SN74151AN: Help understanding HIGH/LOW selector inputs.

Thread Starter

atrumblood

Joined May 13, 2012
59
Hello everyone, I am new to the allaboutcircuits forums. I am here in hopes that I can gain some better understanding on how to use multiplexors.

After trying to teach my self how to use one of them. Specifically the SN74151AN.

I find that I don't quite grasp the high/low logic terminology. I get that high and low refer to the voltage, but I feel I am missing something.

The selectors on the 74151 operate on the high/low logic. but is it simply a matter of connecting them to either Vin or ground to get high or low status on the pin?

For your reference I have included a link to the datasheet.
http://www.alldatasheet.com/datasheet-pdf/pdf/27382/TI/SN74151AN.html


If anyone can help me understand this I would be appreciative.
 

MrChips

Joined Oct 2, 2009
30,621
Firstly, you have to begin with basic digital logic and boolean algebra.
How much of this do you understand?

Have you read the AAC Tutorials on Digital Logic?

http://www.allaboutcircuits.com/vol_4/index.html

A single digital input can be given either of two values, 0 or 1.
This can be referred to as LOW or HIGH.

In 7400 series digital logic, a 0 is represented by a voltage between 0V and 0.6V
and a 1 is represented by a voltage between 2V and 5V.

Digital inputs usually are fed from another digital output. Sometimes GND and Vcc are substituted for digital logic levels.

The 74151 is an 8-line to 1-line data selector. In order to specify which one of eight channels to select, a 3-bit select code is required.

A 3-bit code can have 8 possible values, 0 to 7. This is called a 3-bit binary coded decimal.
This is where knowing the binary number sequence is important.

(There are only 10 types of people in the world, those who understand binary and those who don't.)
 

ErnieM

Joined Apr 24, 2011
8,377
Actually this chip (on page 4) lists a zero as 0.8 volts max, and a one as 2.0V min.

A high or a low can be obtained from any voltage source that meets this range, meaning ground is good for a zero/low, and Vcc for a one/high.
 

MrChips

Joined Oct 2, 2009
30,621
Shucks, I knew the spec sheet said 0.8V but I wanted to be conservative.
I should have realized that someone on AAC will be looking over my shoulder and checking up on me.

Sorry, I'll stick to the facts next time.
 

Thread Starter

atrumblood

Joined May 13, 2012
59
Firstly, you have to begin with basic digital logic and boolean algebra.
How much of this do you understand?

Have you read the AAC Tutorials on Digital Logic?

http://www.allaboutcircuits.com/vol_4/index.html

A single digital input can be given either of two values, 0 or 1.
This can be referred to as LOW or HIGH.

In 7400 series digital logic, a 0 is represented by a voltage between 0V and 0.6V
and a 1 is represented by a voltage between 2V and 5V.

Digital inputs usually are fed from another digital output. Sometimes GND and Vcc are substituted for digital logic levels.

The 74151 is an 8-line to 1-line data selector. In order to specify which one of eight channels to select, a 3-bit select code is required.

A 3-bit code can have 8 possible values, 0 to 7. This is called a 3-bit binary coded decimal.
This is where knowing the binary number sequence is important.

(There are only 10 types of people in the world, those who understand binary and those who don't.)

Thanks for the reply.

I am very familiar with binary counting and I understand boolean logic a bit since I also program as my other hobby. However I have not read too much specifically on boolean algebra.

000 = 0
001 = 1
010 = 2
011 = 3
100 = 4
101 = 5
110 = 6
111 = 7
and so on....


1 = high
0 = low

So based on what you said about voltage a 1 would be set by a voltage of 2 - 5 volts and a 0 would be0 - .6 or .8 volts as the datasheet tells me.

Those 1s and 0s on the selectors tells the mux which input to allow data through?

Now the other thing I have had trouble understanding is; what is the strobe pin for?
 

Ron H

Joined Apr 14, 2005
7,063
Shucks, I knew the spec sheet said 0.8V but I wanted to be conservative.
I should have realized that someone on AAC will be looking over my shoulder and checking up on me.

Sorry, I'll stick to the facts next time.
Quote:
Originally Posted by Ron H
11.8kHz vs 31kHz is not just a component tolerance problem.

Geez Ron, somehow I suspected you will notice this.
Busted again!:D
 

MrChips

Joined Oct 2, 2009
30,621
If you look at the function table in the data sheet you will see STROBE labelled as G', i.e. G with a line over the G.
(Sorry, I don't know how to apply overstrike so I will use G').

This means that G' is an ACTIVE LOW signal.

The function table also shows that when G' is HIGH, the outputs are Y = L and W = H regardless on the SELECT INPUTS.

Y will follow the selected input only when G' is LOW.
 
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Thread Starter

atrumblood

Joined May 13, 2012
59
So if I understand the active low thing correctly with the strobe. When the strobe pin is HIGH the output of Y goes low it reverses the output. Do I have this right.

In other words 1011 becomes 0100 ? Or to put it another way it inverts the output?

Sorry if these questions are a little annoying. I have to put things into a way that I can make sense of them before I can say I have understood something.
 

Thread Starter

atrumblood

Joined May 13, 2012
59
Here is how I have the 74151 hooked up on my breadboard ( see attachment) to try and test out the functionality. I am having an issue getting any signal to come through. Unless I am completely wrong in my understanding.


Shouldn't the way I have that hooked up light up the LED? (ignore the fact that the schematic has no resistors that was intended for simplicity).

Thanks,
Atrumblood
 

Attachments

MrChips

Joined Oct 2, 2009
30,621
So if I understand the active low thing correctly with the strobe. When the strobe pin is HIGH the output of Y goes low it reverses the output. Do I have this right.

In other words 1011 becomes 0100 ? Or to put it another way it inverts the output?

Sorry if these questions are a little annoying. I have to put things into a way that I can make sense of them before I can say I have understood something.
Nope.

When G' is HIGH, the multiplexer is disabled. The outputs Y and W do not change.
Y = 0
W = 1
 

ErnieM

Joined Apr 24, 2011
8,377
I'm in trouble now. I've got both ErnieM and Ron_H checking up on me, keeping me straight :eek:
Tis a minor point, and while your levels are more conservative I was assuming the OP was intently reading the spec sheet and I was attempting to minimize his confusion.

All is good. :D
 

MrChips

Joined Oct 2, 2009
30,621
Tis a minor point, and while your levels are more conservative I was assuming the OP was intently reading the spec sheet and I was attempting to minimize his confusion.

All is good. :D
Whoa! When do beginners every read spec sheets? That would be so delightful!
 

Ron H

Joined Apr 14, 2005
7,063
Here is the pic you requested.

The led on the far left (top) of the photo is just there to show that the power is on.
1. It looks like you are using the wrong bus for ground, unless the buses are in parallel internal to the BB.
2. Why do you have a resistor in series with the power pin? This won't work. It needs to be a wire.

See attachment.
 

Attachments

Thread Starter

atrumblood

Joined May 13, 2012
59
Sorry about the late response.

So I made some changes to my bread board configuration. My power supply was not exactly 5 volts it was more like 12V! lol. Ya that was a face palm moment. So I hooked up an LM317 to my bread board and dropped the voltage down to 5v.

This is why I had the resistors.

I must have been very tired when I tried to throw this project together.

Any way, here is a pick of the new set up.
This time I have the ground connected to the right rail. :p.


The issue now is that the LED which is tied to pin 5(Y) stays on no matter what..

I have G tied low and A,B, and C tied low. So going by the function table (000) D0 should be allowing a signal through, but as I stated above the led will stay on whether D0 is HIGH or LOW.

Sorry to be a pain.

Atrumblood.
 

Attachments

MrChips

Joined Oct 2, 2009
30,621
What is the purpose of that potentiometer?
You cannot use a pot of that kind to drop power.
If you had 12V connected to your chip, the chip is probably dead.
 
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