1 Second Clock with 5" LED Displays Circuit Evaluation

Thread Starter

CoachKalk

Joined Sep 20, 2011
139
Latest circuit drawing per SgtWookies solution to 24V supply for 5" LED displays causing a gianormous pain in my *$$!



Hopefully I didn't screw anything up in the transfer to my file.
 

Wendy

Joined Mar 24, 2008
21,914
I'm about to mess up your day even more. :D

You will note I put designators for parts on my schematics. Every time I haven't it turns out I need them later. Things like parts lists later, for example.

I strongly suggest you get into the habit, it also makes discussing a circuit much easier later down the road.
 
Last edited:

SgtWookie

Joined Jul 17, 2007
22,210
Bill means "You really need to use reference designators on your parts" ;)

Right now, you only have U2a and U2b - the rest of the parts have no reference designators. This makes it awkward when trying to explain to someone which part connection(s) might have a problem, particularly if there are more than one part with the same part number. All reference designators must be unique, and the reference designator prefix should be conventional.
Capacitors - C
Inductors - L
Resistors - R
Diodes - D (you can use ZD for Zeners; CR is an obsolete way to indicate diode)
Transistors - Q (Sometimes TR is used, but Q is preferred)
MOSFETs - Q or M
IC's - U or IC
Jacks - J
Plugs - P
 

SgtWookie

Joined Jul 17, 2007
22,210
you guys are bouncing ideas off each other like its playing cards!
Well, when a problem comes up like this, it's time to solve it! Having several people throw out different ideas helps stimulate more ideas.

Who would be willing to explain the Zener diode config for me? I had already read, and just reread the Zener diode portion on this site, but SgtWookie has me all turned around (not difficult to do).
I'll take a stab at it. ;) By the way, Zener diodes were named after Clarence Melvin Zener, who discovered the effect - so Zener is a proper name, and should be capitalized.
http://en.wikipedia.org/wiki/Clarence_Zener

If not reversed biased, the Zener acts like a reg. diode (-.7V) across. When reversed biased, assuming the supply voltage exceeds the Zener voltage, the zener only allows the zener voltage through.
The Zener breaks down at it's rated voltage, ±a few percent. It acts as a voltage clamp. Once the Zener voltage has been reached, the diode will start conducting in order to keep the voltage from rising higher.

So, in SgtWookies circuit, the 5.1 zener comes before the 12 zener correct?
In this case, the order does not matter. Basically, all we're looking for is a way to make up the difference between the 23.5v where the PNP transistors' base will be in cutoff, and the 9v where the DSx outputs of the 4553 will be. 23.5v-9v = 14.5v. The 5.1v and 12v Zeners together are 17.1v; so we have a 2.6v margin for error - which is good.

When a DSx output goes to 0v then, the high side of the Zeners in series will be at ~17.1v. The PNP transistor starts turning on when it's base is 24v-0.7v, or 23.3v. If you're going to have 15mA per LED segment, then you will need 15mA*7 = 105mA current through the PNP collector, so you'll need 1/10th of that through the base.
23.3v-17.1v = 6.2v
So, what resistance is needed to get 10.5mA when 6.2v is across it?
6.2/10.5mA = 590 Ohms
Except, I calculated for 25mA*7 segments, so I wound up with 175mA total collector current, and 17.5mA base current required, so a 354 Ohm base resistor. Now, why did I come up with 240 Ohms? I can't answer that at the moment. It was late.

You need to tell me how much current you will be running per segment on the 7-segment displays so that I can re-calculate the resistors.

And they are both reverse biased based on the 24V supply.
Yes.
So wouldn't the 5.1 reduce the voltage that the 12 sees down to 5.1V, therefore causing the 12V zener to act like a regular diode (-.7V)? Am I all horsed up or is that the desired effect?
No, the 5.1v Zener drops ~5.1v across itself, so 5.1v is simply subtracted from the 24v. Same thing with the 12v Zener; 12v is subtracted.
 

elec_mech

Joined Nov 12, 2008
1,500
Who would be willing to explain the zener diode config for me? I had already read, and just reread the zener diode portion on this site, but SgtWookie has me all turned around (not difficult to do).
I second this request if I may. If I'm reading SgtWookie's circuit correctly, R2 is a pull-up resistor designed to keep the PNP transistor based pulled high and thus off whenever the DS output is high (off).

After reading the zener diode section here as well, I assume putting two zener diodes in series is equivalent to putting two resistors in series - you add the breakdown voltages together. So, as shown, there will be 12 + 5.1V = 17.1V between the junction of ZD1-R1 and ground when the DS output is low. Subtracting this from the 24V connected to the base of the PNP through R2, we get 24 - 17.1 = 6.9V. Alternately, if we use a single 16V zener, we get 24 - 16 = 8V. This is the voltage left that would pass through the 4553, correct?

Currently, the logic is powered with 9VDC. So using the zeners as shown drops the voltage from 24V to 6.9-8V to fall just below the supplied 9V logic, thereby preventing damage to the 4553. Is this close or am I way off base?

I assume R1 is a current-limiting resistor.

Alternately, if the DS output goes low, 17.1V is dropped across the zener diodes and 6.9V passes through the 4553 via the DS pin to ground, pulling the base of the PNP low and allowing the digit to receive power. When DS is high, the output is approximately 9V. R2 is pulling the PNP base high, thus providing 24V. This 24V still goes through the zeners to DS. There is a voltage difference between the DS output (9V) and the voltage at the base of the PNP (24V), thus 24-9 = 15V. However, with the addition of the zeners, there must be a voltage greater than 17.1V before they will conduct. Thus 17.1V > 15V and the DS pin is safe from the 24V when high.

Okay, so if I'm not totally off base in my understanding, are the zeners used to protect the DS pin and thus the 4553 when high or low or both?
 

SgtWookie

Joined Jul 17, 2007
22,210
Let's try a water analogy to a Zener diode.

You're running water in a bathtub, and the drain is open. The water runs right down the drain. No water depth is allowed to accumulate to build pressure; the drain is essentially a "short" to ground.

You want 5.1" of water in the tub, so you take a length of pipe and a connector that fits the drain exactly, and plug it in. Now the water in the tub fills up to 5.1" before it starts spilling into the end of the length of the pipe; the water can't rise any further. You now have 5.1" worth of water pressure at the bottom of the tub.

Now you add a section of pipe 12" long to the original piece of 5.1" pipe using a connector; you now have a total of 17.1" of pipe above the drain. Hopefully your tub is at least a bit more deep than 17.1"... but anyway, the water keeps running out of the faucet and fills the tub to a depth of 17.1" before it starts spilling into the end of the pipe; the water can rise no further.
 

Wendy

Joined Mar 24, 2008
21,914
Zeners are a favorite of mine. They are very analogous to a pressure regulator, once you exceed their voltage setting they start to conduct, and will burn themselves up if you force the issue and feed them too much current. Below their voltage set point they stop conducting. Like most things in life, they aren't perfect. The lower the setpoint voltage the sloppier they are (their setpoint isn't as stable), for example. Below their setpoint they act pretty much like a standard diode.

They are the heart of many simple voltage regulators.

Here is a little table for 1W Zeners I made a long while back...

.
 

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elec_mech

Joined Nov 12, 2008
1,500
Sorry, SgtWookie replied as I sent my last response.

I'll remember to keep the Z capitalized in the future.

Very interesting. I'll take some time to digest this. Thank you!

P.S. I'm using the 5" Futurelec displays in a project I'm about to send off. Because they are used with a circuit designed to work with the 4" displays, the current draw is less than planned for due to the sizing of the resistors, but I measured 136mA pulled when 00 is displayed - 136mA / 12 segments = 11.3mA. Voltage was about 17.3V, I think. Put behind a dark red 0.125" thick piece of acrylic, it's real easy to see.

I'd recommend planning for 17.6V, 20mA max. and 17.3V, 11mA minimum. 15-20mA would be good in my opinion.

Just remember to add a filter over the displays like red acrylic (assuming you're using red displays) - it really makes all the difference.
 

Wendy

Joined Mar 24, 2008
21,914
I don't remember if I pointed this out, I probably did, but at the risk of repeating myself I have a set of templates (and standards, like the zener table) I call PaintCAD. I use copy and paste to make all my drawings.

It is available for download here...

Introduction and PaintCAD
 

SgtWookie

Joined Jul 17, 2007
22,210
I second this request if I may. If I'm reading SgtWookie's circuit correctly, R2 is a pull-up resistor designed to keep the PNP transistor based pulled high and thus off whenever the DS output is high (off).
Yes, that's correct.
After reading the zener diode section here as well, I assume putting two zener diodes in series is equivalent to putting two resistors in series - you add the breakdown voltages together. So, as shown, there will be 12 + 5.1V = 17.1V between the junction of ZD1-R1 and ground when the DS output is low. Subtracting this from the 24V connected to the base of the PNP through R2, we get 24 - 17.1 = 6.9V. Alternately, if we use a single 16V zener, we get 24 - 16 = 8V.
Yes, very good.
This is the voltage left that would pass through the 4553, correct?
When the DSx output is high, there would be very little current flowing into the 4553's DSx pin. When the DSx output is low, the current sunk depends largely on the Zeners and R1, and somewhat on R2. This is assuming that the 24v and 9v supplies are as advertised; always their respective voltages.

Voltage doesn't "pass through" things; voltage is dropped across things. Current passes through things.

Currently, the logic is powered with 9VDC. So using the zeners as shown drops the voltage from 24V to 6.9-8V to fall just below the supplied 9V logic, thereby preventing damage to the 4553. Is this close or am I way off base?
It's not to protect the 4553; it's so that the PNP transistors' base can rise to be within 0.5v of 24v, allowing it to turn off. Otherwise, the PNP transistors would be stuck "ON" all the time.

The idea is to allow the current flow through R1/R2 to fall low enough that R2 can return the base to nearly 24v.

I assume R1 is a current-limiting resistor.
Yes.

Alternately, if the DS output goes low, 17.1V is dropped across the zener diodes and 6.9V passes through the 4553 via the DS pin to ground, pulling the base of the PNP low and allowing the digit to receive power.
Yes.
When DS is high, the output is approximately 9V. R2 is pulling the PNP base high, thus providing 24V. This 24V still goes through the zeners to DS. There is a voltage difference between the DS output (9V) and the voltage at the base of the PNP (24V), thus 24-9 = 15V. However, with the addition of the zeners, there must be a voltage greater than 17.1V before they will conduct. Thus 17.1V > 15V and the DS pin is safe from the 24V when high.
Ahhh, "safe from the 24v" isn't how I'd put it. There would be very little current flow through the Zeners, as 15v is less than 17.1v, so neither of the Zeners would be in breakdown. There would be a small amount of current, but R2 would still return the base to nearly 24v.

Okay, so if I'm not totally off base in my understanding, are the Zeners used to protect the DS pin and thus the 4553 when high or low or both?
As mentioned already, it's really not protection-oriented.

However, you do bring up a good point; power supply power-on sequence - the 24v supply should not be powered up on it's own. 9v should be available when the 24v supply has reached 12v.
 

elec_mech

Joined Nov 12, 2008
1,500
Voltage doesn't "pass through" things; voltage is dropped across things. Current passes through things.
Good point. I meant to say protect the 4553 from 24V (over-voltage condition and potential damage to the IC). I guess that is not the concern so much as:

It's not to protect the 4553; it's so that the PNP transistors' base can rise to be within 0.5v of 24v, allowing it to turn off. Otherwise, the PNP transistors would be stuck "ON" all the time.The idea is to allow the current flow through R1/R2 to fall low enough that R2 can return the base to nearly 24v.
Excellent! This makes sense to me. Thank you!

So CoachKalk, you'll want to:

1) Add designators to all of your parts in your circuit (R1, C1, U1, etc.) as Bill mentioned earlier.

2) Tweak the values of the zener diodes if you end up using the 20VDC power supply you mentioned earlier and/or if you change the logic to something other than 9V, such as 12V or 15V.

3) The transistors in your latest circuit should be labeled 2N2907 not 2N3907.

4) You may want to add a switch debouncer between your switches and the CD4013. I'd suggest using a CD4093 with a couple of resistors and capacitors. I can provide an example if you'd like.

5) Add a 10kΩ pull down resistor between pin 12 of the CD4060 and ground - assuming the 4013 outputs a high signal to start the CD4060. If it outputs a low signal to start the CD4060, then put the resistor between pin 12 and Vcc.

6) I recommend labeling all of your IC pins with their respective pin numbers. When you go to troubleshoot your circuit, you won't know what pin LE is, you'll need to know what pin number you need to look at.

7) You mentioned earlier that many of the schematics you've come across don't show Vcc or ground. Many don't include filter capacitors as well. This is not a standard practice so much as an attempt to keep things looking neat and clean (or laziness, but we'll give them the benefit of the doubt). For the same reason, people often draw their ICs, as you've done below, with the pin locations all around the IC, not as you would see them on the actual part. It is up to you how you wish to draw your schematics, but I always try to draw the ICs to match the physical part and always include everything I put on my circuit, including Vcc, grounds, filter capacitors, etc. The schematic doesn't always look as pretty as others, but you have everything in front of you and because the ICs drawings match the physical ICs, I can more readily see how I need to lay out my board be it a bread board, protoboard or custom PCB. Just my two cents on the matter.

I'll have to look at the start and stop functions as shown in your diagram for my own benefit. I'm not familar enough with the 4013 to say if this will work or not. Has Bill blessed this portion?:)
 

SgtWookie

Joined Jul 17, 2007
22,210
As far as point #7), it's important to be able to grasp the function of the circuit quickly. If you go strictly by the pin layout of the device, this becomes much more difficult; the layout in the schematic becomes awkward and you wind up either running wires all over the place, or using lots of labels with interrupted signal wires (which is just about as bad).

It should be a given that bypass capacitors are ALWAYS required for EVERY IC per supply rail, and each board needs at least one larger electrolytic cap per supply rail. The reason for omitting them on a casual schematic is to help the viewers grasp the circuit operation as quickly as possible.

As far as parts placement, that's one of the functions that a PCB layout is used for. One needs an easy-to-understand schematic, and a parts placement diagram, which a PCB layout can serve as. If one is going to go into production with a board, they'll have to generate both a schematic AND a PCB layout anyway; so it makes sense to make both as easy to understand as possible; the former for humans, the latter optimized for signal path. If both a schematic and PCB layout are being generated, pin numbers are not so critical on the schematic. It's nice to know both the pins and their function, if it doesn't make the circuit look too cluttered.
 

Thread Starter

CoachKalk

Joined Sep 20, 2011
139


So CoachKalk, you'll want to:

1) Add designators to all of your parts in your circuit (R1, C1, U1, etc.) as Bill mentioned earlier.
Done - Started doing it then got lazy. Back on the straight and narrow path now!

2) Tweak the values of the zener diodes if you end up using the 20VDC power supply you mentioned earlier and/or if you change the logic to something other than 9V, such as 12V or 15V.
I am waiting patiently for my components. Emailed shipping info almost a week ago, now I am like a kid @ Christmas time. I have already noted the possible changes to the Zener's and resistors when it is GO TIME!

3) The transistors in your latest circuit should be labeled 2N2907 not 2N3907.
Fixed - Not sure when I horsed that up, maybe wrong/typo from the beginning.

4) You may want to add a switch debouncer between your switches and the CD4013. I'd suggest using a CD4093 with a couple of resistors and capacitors. I can provide an example if you'd like.
Your suggestion for a debouncer is the second time I have read that. The last poster mentioned it should work without it, but I may have to hit the button twice? I was even more overwhelmed then than I am now, so I figured hitting it twice was doable. Now that I am slowly putting the picture together, I would be very interested in a debouncer circuit.

5) Add a 10kΩ pull down resistor between pin 12 of the CD4060 and ground - assuming the 4013 outputs a high signal to start the CD4060. If it outputs a low signal to start the CD4060, then put the resistor between pin 12 and Vcc.
I have noted this point. I have been printing all Datasheets for the IC's I am using so I will take a look and add the resistor (after of course I check with you guys here to make sure I understood the datasheet correctly:D).

6) I recommend labeling all of your IC pins with their respective pin numbers. When you go to troubleshoot your circuit, you won't know what pin LE is, you'll need to know what pin number you need to look at.

7) You mentioned earlier that many of the schematics you've come across don't show Vcc or ground. Many don't include filter capacitors as well. This is not a standard practice so much as an attempt to keep things looking neat and clean (or laziness, but we'll give them the benefit of the doubt). For the same reason, people often draw their ICs, as you've done below, with the pin locations all around the IC, not as you would see them on the actual part. It is up to you how you wish to draw your schematics, but I always try to draw the ICs to match the physical part and always include everything I put on my circuit, including Vcc, grounds, filter capacitors, etc. The schematic doesn't always look as pretty as others, but you have everything in front of you and because the ICs drawings match the physical ICs, I can more readily see how I need to lay out my board be it a bread board, protoboard or custom PCB. Just my two cents on the matter.
I quickly realized pins were sometimes left off and/or placed in differing spots. I do have the datasheets, but I will look at adding pin notations as well. I think (crossed fingers) I have all the necessary bypass capacitors inplace. I am trying to get my head around the placement of resistors (the ones that all of you "know" should be placed between inputs and ground) but I don't even think about them until it is noted that I missed them. One day, one day I say ... I will get it!

I'll have to look at the start and stop functions as shown in your diagram for my own benefit. I'm not familar enough with the 4013 to say if this will work or not. Has Bill blessed this portion?:)
Like Bill said, He did it. I am just a good old-fashioned copy-catter!
 

elec_mech

Joined Nov 12, 2008
1,500
Your suggestion for a debouncer is the second time I have read that. The last poster mentioned it should work without it, but I may have to hit the button twice? I was even more overwhelmed then than I am now, so I figured hitting it twice was doable. Now that I am slowly putting the picture together, I would be very interested in a debouncer circuit.
You shouldn't have to hit a button twice. Because switches are mechanical, they can have "bounce", meaning a bunch of high and low signals (noise) is generated as you press the switch due to mechanical action. ICs operate faster than we can blink, so having a bunch of high and low transistions can "fool" the IC into thinking it is receiving a lot of pulses in short amount of time even though you're just pressing the switch once. The result is your display may appear to jump to some random number when you press the switch instead of just incrementing by one. You can use a resistor and capacitor combo (better than nothing) or an IC such as the CD4093 with a 0.01uF cap and 10kΩ resistor.
 

Wendy

Joined Mar 24, 2008
21,914
Take another look at that flip flop, it is an excellent debounce unto itself. No other debounce needed. The flip flop will change states the moment the button makes contact, and stay that way.
 

elec_mech

Joined Nov 12, 2008
1,500
The flip flop will change states the moment the button makes contact, and stay that way.
Ooo, I didn't know a flip flop could work as a debouncer - nice. :D

The CD4093 I proposed earlier is good if you need to debounce a momentary signal, i.e., you want to sent a single high signal for as long as the switch is pushed. Releasing a switch connected to a 4093 debouncer will immediately put it back to a low state, at least in the circuit I've used. So, you may want to use a 4093 to debouce the reset switch, but you're set on the start and stop switches.

You may not need a pulldown on pin 12 of the 4060 since the 4013 will always output a high or low signal - in other words, pin 12 should never float and I think the output of the 4013 takes care of that.

I missed this before, but be sure to add junction dots anywhere connections cross that are supposed to be joined. Generally, you either use a) arcs over intersecting connections that should not be joined and not use dots OR b) omit arcs and use dots whenever two intersecting connections are supposed to be joined. This way you know what signals are connected to each other and which are not.
 

Thread Starter

CoachKalk

Joined Sep 20, 2011
139
Good morning everyone. I have been trying to get my mind around all of the items I had been working on last fall and I am slowly coming along.

I have recently been working on getting the 5" LED displays powered up. I have included a circuit that is a simplified version of what some of you had helped me with last fall. At this time, I am simply trying to get the 3 5" segments powered up and respond to manual hits at the 4553 clock pin. I realize the counts may be jumpy and in the end I will have the time portion added, but I am trying baby steps.

Here is the circuit as I "think" I wired it up.



I have a few comments/questions about what I have noticed and then tried. Hopefully you can suggest some areas for me to troubleshoot.

First, I will say I am using a 24V supply for the 5" Segments, but the Vcc is measuring around 22.2V (FYI). The supply is rated at 400mA.

I have also pasted the specs from the Futurlec site for the segments below.
Absolute Maximum Ratings
Pulse Forward Current (Ifp): 150 mA
DC Forward Current (If): 25 mA
Reverse Current (Ir): 20 µA
Power Dissipation (If): 60x36 mW
Operating Temperature (Top): -40 to +80 oC

Electro-Optical Data (@ 20 mA)
Forward Voltage - Typical: 9.25 V
Forward Voltage - Maximum: 11.00 V
Luminous Intensity Minimum Per Segment: 288 Mcd
Luminous Intensity Typical Per Segment: 432 Mcd


But, after rereading this thread, elec mech mentioned actual values for the exact segments he used in a different project - 136mA with 00 (12 segments) displayed or approx. 11.3 mA per segment.


So, here is what happened.

When I powered everything up, the segments on the middle digit flickered for a few seconds and then went dark.


After/when I checked my solder joints, I realized the pin assignment I found online was not correct (nothing provided by Futurlec -
) I had to change one segment. I had incorrectly used the decimal point pin for segment d.


After correcting the "d" pin, I powered each segment separately to verify all segments were lighting. By separately, I mean I removed each segment from the 2N2907's and went through each segment. All segments worked great.


One note here. At first, I quickly checked the segments directly on the back of the display (no resistors in play). Not surprisingly, each segment was well lit. I then connected directly out of the 2804 (resistors in play now) and the correct segment lit up (obviously not as bright as previously) BUT the segments on either side of the segment I was testing lit up very soflty as well.


I then conneted the segments back through the 2N2907's and manually checked the segments again. Again, each segment lit up individually, but nothing is on if the 24V supply is conneted to ground.


I checked the zener diode portion and measured the correct/expected drop after each one.


So, I come asking/begging for suggestions.
 

elec_mech

Joined Nov 12, 2008
1,500
Hey CoachKalk,

Good to hear from you. A few notes first:

Yes, the Futurelec website doesn't do a great job with their LED display datasheets. Note that the forward voltage they post for the 5" display is for the decimal point only. The forward voltage of the segments is 18VDC nominal, 22VDC max. Whenever you get a display, the manufacturer usually prints their part number on the side. I suggest always looking it up before you wire anything to make sure you have the right datasheet - I learned this the hard way. I've dealt directly with this manufacturer before for a past job. You can find them here: http://www.nb-flying.com/. And I belive this is the datasheet for the digits you have:
http://www.nb-flying.com/images/NFD-50011ABx-11.pdf.

Double-check the part number printed on your displays with one in the datasheet to make sure this is the right one.

I think your power supply is a bit low. Assume worst case scenario, all segments lit, 7 segments x (let's say) 20mA x 3 digits = 420mA. Now, I could be off on this as the datasheet seems to infer each strand of LEDs takes 25mA, but each segment would then consume 50mA. If this is correct, then we're talking 7 x 40mA x 3 = 840mA. I'm hoping Bill will chime in with his thoughts regarding what the true current per segment should be.

The 4553 is multiplexing the segments. That is, it is providing power to each segment through the 2907s about 1/3 of the time. The current values above assume power is provided 100% of the time. When you provide power for a fraction of the time at the same current values, the displays will appear dimmer. To compensate, you need to provide more current so the display is just as bright despite getting power a fraction of the time. Again, I hope Bill will see this and correct me if I'm wrong. But let's worry too much about this at the moment.

So, what exactly is happening when you connect the 5" displays to your circuit? Are they dim? Are only certain segments lighting up? Are certain segments brighter than others? Do all the digits appear to be off?

What is the voltage across the C.A. pin and one of the segment inputs when your circuit is on? Do this for several segments and for all digits.

What is the current going into each segment? Same as before, please do this for multiple segments for each digit. Let us know if you're unsure how to do this with your meter and I'll explain. If there is any doubt, please ask first because it is possible to short something out if this is done wrong.
 
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