First Electrical Engineering Lab, Need Help!

Discussion in 'Homework Help' started by Hunter Neumann, Sep 3, 2015.

1. Hunter Neumann Thread Starter Member

Aug 24, 2015
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3
Today in my first electrical engineering lab we used the chips 7404, 7408, and 7432 to perform the equation

2. GopherT AAC Fanatic!

Nov 23, 2012
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Welcome, Hunter. You should ask the Moderator in duty to move your question to the Homework section. People there will guide you through - but you need to show some attempt to get thud solved (wrong or right, get started).

3. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
3

How do I do that? I have no problem with a moderator moving it, I thought this was the best spot for the question. As far as showing an attempt to solve the equation I will be uploading the truth tables and circuit diagram of the equation that I have completed later tonight or this weekend. I can easily do all of it on paper but I didn't get to learn much about the hands on stuff such as using the wires to connect from pin to pin and gate to gate to solve the equation. Thanks for the response

4. GopherT AAC Fanatic!

Nov 23, 2012
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Use the REPORT button on any post. I already did it for you.

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5. Papabravo Expert

Feb 24, 2006
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In the algebra of real numbers you learned how to break down a complicated expression by by starting with the expressions inside of parentheses. You also may remember that multiplication should be done before addition. In Boolean algebra we have only the two elements 0 and 1, and the operations of AND (similar to multiplication) and OR (similar to addition). You also have NOT or inversion which has no analog in the algebra of real numbers. If you count the operations in your expression, you have two AND operations and two OR operations and one NOT operation. Each operation has two inputs and the physical gates each have two inputs. In the three chips you mentioned there are six inverters, four 2-input AND gates, and four 2-input OR gates. You should be able to implement this function with gates to spare in each of the packages. I look forward to your solution.

6. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
3
these are the pictures of the lab that we did

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7. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
3
I posted the lab above

8. DerStrom8 Well-Known Member

Feb 20, 2011
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Hello Hunter, welcome to AAC!

Just a couple of notes:

X + Y : This means X OR Y. In other words, X is put on one input pin of an OR gate, and Y is put on the other input. X + Y is on the output.
XY: This means X AND Y. In other words, X is put on one input of an AND gate, and Y is put on the other input. XY is on the output
X': This means NOT X. In other words, X is put on the input of a NOT gate. X' is on the output.

Hope this helps a bit!
Matt

9. Papabravo Expert

Feb 24, 2006
10,340
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I didn't see the explicit use of an inverter in your diagram. Each of the symbols looked like a 2-input gate which must be wrong. A freehand drawing is just about the worst way to make a logic schematic. You need to use a drawing program or at least get a template for the hand drawn schematics, and make them large and well organized so it is possible to follow the logic.

Edit: I went back and looked again at your drawing and found the inverter.

Sugggetion:
LTSpiceIV is a free download from Linear Technology. Besides using it as a simulator, it has an adequate schematic drawing package that can be used for quick schematic interchange.

Last edited: Sep 4, 2015
10. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
3
Due to my professor I had no choice but to handwrite this lab, I can see how a simulator would really help out. Im downloading it right now to check it out

11. shteii01 AAC Fanatic!

Feb 19, 2010
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Yeah, your writing will need a lot of work. Don't be surprised if your early lab reports will have a lot of red.

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12. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
3
these aren't due until next Wednesday, do you have any suggestions on what I could change?

13. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
3
met to reply, see post above^^

14. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
3
okay so this is a diagram on ICircuit of just the basics so far, I'm going to attempt to diagram the rest of the equation this weekend. Just looking on some feedback of what I have so far and how to put the led and switches using Icircuit. Thanks for your time. I also ordered wires, 7400, 7402, 7404, 7408, and 7432 chips to try on my own CADET system I ordered earlier in the week. Any certain type of wire I should have gotten? I just got a jumper wire cable kit for solderless breadboard. All feedback is appreciated

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15. MrAl Distinguished Member

Jun 17, 2014
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Hello there,

Perhaps the only thing missing from your understanding of going from equation to circuit is the concept of what is known as a *cascade* of devices.

When we cascade two devices that means we connect the output of the first device to the input of the second device. The input of the first device then becomes the main input, and the output of the second device then becomes the main output. So we end up with one input and one output, but we've connected two devices in cascade to achieve this result.

You seem to already know that the addition sign in boolean logic means "OR" and the multiplication sign (or no sign at all) means "AND". The single apostrophe signifies logical inversion, the "NOT" function.

With all this in mind, lets consider a simple expression like D=(A+B)*C.
Here we have two logical operators, OR and AND. We first OR the A and B inputs, then AND that result with the C input. That gives us an output we are calling D.
The cascade then is made from the OR gate followed by the AND gate. The output from the OR gate goes to one input of the AND gate (cascade) and the other input C goes to the other input of the AND gate. Thus we need a two input OR gate and a two input AND gate.

Inversion is almost the same. Say we have D=(A+B)'*C.
We still need the OR gate, but because now the output of the OR gate is inverted, we need to place an inverter at the output so that the output of the OR gate goes to the input of the NOT gate (inverter). The output is then inverted logically, and then that new output from the NOT gate goes to the input of the AND gate, and then again the third input C goes to the other input of the AND gate.

That was an example of a direct implementation, although we could apply some boolean logic first before we choose the gates. This helps reduce the complexity sometimes.

To try to reduce that last expression D=(A+B)'*C we could try applying the inversion first:
D=A'*B'*C

and now we have a single AND gate with inputs A', B', and C. We would need two inverters for A and B inputs unless we already had the inverted signals available in the system already. If we did have them then all we would need is a single three input AND gate.

You could try a couple simpler examples first before doing your assignment and we could check them here. For example, D=A*B+C. Can you find the gate arrangement for this expression?

16. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
3
I will try to diagram the equation you gave me tomorrow, thanks for your response. I tried to clean up my diagram of the equation we used in the lab by using Icircuit but I know its not right. Im not sure what to use for the LED or the variable switches or if I connected them wrong. Im new to the app so I'm not quite sure how it all works. I just used the analog switch and labeled them as the variables. I also used the lamp as my LED and tried to connect everything together to get the right output. Would love some feedback.

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17. GopherT AAC Fanatic!

Nov 23, 2012
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Hunter, your drawings are kind of a mess.

Also, your X, Y and Z are not set up correctly.

Here is an attempt to help you walk before you run. A single AND gate set up with two switches. Note, the way the switches are set up. When open (not connected) they are connected only to ground (through a 10k resistor), (a logic 0). When the switch is closed, the point is connected to +5V (logic 1).

I have all four combinations of switches. Note, the simulator will light the LED when the gate's output is high.

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18. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
3
I guess I am getting ahead of myself trying to implement my lab equation onto ICircuit. I have modified it to look like this(see picture below) but I am going to save it at that until after I get comfortable playing around with each gate and smaller equations. Thanks for the feedback.

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19. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
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Thanks for your reply, here is my Icircuit diagram of what I think I should start out with for the equation you gave me. Looking forward to your response.

Edit: I now know (or think I do anyway) to change the switches to spst

• Screen Shot 2015-09-05 at 2.05.54 AM.png
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Last edited: Sep 5, 2015
20. Hunter Neumann Thread Starter Member

Aug 24, 2015
53
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First chip I think I got to work on the simulator! good way to end the night. Ill be doing more tomorrow with the other gates/chips (I know this is ridiculously basic but at least its something)

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