Question on connecting 7-Segment LED

Ok so my group has this big project that we're doing and something went wrong with the 7-Segment LED circuit.We tried to display a number 4 to it,be connecting all the required pins directly to our voltage source.(Eg, pins representing b,c,g and f).

As our project involes connected circuits with a PLC,all of our circuits will be using the 24V DC power supply.So we constructed the circuit,obviously using resisters for all pins of the 7-segment display.We connected the circuit and it didn't light up.We went to google for 7-segment display circuits and found that alot of them used counters.Is there a way to make ours work without using a counter?We just want to show a number 4 when the current passes through the circuit.

Another problem we faced was that the power supply had no ground.So what should we use as a ground?

We also want to know what values of resistors we should use.We're currently using 2.7k ohm resisters but we're not sure if its okay.

Thanks for the help!

Post a schematic.

What is the model of the 7-segment display? Is it anode-common or cathode-common?

What do you mean there is no ground? You can't power the circuit if you don't have a ground or common (return path) for the current.

To save you any extra heartache, I would use 1.5k resistors. You will have 24-1.7V across the resistor and the LEDs need about 15mA through them to give them half decent brightness. Start with 1.5k resistors and this is one less thing to stop you in your tracks

We are using common cathode displays.

there is only +24V and -24V from our power supply. we are not sure as to where our ground is but we have been told that there is no ground for our power supply. so we were thinking if it is possible to connect the -24V to the ground, using the -24V as a ground.

@simonbramble Oh ok thanks alot

mngeow said:

there is only +24V and -24V from our power supply. we are not sure as to where our ground is but we have been told that there is no ground for our power supply. so we were thinking if it is possible to connect the -24V to the ground, using the -24V as a ground.

Click to expand...

Simple arithmetic should tell you the answer, which is a definite "no": 24 - (-24) = 48.

Sounds like you need to find a single-ended (V+ and ground) power supply to power your circuit.

Hey, watch the markings. This https://www.electro-tech-online.com/custompdfs/2011/08/mdr10.pdf supply is labeled +V and -V and it isn't a dual supply.

I actually suspect yours isn't either. it's just marked in an unexpected way.

Read it as the minus terminal of the 24 VDC supply.

[Voltmeter to the rescue]

So what would one do if all one had was a 24-volt power supply and needed to power some chips (counters, etc.) that run on, say, 15 volts max? Linear regulator (78xx?) DC-DC converter?

The choice of linear vs. switching and isolated vs. non-isolated would come from other factors. Linear vs. switching would primarily come from how much power to I need? How much will be wasted?

mngeow said:

We are using common cathode displays.

there is only +24V and -24V from our power supply. we are not sure as to where our ground is but we have been told that there is no ground for our power supply. so we were thinking if it is possible to connect the -24V to the ground, using the -24V as a ground.

Click to expand...

As KISS noted, the -24V connection likely can be used as the common. Use a voltmeter to determine this.

So if you connect the display common cathode to the power supply common and one of the display resistors to the +24V, the display segment should light.

Back to basics. Def: Common Cathode
Vf (forward voltage drop) It's different for different colors. At 24V we can probably neglect. In the range of 1.2 to 3.2 Volts.
If (Forward current) In the range of 1 to 50 mA for displays and probably single LED's too.

It's best to put a series resistor for EACH segment. Rs = (V-Vf-Vother)/If. Vother can be Vce(sat) for instance.

Common Cathode means that that the segments common connection is the (-) of the supply. Assuming the supply has (+) and (-) terminals.

If you can get a single LED to light, your half way there. The cathode of a satandard T 1 3/4 LED is the side with the flat. Put a resistor to +24, the LED, the Flat portion to (-) and the LED should light.

Now extrapolate that to the display case.

what is the PLC you user (or more precisely what is the output card you plan on using). are outputs PNP or NPN or Relay?
how fast you want to switch the display? relays are slow and not meant for frequent operation. also there is minimum current requirement to maintain contacts in good order.
small PLCs have only few outputs so this is not economical. this makes most sense if you have high density transistor output cards (like AB 17xx-OB32 etc.).
you would need individual resistor per segment (2k2 0.25W for example).

if the PLC outputs are PNP (like in 17xx-OB32 etc. common in western PLCs) then you need 7seg display that has common cathode.
if the PLC outputs are NPN (common with Japaneese PLCs) then you need display with common anode.
if you plan on using multiplexing, you would need both PNP and NPN ouputs but limitation is going to be scan time. most PLCs have scan time in order of 10-50ms, fast ones get down to 1ms range.

alternatively, use assembled displays and drive them using BCD lines or through serial port for example (right now i am using https://www.futurlec.com/4x7_Segment.shtml)