BreadBoard

1. The supply voltage to the 7404 is rated for 4.75 to 5.25 VDC. 9 VDC is too much.

2. Each gate inverts the state at the input. When its input is at a logic 0 (low) the output is a logic 1 (high).

3. Driving an LED from this gate requires a current limiting resistor, which you need to calculate for the specific LED you are using. The low level output current for each gate is rated at 16 mA, which you need to consider in your design.

Go spend some time and look at the data sheet to understand how this works. If this is a homework assignment, you should try to first understand the problem as best you can, then write down your specific questions and ask the instructor.

No one here will do your homework, but people will be glad to help educate you if you demonstrate that you are willing to make the effort to learn. It is all about asking the right questions.

If this is not a homework assignment, you should first briefly and concisely explain what you are trying to accomplish (you have not), then explain how you are planning to execute that idea.

I agree

Hello, Thanks for the reply.

First of all i am new to breadboarding and also circuits .

2nd this is not a homework , i just want to learn circuits and breadboarding.

3rd The LED that i am using can handle the 9volt charge. without resistor the LED is big.

The only question i will ask:

how can i connect the INPUT/Ouputs of the IC 7404 the NOT Gate.

here's the diagram i make.

Okay. Each gate inverts the state of its input, so, even though you have daisy chained three gates, the net result is inversion. You can make a truth table that looks like this:

INPUT OUTPUT
H L
L H

To invert the state of the input only requires one gate, not three. The other 5 gates should have the inputs tied to ground to prevent oscillation.

The real problem is driving the LED. LEDs are rated for current (not voltage), usually, so I need to see the part number of the LED.

In your case, the output of the 7404 really should drive a transistor that switches the LED and not drive it directly, I think. I still need to see what the LED is that you are using.

The outputs on the 7404 do a better job sinking current than providing it. So, you might connect the LED to +5VDC and the other side of the LED to the output of the 7404 if the LED does not draw more than 16 mA.

Note, there is a polarity to the LED (Anode and Cathode). The cathode should go to +5 VDC and the anode to the output of the gate. You may still need a current limiting resistor. Again, it depends on the LED.

Plugging in an LED backwards will not harm the LED. It simply will not work. Just reverse the leads and see if that works.

Correction

Anode should go to +5 volts and a resistor that limits the current through the LED should be in series with the LED. This series resistor is for the basic LED but there are LEDs available for different voltages with an integral resistor.

Gajanan Phadte

Good catch, thanks.