7 segment decoder

[koolguy]

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??

 

[Ian Rogers]

Why didn't you use the constant current device... The ULN2003 sinks current when you switch on the darlington's inside they will provide a negative output..

If you use the TLC5925 ( or equivelent ) you can serially load 16 segments and set the current so all the LED's in a string get the right current...

 

[koolguy]

Anyway, rightnow i am using uln2003 the pin 8 E is at GND and COM is 12V vcc

 

[koolguy]

The TLC5925
work at 3.3-V to 5-V Supply Voltage
ok ok it is sink...!!
but 16-bit shift register

 

[koolguy]

Why didn't you use the constant current device... The ULN2003 sinks current when you switch on the darlington's inside they will provide a negative output..

Yes i have connected all anode to Vcc 12V with 120ohms.

 

[koolguy]

Hello,
please tell suitable circuit of CD4511 and ULN2003

 

[KeepItSimpleStupid]

I need to look at the datasheets again, but anyway I was thinking of a high e.g. https://www.google.com/url?sa=t&rct...=utFr6btCnp4FmvCafpetXw&bvm=bv.91427555,d.cWc and low side driver, but you may have to convert the output to a logic level.

The 2803 adds another 1.2 volts. The internal high side driver probably has a voltage drop too. You could simulate the worst case by using 3 diodes in series with your string.

You seem to be on the hairy edge of everything.

I'm thinking, probably in terms of a multiplexed display outside of the normal limits where you would need a high side driver for the segments and a low side driver for the digits.

In that case, you would have to account for a possible voltage drop for the segment driver and the voltage drop for the digit driver. If you use a darlington like the ULN parts, that would be another 1.2 V to account for.

If you break the segments into two, then you also have 2x the current to deal with.

Can you do the following. Use a 12 V supply on one segment with 3 diodes like the 1n4001 in series and see if it works. You might have to change the series resistor.
Measure the voltage across the series resistor and translate that to a current so, we can see if everything is in range. For fun, you can also measure the voltage drop across each LED to get an idea of the maximum and average drop.

The design may be too close to the limits. If my memory serves me, so parts have a 20 mA limit and some have 40.

Meanwhile I'll look at the datasheets.

 

[ChrisP58]

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 you are driving each LED string directly, you need individual current limiting resistors for each string of LEDs.

 

[KeepItSimpleStupid]

Note: The ULN chip does not have a power pin. It just needs ground to operate.

The COM pin just connects the reversed biased diodes to +12. The COM connection is needed for RELAYS ONLY.

8+5 is 12, so about 2.4 for the led and 4*0.6 for the chip. A little higher than expected.

 

[KeepItSimpleStupid]

The CD vcc pin to ground should be 12. For the CD chip ALL unused inputs must not float.

 

[KeepItSimpleStupid]

I looked at the 4511 and it looks like you have 12-0.6V and 20 mA if you go direct. You can go to 15 V at 20 mA/

If you go the ULN200x route, you have V(display)-1.2 available to work with and I think 100 mA.

So, questions are, do you have other voltages available like +5, +12, +15 or +24 with 12 through 24V available to drive the LED's.

You can use the ULN2003 if you have 5V available otherwise you should use the ULN2004 if you want to operate the CD4511 at 12V.

You will lose 1.2V in any event when using the ULN2x00x drivers. You can simulate that with two 1n400x diodes in series.

Summary

12-15 V logic ULN2004
5V logic ULN2003
You can do 100 mA without much effort.

LED supply can be nearly anything.

5 LEDS with a drop of 2.4 V each and another drop of 1.2V = =5*2.4+1.2 = 13.2 SUGGESTS that a 12 V design won't work.

You can out the driver on the display board and break the single 5 LED segment into multiple ones. One 3 LED segment and one 2 LED segment would work just fine on 12 V. Connect two of the UL inputs together and use 2 resistors of different values for each string.

You probably have 5V available. You may even have 24 V available since you mentioned PLC. If you have 24 V availabel, then you can use a 5 LED string.

So, what supplies do you have available?
Can you mount the ULN chip in the LED box?
The COOL thing about the ULN chip is with the inputs:
OPEN = OFF
LOGIC high = ON
LOGIC low = OFF

With any CD device, all inputs MUST NOT be allowed to FLOAT!

 

[koolguy]

Hello,
The 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.
But why ?
You will lose 1.2V in any event when using the ULN2x00x drivers.

 

[koolguy]

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??

 

[koolguy]

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??

 

[koolguy]

http://www.thelearningpit.com/lp/doc/7seg/7seg.html

 

[KeepItSimpleStupid]

What I'd like you to do is operate the 4511 on +5 like the schematic above suggests, but loose the resistors and the 7-segment display part of the schematic.

Connect the a-g outputs to the inputs of the ULN2003.

IF you were operating one LED per segment, connect the +side of the LED to a resistor and to 24 VDC. Connect the - side of the LED to the corresponding output port on the ULN2003. pin 8 of of the ULN2003 connects to ground. Pin 9 can be left open.

the resistor would be R=(24-1.2-n*2.4)/10E-3;n is the number of LEDs. 10e-3 is 10 mA or whatever you want to operate at.

So, something around 2K for 1 LED for each of the 7 segments.

For your 5 LED string, R would be around 1K for each of the 7 strings.

 

[koolguy]

The Display is at 12V from transformer 0-9V at AC then DC at 12V (9x1.45) peak.
Vin = 24V
R1 = 10K
R2 = 4.7K
Vo = 24V*(4.7K/(10K+4.7K)) = 8V.
Will 8V is enough??
C4511 is at 12V from Transformer external source not from PLC!

and Both PLC and led should be gnd for reference??

 

[KeepItSimpleStupid]

if your going to power the ULN2004 with a source greater than 6V, you need to use a ULN2004.

The voltage divider for 8V is not likely to work, especially with a ULN2003.

So, you made an unregulated 12 V supply. How much capacitance?

Well, what I said was to try with a 5V supply. A 7805 should work because the 4511 would not draw much even with 24 V as input.

yes the commons have to be shared.

Because you mentioned a TTL part, I though you might have 5V available.

 

[koolguy]

Hello,
What are you saying man, I am not using ULN2003 or other.
CD4511 directly to 12V and input and pull down resistance of 10K both!

 

[Ian Rogers]

Hello,
What are you saying man, I am not using ULN2003 or other.
CD4511 directly to 12V and input and pull down resistance of 10K both!

So the LED string isn't a string but a simple 7 segment display??

Your first post implies you are using multiple LED's... Are you trying to put a string of LED's directly to the 4511?? If so how much current can the 4511 sink? This is why people use drivers so they can "Up the current" so to speak...