Hello,
Here is Display with 12V directly 120Ohm series taking 20mA each segment!
9.5V

 

The 74HC/HCT4511 work in 5V and i will be using above it with uln2803.
The ULn2803 will convert the input 5V to GND
http://thetechnicalbrain.com/electronics/images_stepper/uln2803.jpg

If your inputs is the scan signal then you can use uln2003 or uln2803, 2003 is less one channel than 2803, cd4511 is right.

 

The 4511 takes 0000 and give 1111110 //0
and the ULN2003/8 will convert 1(high input) to 0(low output) and 0(low input will be give what) will convert x??

 

The 4511 takes 0000 and give 1111110 //0
and the ULN2003/8 will convert 1(high input) to 0(low output) and 0(low input will be give what) will convert x??

If the CD4511 is not using scan signal to drive then you don't need the uln2003/ul2803, only you want to using scan signal and used the common cathode led then you will need the uln2003/2803 to drive the leds, the uln2003/2803 is a Darlington drivers, so you can treat them as a inverter has big current, when the input is high(1) then the output will be low(0).

The MC14543 and MC14553 have internal scan signal, so they just need less parts.

 

Hello,
Just now i used CD4511 with 2 series 120ohms to drive single led at output.
the vcc was 12V, voltage across 120+120 was 8 V the leds 2V but how the CDVcc and gndd was showing 2V not 12V supply??
should i connect the series resistance indival at all segment pin??

If the CD4511 is not using scan signal to drive then you don't need the uln2003/ul2803,

I am using decoder individual no scanning just driving!

 

12V → CD4511 → 620Ω → (+)LED(-)(2V/16mA) → GND
R= (12V[4511Vout] - 2V[led])/16mA[80% of 20mA]
R= (12V - 2V)/16mA = 625Ω, choosing 620Ω

For the using life and light fading, so it just use 80% of rating current, I = 20mA*80% = 16mA.

If you don't need the dot then you just using 7 resistors in series with 7 segments leds for each digit.

 

OK, I have connected the CD4511 with ULN2003.
The pin 8 E to GND and pin 9 COM to Vcc 12V.
and CD4511 120Ohm in between of ULn2003 is this correct, or should the output segment pin drive the ULN2003 with series resistance!

 

Hello,
If CD4511 is at 12V and uln2003 is connected via 120Ohm then what is the ULN2003 driving current?
The min input ULN2003 required is 1.5mA...

 

Hello,
please tell suitable circuit of CD4511 and ULN2003

 

First, the base transistor between the 4511 outputs and the ULN2003 inputs depends on the current requirement of your 7 segment.

A quick calculation shows that with a 120 ohm resistor, the transistor is being driven with 100ma. Certainly that is too large. (Current=12V / 120 ohm = 0.1A or 100mA)

Using the rule of thumb of one tenth the load current, a 7.5k ohm resistor will give you 16mA to drive your ULN200x, which drives your 7 segment device.

 

First, the base transistor between the 4511 outputs and the ULN2003 inputs depends on the current requirement of your 7 segment.

A quick calculation shows that with a 120 ohm resistor, the transistor is being driven with 100ma. Certainly that is too large. (Current=12V / 120 ohm = 0.1A or 100mA)

Using the rule of thumb of one tenth the load current, a 7.5k ohm resistor will give you 16mA to drive your 7 segment device.

a 7.5k ohm resistor will give you 1.6mA to drive your 7 segment device.

 

OK, I have connected the CD4511 with ULN2003.
The pin 8 E to GND and pin 9 COM to Vcc 12V.
and CD4511 120Ohm in between of ULn2003 is this correct, or should the output segment pin drive the ULN2003 with series resistance!

The common pin was used for the inductive components as relay or motor to protect the bjt from back emf, so you can ignore it here.

When the uln2803 get into the saturation status then the Vce will be about 1V, it is not as one bjt Vce=0.2V.
12V → CD4511 → 560Ω → (+)LED(-)(2V/16mA) → Vce(uln2803) 1V → GND

 

a 7.5k ohm resistor will give you 1.6mA to drive your 7 segment device.

The 7.5k ohm resistor is a base resistor feeding a transistor in the ULN2003 array. The transistor then switches ~ 16ma to the LED. I'm on my phone; this thread is screaming for a schematic!

 

Hello,
I have just connected the C4511 with 12V, an the common cathode led segments.
4511 is working normally the oval led segment also bright in few mA.
I have kept the circuit on power for whole night chip was not fry .
so, should i use directly 4511 with common cathode ?

as ul2003 is not function at 12V.

 

Hello,
I have just connected the C4511 with 12V, an the common cathode led segments.
4511 is working normally the oval led segment also bright in few mA.
I have kept the circuit on power for whole night chip was not fry .
so, should i use directly 4511 with common cathode ?

as ul2003 is not function at 12V.

Normally, I used 4511 and the 7-seg led with common cathode, and connected the common cathode to ground.

 

Hello,
Is is fine??

 

That is depends on the current of your 7-seg led, if your feel the brightness is enough then it's ok.

 

One thing more if PLC operates at 24V. at what value it will give logic signal 5V or 12V or 24V
because 4511 is operating at 12V, an pull up with 10K.
so, what logic input is require for working on it??

 

If you want to send a 4 digits signal to CD4511 from PLC as 24V level then you can using two resistors 12K and 10K to be as a voltage divider, the 12K is connecting to 24V and 10K is connecting to ground, the common pin will be the output signal then it will sending about 10.9V to cd4511.
Vin = 24V
R1 = 12K
R2 = 10K
Vo = 24V*(10K/(12K+10K)) = 24V*0.455 = 10.92V.
10.9V is enough to give a high level to the CD4511.

 

OK, that mean 10k pull down to all input!
an input must be 10k or 12K to plc output.
one thing more does plc give 24V for input output?
an if we are operating 4511 at 12V an 5v as input will this not work??