can any one give me the scematic for this??

[thelma]

betwixt,its amazing to read your explanations.i'm also a person who is looking for the same thing wat irshad is doing except the controller because am not yet clear wat setup am asked for.
i would be really thankful if you could suggest based on the below articles which i'm bought to make a led to glow.
1)2 tap transformers
2)4 optocouplers(mct6)
3)CD4053(DIP),74HC14(SOIC)
4)CD4543B
5)ULN2003A

these are the above am asked to use to burn an common anode led 7 seg display.how should i proceed.schematic would be really gud along with the usage of the IC's which am using.please help me.

 

[betwixt]

Sorry I did not reply sooner Thelma, I have been away from home and not able to use the Internet.

You can drive a 7-segment LED directly from the CD4543 and some current limiting resistors to get a number to display. You would not need any of the other components to do this.

I assume this is a college assignment, is it necessary to use all the parts or can you just use some of them?

Brian.

 

[thelma]

no i 've to use all these and work on it. by d way if you are online can you clarify my doubt imiedietly??i need a response from you now so that i can proceed my question.

 

[betwixt]

They are a strange assortment of parts to make anything.

I am going to suggest the ULN2003 is used as a current booster for the seven segments of the LED. This will make the LED brighter. The CD4543 is a binary to 7-segment decoder so we can use that to provide the signals to drive the ULN2003.

The next part is more difficult. Not many schematics use the remaining components.
All I can think of is to make a binary to digit decoder with the option to use normal or inverted inputs and to opto-couple the input signals. I can't think of a practical use for it but it would use all the parts. Would that be OK?

I am in meetings most of the time so although I stay logged in to EDAboard, I can not be here to answer messages straight away.

Brian.

 

[thelma]

k thanks brian,first i taut of using optocouplers but now i made up ma mind to do this using decoder and uln2003. even if i use decoder limiting resistors are required ????if so wat is the range of resistor i can use??

 

[betwixt]

You MUST use current limiting resistors, one in each connection to the LED segments.
The value will be (VCC - Vled)/Iled where Vled is the forward voltage drop of the LED segment and Iled is the current you want to flow. Most LED displays use a current around 10mA but the voltage will depend on the number of LEDs that light the segment. Some 7-segment displays just use one LED, some use two or more so you will have to find out from the data sheet.

Brian.

 

[thelma]

can you guide me of how should i start with datashet? brian i'm very new to this kinda work.i've the habit of doing only codes.so far i've not assigned any designs on the board on my own.if i 've the schematic,i mount it else i dono wat current requirement i need and how much volage i should send?..i'm totally blank..how to read all these things and give me the basic process of how should i start??

Added after 9 minutes:

i went through the datasheet of uln2003a.i dint understand two parameters.its given like 7 drivers per package and input compatibility is cmos and ttl.wat is this mean?

 

[betwixt]

We all start somewhere.

You will find most of the information you need in data sheets, they are your friend and helper. I agree though that trying to understand them can be difficult at first. It's a bit like being given precise information in Hebrew and a book to translate Swahili to Russian to help you decode it :|. You know what you want is there but finding it is almost impossible.

You can start almost anywhere but I would suggest you begin by building something that actually gives you a display on the LED, then you can test it in stages as you add more parts to it.

If we start with the LED and ULN2003A, the LED will light up if you pass current through it, and the ULN2003A is a current amplifier. Unfortunately in this case, the '2003 is a very good amplifier and for a tiny current flowing into it's input pin, a large current will be allowed to flow at the out pin. If you just connected the LED directly to the '2003, the current would be so large that the LED would be damaged. That is what the resistors are for, they reduce the LED current to a level high enough for it to light up but not so high that it gets damaged.

Back to the data sheet.

Look at the figures for the LED, you will see several items like 'forward voltage' and maximum current. To find out the resistor values we need to take into account three things, the supply voltage you are using, the voltage across the LED and the current you want to flow though it.

You choose the supply voltage, most people use 5 Volts because that's what most logic and microprocessor ICs like to run from.

You choose the current, the data sheet will probably have a recommended current on it. If you can't see it, find the "absolute maximum rating" current (this is the current that will start to damage the LED) and use half that figure.

Now look for the forward voltage drop (Vf), this depends on the LED color and how many LEDs are physically present in the LED segment. Usually the figure is around 1.5V to 2.5V. This is the voltage that you could measure across the anode to cathode of one segment of the display, in other words the voltage dropped across the LED when current is passed through it.

With this information, you can calculate the resistor value. The resistor needs to drop the excess voltage between the LED voltage and the supply voltage at the current you chose. Using Ohms law R=V/I you can now substitute real numbers. V is what you want to drop and I is the current so R = (Supply voltage - LED forward voltage) / Current. For example, if your supply is 5V and the LED drops 2V and the current is 10mA, the resistor would be (5-2)/.01 = 300Ω.

You need to put one resistor in series with each LED cathode connection so you need 7 resistors. The supply voltage goes to the common anode pin of the LED and the end of the resistors NOT connected to the LED goes to the output pins of the ULN2003A. The only other connection you need at the moment is to wire the ground pin of the ULN2003 to your 0V (negative) side of your 5V supply.

At this point, if you switch on, nothing should happen but don't worry!!

Now using a wire, connect one end to the 5V supply and touch the pther end to the input pins of the ULN2003A. As you connect the input pin, the output pin should conduct current to ground and make one segment of the LED light up.

You might have noticed that you could have done this without the ULN2003 at all. The reason it is used will answer the second part of your question. The LED is conducting a relatively high current, 10mA or whatever you chose earlier but CMOS and TTL IC's can't supply enough current to drive the LED directly. The '2003 is acting as an amplifier to boost the current.

TTL and CMOS are two different methods of making ICs and although they do the same job, the way they connect to other circuitry is quite different. The ULN2003A is designed so it will work equally well with both types, that's why it says it is 'compatible'.

TTL = "Transistor-Transistor Logic" uses normal transistors inside the IC. It allows fast operation but uses a lot of power to operate. They were used in almost all electronic circuits for many years but are now losing popularity as new technology takes over.

CMOS = "Complimentary Metal Oxide Semiconductor", uses FET transistors and uses far less power than TTL but has the drawback of not being able to provide as much power from its output pins. For example, the CD4543 is CMOS and it uses almost no power to operate it's decoding circuits but if you used it to directly drive the LED it would be very dim and difficult to see.

Are you with me so far?

Brian.

 

[thelma]

brian i made up my mind to use cd4543 and am planning to use the current limitting resistors but in the datasheet i found a pin named latch disable???should i ve to keep that high or low??

 

[thelma]

if i mount a ic on gpb ,wat are all the parameters should i check??

 

[Audioguru]

brian i made up my mind to use cd4543 and am planning to use the current limitting resistors but in the datasheet i found a pin named latch disable???should i ve to keep that high or low??

You should not build a circuit without having the datasheets handy.
The datasheet tells you what each pin does and how to do it.
I get datasheets from www.datasheetarchive.com .

 

[thelma]

brian how to measure current output of the ic once the supply is given to ic?

Added after 6 minutes:

audioguru thanks soo much. i'll be really thankful,if yu can answer my silly doubts.actually we 've to use current limittin resistors. here i've lot of confusion .as yu said i took the printouts of the data sheet handy.now the formuls fr resistor finding is R=(supply voltage- voltage drop accross led)/I(led).

now in the datasheet of led i've drop down voltage as 2.5 and my supply voltage is 5v and hence the current ,wat should i use? should i ve to use the max current that led can withstand that is given in the led datasheet as 30 ma or should i ve to use other value. how should i find the current flow in led??

 

[betwixt]

Hi Thelma.

The IC (ULN2003A) doesn't have a supply pin so it can not supply current to the LED. All that is inside the IC is seven identical transistor networks that can SINK current down from the output pin to the ground pin. You have to provide the current from somewhere else, in your case, the common anode pin on the LED. What the IC does is complete the circuit so current can flow through the cathode (segment connection) of the LED to ground, passing through the resistor as it does so.

You might like to think of the ULN2003 as being seven switches with one side of them all being connected to the ground pin and each switch being turned on or off by a small current going in at the input pins. That is a bit simpler than it really is but it gives the general idea.

It is up to you to decide how much current should flow through the LED. 30mA is probably the 'absolute maximum' that it is rated at so you want to use less than that or it might be damaged. On the other hand, if you decide a current too low, the LED will not be bright enough. I would suggest you use 10mA, that should make plenty of light and is well inside the safe limit. So your resistors would be (5 - 2.5)/0.01 Ohms or 250 Ohms. That is not a common value you might find in a market but if you use the common values each side of that, 220Ω or 270Ω the difference is not going to be noticed.

When you have the LED working, we will look at the CD4543 circuit.

Brian.

 

[Audioguru]

But the ULN2003 uses darlington transistors that have a saturation voltage loss of about 1V at 10mA.
Then if you use 220 ohm current-limiting resistors with a 5V supply and 2.5V LEDs, the current in each LED is (5V - 1V - 2.5V)/220 ohms= 6.8mA which will be dim. Use 68 ohms for a current of 22mA.

 

[Audioguru]

Thelma, are you using the CD4543 to drive the LEDs through current-limiting resistors?
The datasheet from Texas Instruments has a graph that shows the typical output current with a 5V supply is only 3mA into an LED without a current-limiting resistor. A resistor will reduce this small current. 3mA is dim.

 

[betwixt]

Thelma is tasked with finding the best use for an almost random selection of components handed to them. One of the parts is a ULN2003A and as I can see no other use for it, I have been advising it is used for driving the 7-segment LED.

I am advising that the CD4543 is used to drive the ULN2003A. the other parts are a weird mix of multiplexer, hex inverter, transformers and opto-couplers so finding a way of using them all to do something meaningful is a challenge!

Audioguru the 1 V drop in the ULN2003A is noted - I'm trying to keep it simple for Thelma but you are quite right. My house lights are LED with ULN2803A drivers to them all!

Brian.

 

[thelma]

hey brian, i think i got a clear view. am using cd4543 now. first i gave 5 v supply and 150 ohms as ma limiting resistors but the current passed was only 3 mA. hence now i'm going to give 5v supply and place 1 k ohm resistor,so tat i get 9-12 ma current which can easily drive ma ic. will this wrk??

Added after 10 minutes:

hey brain, now i got clear view. as u said i'm using cd4543.now in teh graph,they had shown for 5v there is 2.5 ma of current.is tat with using limitting resistor?? because already the current output is low ,then y shud v use resistor,if am using cd4543. just clarify this please??