connect 74LS47 to seven-segment display does not work?

[Shank747]

That was enlightening! I never would have guessed the theory of the variation of current based on the working segments. Thank you, Tom.

I'm trying to build this circuit on a breadboard, and i only have a 9V ordinary DC portable battery (no lab power supply as in the multisim simulation). I've tried to simulate the circuit using the 9V battery but it isn't displaying anything. Basically, i just replaced the VCC 5V with the +ve battery terminal and all ground lines now go to the -ve battery terminal. Simulation gives me an error, so i don't know what is wrong. I have attached the multisim file.

Also, i've been searching for a DIP resistor pack component to accomodate the 2 sets of similar resistors on the breadboard but i was unsuccessful in tracking down the serial number of the IC required for a pack of 7 or 8 resistors of 100 ohms and another pack of 8 resistors of 10k ohms. If such a thing exist, it will take much less space than the current 15 separate resistors. Google helped with some lab document about some 4.7 kohms DIP resistor pack having serial 472 but i think it's not a complete serial or maybe it's just a custom code for lab use only?

[Tom__Sedlack]

You need to have a ground connection for the simulator to use for a reference.  Just add one back and connect it to the 9V battery negative side.  There are a few places that sell components such as Digikey, Mouser, and Newark.  If you are going to buy components to build your circuit, I recommend getting a 5V regulator for use with your battery because the 9 volts will probably destroy the 5V components you are using in your circuit.

[Shank747]

The simulation works with one ground, as you said. Thank you for your help.

[Shank747]

Hi Tom

I am building the actual circuit using the ICs 7448, 74148 and 7404. Together with the exact same resistors as you prescribed; 9 x 10K and 7 x 100 ohms.

I have been working on the circuit since this morning (and it's 6 PM now) but i do not understand what is wrong. I dismantled and rebuilt it several times, even spreading the entire circuit over 2 breadboards. I am using 3 x 1.5V regular dry DC batteries to power the circuit with the -ve acting as ground and the +ve terminal as VCC.

Will any actual pictures help? I am getting 8 when all the switches are open. It should be zero. Maybe if i tabulate the actual results against the expected ones, it will help you to diagnose the problem?

Actual vs Expected results

All switches OFF: 8 - 0

Switch 1 ON: 9 - 1

Switch 2 ON: c - 2

Switch 3 ON: (segments g, c,d looks like the vertical reflection of letter c) - 3

Switch 4 ON: (segments b, g, f) - 4

Switch 5 ON: (looks like 5 without segment c) - 5

Switch 6 ON:  (looks like the small alphabet t)- 6

Switch 7 ON:  (no display)- 7

[Shank747]

I have attached pictures of the circuit using the components described in my post above. I hope that you'll be able to advise on what is wrong and any possible solution/s to make the actual circuit work just like in the simulation.

I've been struggling with this for so many hours, i'm not sure if the simulation "assumed" some values (high or grounded) for some pins or maybe different resistors which i should have used instead of the current 10K and 100 ohms ones?

I have attached a zip file containing some pictures of my work spread across 2 breadboards to make the circuit more understandable and less crammed.

[Tom__Sedlack]

It sounds to me that you did not ground the 'D' input on your 7448 LED Driver.  I'm not sure if I mentioned it here but I know I have in other threads.  The simulator will apply 0 volts to an un-terminated input.  In real life, an un-terminated input is (unless the spec sheet says otherwise) in an un-determined state.  In your case, I bet it is floating to a logic high which would give the display you describe. To verify, wire the 'D' input in your simulator to +VE and see what the display looks like.

Get in the habit of always terminating un-used inputs.  Either pull them high or ground them. Never let them float.

[Shank747]

Oh, wow! It's incredible! Tom, you are a genius! I've been ramming my head into the wall trying to figure out why the circuit wouldn't work. But your suggestion was absolutely head-on! It fixed the circuit and the LED now displays the expected value. I am now going to port this breadboard arrangement into a permanent setting onto a stripboard/Veroboard.

THANK YOU SO MUCH!

[Tom__Sedlack]

It's a nice way to start a weekend 

[Shank747]

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@Tom Sedlack wrote:

It's a nice way to start a weekend 

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I couldn't agree more.

But i have a question about the resistors used in the circuit (with IC 7448). Why choose the set of 7 x 100 ohms connected to the 7 segment? In several online notes, i've seen a set of 7 x 330 ohms resistors which are often used instead of the 100 ohms. Is there any rule of thumb or calculations behind the use of any specific resistances to protect the 7 segment from being overdriven by limiting the current? Is it based on the VCC voltage? This question also applies to the 8 x 10k ohms which are used across the inputs of the IC 74148. I'm just curious to know why some designs prefer 330 ohms instead of 100 ohms or other resistance values. Do you think i should use 330 ohms instead?

[Tom__Sedlack]

The resistor value depends on the supply voltage, the voltage drop across LED segment, and the desired operating current of the LED.  The more current, the brighter the display.  Get the spec sheet for the display you are using and find out what the typical voltage drop and current limits are.  Use Ohm's law to calculate the resistor value.

As for the 10K pull-ups, I'm not sure if there is a "formula" for using a certain value.  I use 10K since I have so many 10K resistors laying around.