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.
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 :?
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.
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.
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.
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.
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.