Help with a project I'm doing!

[MilanoChris]

Hi guys, I'm new here so if this is in the wrong section then apologies in advance.

I'm currently doing CSL as part of my HNC in elec engineering. We've got a project to do and I've gotten so far but I'm a bit stuck on the main part and some advice/hints would be greatly appreciated.

Basically I'm to build a logic circuit using Circuitmaker 2000 using the following:

Encoders; Decoders; Buffers/Drivers; Mux's or Discrete Gates; Inverters and common anode 7 segment displays.

The circuit is to work out and display the value of Y when X is any whole number between 0 and 7. The equation being Y=X(X) + X - 2. The circuit is also to display the binary input (0-7) using three 24V lamps.

It needs to be done using 3 displays, one showing minus if applicable then blank for the rest, whist the other 2 display tens and units.

I've got the first part done and I'm a bit confused as how to do the second part. I know I need to use multiplexers but I can't get my head round where to start off.

I can upload what I've done so you can where I am at the moment if need be.

Cheers,

Chris.

 

[MilanoChris]

Anyone please?

 

[MilanoChris]

OK I've fot the tens and the units counter working, but how do I get the minus display when the input logic is zero?

I'm all out of ideas!

**broken link removed**

 

[ljcox]

I don'r understand your equation. What is X(X).

Do you mean X multipled by X? Or X as a function of X?

 

[ljcox]

I've got the first part done and I'm a bit confused as how to do the second part.

In addition to posting your circuit, it would help me if you posted a block diagram.

I don't have the time to analyse your circuit. It is obviously based around 3 MUXs but what part of the eqaution are they calculating?

 

[ljcox]

Well I could not help myself, I did analyse the circuit and drew the attached block diagram and truth table.

Please check these for errors.

I cannot see what it is doing.

I assume the equation is a quadratic. ie. Y = X^2 + X -2

Is this correct?

 

[MilanoChris]

I don'r understand your equation. What is X(X).

Do you mean X multipled by X? Or X as a function of X?

X squared. When Y (the input) is = to 0 then the output display needs to show - 0 2.

As you can see I've got the 0 and 2 showing but I need to add another 7 segmen Common Anode display and get it so that whan Y = 0 it shows -. Ie only segment G is displayed and only in that condition.

It needs to be blanked for the rest. I have no Idea that I need from a third 7447 in order to drive it that way :?

 

[MilanoChris]

Just seen your block diagram. That seems to be correct. Please bear with me as I've not really done any digital circuits before.

 

[ljcox]

OK I've fot the tens and the units counter working, but how do I get the minus display when the input logic is zero?

I'm all out of ideas!

There are no counters in the posted circuit?

To set the - sign when X = 0 is easy.
Use a gate to detect C = B = A = 0 and turn on segment f (I think from memory) of the sign display.
You won't need a display decoder/driver if - is the only symbol to be displayed by this display.

 

[MilanoChris]

But surely in a practical application then the 7 segment would just draw too much power? Sorry for using the word counters - should have been display!


My aim was to inroduce another 7447 and Mux to drive a third 7 segment display.

 

[ljcox]

Just seen your block diagram. That seems to be correct. Please bear with me as I've not really done any digital circuits before.

No problem.

I now understand what you a doing, but my truth table does not agree with your equation.

You need: -2, 0, 4, 10, 18, 28, 40 & 54.

But my table shows: 2, 0, 0, 14, 18, 28, 44 & 50.

So is my table wrong or have you made an error or 2?

 

[ljcox]

But surely in a practical application then the 7 segment would just draw too much power? Sorry for using the word counters - should have been display!


My aim was to inroduce another 7447 and Mux to drive a third 7 segment display.

You need to study the data sheet of the gate you intend to use. If it can sink enough current, then no driver will be necessary. If not, then you will need either a transistor to drive the g segment or an IC buffer that can sink the current.

Using another 7447 is one solution, but is an overkill.

 

[MilanoChris]

Just seen your block diagram. That seems to be correct. Please bear with me as I've not really done any digital circuits before.

No problem.

I now understand what you a doing, but my truth table does not agree with your equation.

You need: -2, 0, 4, 10, 18, 28, 40 & 54.

But my table shows: 2, 0, 0, 14, 18, 28, 44 & 50.

So is my table wrong or have you made an error or 2?

Sorry, think I've had a few too many beers tonight :lol:

As you state, I need -2, 0, 4, 10 etc.

The thing I don't understand is how do I get a 7447 to put pin G low only when Y=0? It needs to be blanked for the rest of the time.

 

[ljcox]

The simplest and most elegant way to control the - display would be to use a 4 input NOR gate.

This will give you a high when Y = 0.

You could then use the other half of the IC as an inverter and, assuming it can sink enough current, drive the g seg directly.

If not, connect output of the NOR to the base an NPN transistor (via a resistor) and use it to drive the g segment.

If you want to use the 4774, then connect the output of the NOR to A0 and connect the b or c output of the 4774 to the g segment.

This means the 4774 is simply being used as a driver - which is not very elegant.

I'm about to go out for a few hours, so I'll tune in again when I return.

 

[ljcox]

Here is an alternative using NAND & NOR gates.

You don't need a third display for the - sign, it can be done as in the attachment,
ie. when the input is 0, blank display 1 and turn on segment g with a transistor.
I don't have a data sheet for the 74LS47, so I don't know whether the RBI input is active high or active low.
So I've indicated both possibilities on the "final.gif" attachment.

I suggest you check my logic as I may have missed something.

There is a minor error in "final.gif". * Display 1 RBI should be * U9 RBI.

 

[MilanoChris]

Thankd for you input so far!

I'm nto allowed to use any transistors unfortunatly.

 

[ljcox]

I have thought of an even simpler solution. You don't need U1.

I'll draw it up and post it in few minutes.

As for the transistor, then you will have to use another 74LS47 and only use one input and one output. If you use the A3 input, it will output 0 if A3 is low or 8 when it is high. So if you connect the g output to g of Disp 1, and blank display 1 when the input is 0 (ie. I0 = 1) then Disp 1 will show a - sign at 0.

 

[ljcox]

Here is the alternative to my previous.

The only disadvantage is that switch I1 is not connected, so the display will show 00 if all switches are off and regardless of the setting of I1.

Edit: This disadvantage is common to your original circuit and both of mine. However, it can be solved as in the next post.

I can't see any easy way to correct this, it would require extra gating.

But does it matter anyway?

Note that switch I0 is wired opposite to the others, this could be corrected (if necessary) by inserting the unsued gate (IC4c) as an inverter.

 

[ljcox]

This is a method to blank the displays when all switches are off.

It needs to be combined with my previous circuit.

Note that the 74LS48 has internal pull up resistors so you don't need to include internal ones. I have not shown pull up resistors in my drawings.

 

[MilanoChris]

I see the logig in your circuit but I definatly need to have the minus one shown on one seperate way. I'll take a look at it tomorrow and post up what I've done.