I've wired a few very simple things but I'm expanding my horizons and working on a new project. Router with Analog meter

My instructions tell me to take the 3 volts coming from the router led and invert the signal before using a 741 op-amp to increase it. I don't really understand how the 74hc04 chip works. To invert the signal I would need to wire the gates back to back according to the instructions, but that doesn't make any sense to me. The 74hc04 chip has 6 gates and a Vcc and ground. Here are my questions:

1. Does the 3v + from led connect to the Vcc on the chip or do I need external power to power the 74hc04 chip?
2. How do you wire gates back to back?


Thanks,

Sander

 

I am trying to build something that I found online. It's a network router attached to an analog guage to display the activity. http://www.instructables.com/id/Wire...-met/?ALLSTEPSI figured it was a good project for me to learn on. I've done a few very basic circuits.

I'm taking the 3volts that comes from an LED. My instructions state that I need to invert the signal before running it through a 741 op-amp to increase the voltage. "74hc04 - This guy comes with 6 gates, so if you want to get the same output signal as the input signal you tie to gates back to back."

I'm not really sure how a 74hc04 works.

Is the Vcc on the 74hc04 the 3v from the led or separate a separate power source?
How do I invert the signal by tieing the gates back to back?

 

The 74HC04 is a hex inverter. It contains 6 gates which output the logic inverse of the inverse; i.e., if you input a LOW signal, it outputs a HIGH and vice versa. VCC will be a separate DC supply voltage, usually obtained from the circuit the device is operating in. It is +5 V nominal for the 74HC04. If you have a higher supply voltage in the circuit to be modified, you can use a 7805 voltage regulator to supply the VCC.

You don't need to tie any gates together to invert the signal; that's what the inverter is for. If you wanted to, you can use two inverters to "clean up" the logic signals while keeping the output logic levels the same as the input by feeding your signal into one gate and connecting its output to the input of another gate. You don't seem to need to do this for your application.

 

Thanks!

Correct me if I am wrong here.

My positive from the led should go to 1a and the negative will just be grounded. Then I will take 1y to the input on my 741.

 

I probably should have read the link, too. It looks as though he's using the inverters as a buffer; you don't actually want to invert the signal, you want the output the same as the input. That being the case, you want to take the output from 1Y and feed it to another gate input, say, 2A. Then the output at 2Y will go to the input on the 741. You could simplify it and use a 74HC07 hex buffer. Then you only need a single gate.

 

Ok, I may have fried my 741 or at least it smelled like it, but that's ok I have another one.

I'm not really sure how to draw a schematic so I will try to type out how I have it laid out.



74hc04:
Positive from the LED is going to 1A
1Y jumpered to 2A and 2Y goes to the non inverse input on the 741
I am supplying 5v to the Vcc on the 04 chip and have GND going to my ground.

On the 741:
Non Inverse input is coming from 2Y
V+ is 12v input
V- is grounded
Output is going to a leg of the trimpot

Trimpot:
Right pin is my output from 741
middle pin goes to my analog gauge

Gauge:
One side is grounded and the other is coming from the middle of my trimpot

 

No.

Yes.

Yes.

Yes.

It is not the input, it is the positive of the car battery that charges at about 14V.

Yes.

[/quote]Output is going to a leg of the trimpot[/quote]
I don't see a trimpot but if there is one then you must connect it properly.

Maybe, it depends on how the trimpot is made.

Probably.

The unused inverters in the 74HC04 must be disabled by connecting their inputs to 0V.
Make the circuit like this:

Attached Images

instructable.PNG (78.4 KB, 16 views)

__________________
Uncle $crooge

 

I definately smoked my 741. So I do need to connect the inverse and non inverse? I had not been connecting the inverse on anything.

Just to see if I can get this circuit built without frying everything, I'm leaving the trimpot out. (I hope I am calling that the right word, it's the adjustable one)

Can you clarify if R1 and R2 are the connections on the guage or separate resistors.

 

R1 and R2 form a negative feedback voltage divider into the inverting input of the 741. If you didn't connect those, that's probably why you fried your chip.

 

does that mean R1 and R2 are resistors? I'm not seeing them on the actual picture of the finished project.

Also this is a trimpot in the image correct?

 

The trimpot (yes, that's the correct term) is R1 and R2. In the schematic, it's shown as two discrete resistors.

 

Hmmm... I can't imagine how leaving out feedback resistors can fry a 741.

__________________
Ron

 

Runaway output? Dunno if that's actually what happened; I can't trace his circuit easily from his photo. I'll change my "probably" to "possibly," though.

 

So if R1 and R2 are the poles on the trimpot, how can I tell which is what on the trimpot? I used an Ohm meter to check the legs on the trimpot and the left leg appeared to be a ground and the middle and right were the pot. Does that sound right?

 

The pinout of your trimpot exactly corresponds to your schematic: the left pin (Pin 1) of the pot gets connected to ground, the center pin (Pin 2) gets connected to the inverting input (Pin 2) of the 741 and the right pin (Pin 3) gets connected to the output (Pin 6) of the 741. Adjust the pot so the value of R(1-2) and R(2-3) correspond the the required values of R1 and R2, respectively.