the voltage is being minimized by the inverter; output from AND gate is 4.38V which is inputed on the Inverter but it outputs 0.06V which is not enough to light up the diode.

What I need to do in oder to get an output from the inverter of around 3.19V?

Please help, thanks for your time.
 

bloguetronica

Joined Apr 27, 2007
1,372
The logic inverter inverts the signal. If the input is 5V, the output is 0V. Also, this particular inverter you are using has open collector outputs, which means that you can only sink current from your load. So you will have to use the LED connected to Vcc. You will obtain no voltage across the LED with it connected to GND. You can use a 74LS04 inverter, but even though it only sources 2mA, besides sinking only 16mA.

The best choice here is to use CMOS. If you opt for the 74HC family (CMOS), use the 74HC08 and the 74HC04. You have also the 4000 family, also CMOS, with the advantage of supporting a supply voltage from 3V to 15V (logic levels vary with the supply voltage). In that case use the 4081 AND gate and the 4069 NOT gate.

This page is useful: http://www.kpsec.freeuk.com/components/ic.htm
 

SgtWookie

Joined Jul 17, 2007
22,219
You COULD connect a pull-up load resistor to VCC, then to that your LED, and the negative terminal of the LED to the output of the NTE7406. That way, when the output of the AND gate was high, the output of the inverter would be low, and the LED would be on. The 7406 should be capable of carrying at least 15mA unless it's 74Lx.

To invert the LED light logic, connect the pull-up resistor to VCC, the other lead and the + lead of the LED to the output pin, and the - lead of the LED to ground. In this case, you'll need to increase the resistance of the pull-up resistor to avoid exceeding the sink capacity of the 7406, yet small enough to drive the LED.

There won't be much current left over to drive much of anything else, however.
 

bloguetronica

Joined Apr 27, 2007
1,372
You COULD connect a pull-up load resistor to VCC, then to that your LED, and the negative terminal of the LED to the output of the NTE7406. That way, when the output of the AND gate was high, the output of the inverter would be low, and the LED would be on. The 7406 should be capable of carrying at least 15mA unless it's 74Lx.
That is a bad practice, since you have to resistor that LED to have much less than 16mA, so the resistor alone will draw a maximum of 16mA (when being sinked by the outupt transistor).
Better is to use a CMOS 74HC AND gate to light up the LED directly, by sourcing it (LED will be connected to ground). The gate can source 20mA by itself.
 

SgtWookie

Joined Jul 17, 2007
22,219
That is a bad practice, since you have to resistor that LED to have much less than 16mA, so the resistor alone will draw a maximum of 16mA (when being sinked by the outupt transistor).
Better is to use a CMOS 74HC AND gate to light up the LED directly, by sourcing it (LED will be connected to ground). The gate can source 20mA by itself.
You're absolutely correct - I should know better than to respond to posts when I'm tired! :eek: ;)
 
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