LED sequencing circuit

vne147 · Jan 23, 2010

dougy83 said:
The base voltage would be 12V, the emitter voltage would be ~11.3V & the collector voltage would be 21V. The transistor would drop ~9.7V (VCE).

OK, think I'm starting to get it. So the lower base voltage drops the emmiter to Vb - Vbe and the issue becomes that there won't be enough voltage left for the 5 LEDs in series. Correct? What if there were only 3 LEDs for example. Would it work then?

dougy83 said:
The column (those with emitter connected to ground) transistors will work (they'll sink current). I might reduce the 33k resistor but it's not very important.

Sorry for the confusion, by column I meant the other transistors, the ones with their collectors connected to supply. I still don't understand why they wouldn't have worked if there was only one supply rail. Iignoring the fact that the 4017 can't operate on 21V, if the output of the 4017 was 21V then the voltage at the emitter of the column transistor would be 20.3 V which should be more than enough for the 5 LEDs.

dougy83 · Jan 23, 2010

vne147 said:
OK, think I'm starting to get it. So the lower base voltage drops the emmiter to Vb - Vbe and the issue becomes that there won't be enough voltage left for the 5 LEDs in series.

Yes. Also the fact that the 21V is not utilised/provides no benefit in that configuration (it's the same output even with the collectors connected to 12V).

vne147 said:
What if there were only 3 LEDs for example. Would it work then?

3 LEDs --> 10.5V, which can be driven from 12V-0.7-0.3=~11V. I guess it would.

vne147 said:
Sorry for the confusion, by column I meant the other transistors, the ones with their collectors connected to supply.

My fault, it makes more sense this way.

vne147 said:
I still don't understand why they wouldn't have worked if there was only one supply rail. Iignoring the fact that the 4017 can't operate on 21V, if the output of the 4017 was 21V then the voltage at the emitter of the column transistor would be 20.3 V which should be more than enough for the 5 LEDs.

If the 4017 and 555 could operate at 21V, you're quite right, there would be none of these issues.

vne147 · Jan 23, 2010

OK, think I get it. Thanks for the explainations. I think I'm going to go ahead and mod my circuit to fix the column transistor problems. I don't see why your mod wouldn't work and it seems simplest but I'm going to change mine just for an exercise. I'll post it back when done.

vne147 · Jan 23, 2010

OK here's the latest version. Tear it up!!!

dougy83 · Jan 23, 2010

Looks good to me.

Boncuk · Jan 23, 2010

Hi vne147,

I simulated the circuit using five LEDs in group at a supply voltage of 21V. There is no way to control the transistors with a maximum base voltage (theoretically) of 12 to conduct 21V. (The maximum I got out was 11.6V)

Reducing the number of LEDs to three the group can be well controlled. I used base resistors of 6K8 for each transistor and a pulldown resistor on each counter output of 10K. In practical design the pulldown resistors might be a 10 element SIL-package.

Not using the pulldown resistors the circuit becomes unstable and produces error messages while simulating. With the pulldown I "played" the circuit at least 50 times with no error message, doing what the counter outputs dictated.

Using three LEDs in a group the OP might use just one power supply which saves cost. It also increases the number of flashes.

Two counter circuits won't suffice, but adding a third one shouldn't be a problem, increasing the number of LEDs to 501 total at a spacing of 1.996cm instead of 2cm

BTW, I suggest to connect the MR pins (15) of the counters to ground as well. It won't hurt.

Attached are four screenshots of different groups being lit. None is glowing dimly.

I put the current limiting resistors of 27Ω at the high side. The resulting current for LEDs Vf=3.5V, If=20mA is 18.6mA.

Regards

Boncuk

P.S. The pulldown resistors at U2 weren't changed to 10K, but at 1K they did well.

mbarazeen · Jan 24, 2010

what if we try a MOSFET instead of the transistor? it may excite the source voltage regardless of gate voltage?

BrownOut · Jan 24, 2010

That won't work. In fact, it'll be worse.

Boncuk · Jan 24, 2010

Based on vne147's circuit and my add-ons I have designed a single sided PCB layout which requires 10 wire jumps around the transistors.

Board dimensions are: 160X83mm (6.3125X3.3inches)

There is a total of 27 pieces of 10 pin connectors (10 for the colums and 17 for the rows) which should make wiring of the LEDs easy.

Boncuk

vne147 · Jan 24, 2010

BrownOut said:
The problem is still going to be the column drivers. Although you have them connected to a higher voltage, the voltage delivered at the emitters is going to only be the level output by the IC's minus a diode drop. You're either gonna need a pair of transistors at each comumn (NPN/PNP), or else a row of inverters and PNP pullup transistors. I can't think of which is better right now, cause I'm in the middle of somethings.... but those drivers will need to be fixed.

mbarazeen said:
as brownout told, those transistors wil fuction as emitter followers and only give the output little less than IC high out.
you better shift all LEDs directly from high voltage side in series with the current limiting resistor, then use both transistors for switching with base resistors.

otherwise you have to connect the group LEDs in parellel by connecting a single resistor for each LED. the power consumption of the circuit will increase by this.

Ubergeek63 said:
vne147's circuit does not work - there is no way for the 12V logic source to forward bias a BE junction into 16V of LEDs. easily corrected with an extra set of transistors and resistors but kind of kills the "low componewnts count" bragging point

Hey guys. Does the latest circuit, the one in post # 44 correct the problems with the column transistors that you were talking about?

dougy83 said:
Looks good to me.

Thanks!

Boncuk said:
Hi vne147,

I simulated the circuit using five LEDs in group at a supply voltage of 21V. There is no way to control the transistors with a maximum base voltage (theoretically) of 12 to conduct 21V. (The maximum I got out was 11.6V)

Reducing the number of LEDs to three the group can be well controlled. I used base resistors of 6K8 for each transistor and a pulldown resistor on each counter output of 10K. In practical design the pulldown resistors might be a 10 element SIL-package.

Not using the pulldown resistors the circuit becomes unstable and produces error messages while simulating. With the pulldown I "played" the circuit at least 50 times with no error message, doing what the counter outputs dictated.

Using three LEDs in a group the OP might use just one power supply which saves cost. It also increases the number of flashes.

Two counter circuits won't suffice, but adding a third one shouldn't be a problem, increasing the number of LEDs to 501 total at a spacing of 1.996cm instead of 2cm

BTW, I suggest to connect the MR pins (15) of the counters to ground as well. It won't hurt.

Attached are four screenshots of different groups being lit. None is glowing dimly.

I put the current limiting resistors of 27Ω at the high side. The resulting current for LEDs Vf=3.5V, If=20mA is 18.6mA.

Regards

Boncuk

P.S. The pulldown resistors at U2 weren't changed to 10K, but at 1K they did well.

Boncuk,

Which circuit are you referring to in this reply? I posted about 5 different versions. I think I understand the error in the older versions. But the latest version in post # 44 with the NPN/PNP transistor combination should work with the 2 supply rails, shouldn't it?

I understand your comment about the pulldown resistors and grounding the reset pin. Those are good suggested improvements I think and I'll make the changes. Thanks.

Also, thanks for taking the time to simulate the circuit. BTW, what simulation software are you using there?

Boncuk said:
Based on vne147's circuit and my add-ons I have designed a single sided PCB layout which requires 10 wire jumps around the transistors.

Board dimensions are: 160X83mm (6.3125X3.3inches)

There is a total of 27 pieces of 10 pin connectors (10 for the colums and 17 for the rows) which should make wiring of the LEDs easy.

Boncuk

Boncuk,

Can you post the schematic of this PCB please? Thanks.

BrownOut · Jan 24, 2010

Hey guys. Does the latest circuit, the one in post # 44 correct the problems with the column transistors that you were talking about?

\

You have the right idea, but your base resistors are quite high. A good starting point is to use ib=ic/10, so try this:

Rb = 10*((V(input) - .7)/ic)

So for example, if you want 10mA at the collector, and the input is 10 volts:

10*((10-.7)/10mA) ~= 10K ohms. I've lost track of what you want for current, just substitute your values into the equaiton.

dougy83 · Jan 24, 2010

vne147 said:
I understand your comment about the pulldown resistors and grounding the reset pin.

The pulldown resistors sound like they were put there for some simulator anomaly/quirk. They shouldn't be needed in the real thing. Grounding reset should be done though.

BrownOut · Jan 24, 2010

Refering back to my post #51, in the case of the PNP's, the full VCC minus about a volt will be felt from the base to ground. So, in the equation, you mabe use VCC-1V in place of Vin-.7V.

vne147 · Jan 24, 2010

BrownOut said:
\

You have the right idea, but your base resistors are quite high. A good starting point is to use ib=ic/10, so try this:

Rb = 10*((V(input) - .7)/ic)

So for example, if you want 10mA at the collector, and the input is 10 volts:

10*((10-.7)/10mA) ~= 10K ohms. I've lost track of what you want for current, just substitute your values into the equaiton.

Are you referring to the resistors in between the base of the PNP and the collector on the NPN or the resistors between the base of the NPN and the output of the 4017?

BrownOut · Jan 24, 2010

All of them. I good method would be to start with the PNP's, using the modification of ponst #53, and work back to the NPN resisotrs. If you get really, really stuck, then re-post the current requirement for the LED's and I'll try to run the numbers for you.

Hint, according to one poster, 17.5V is required across the LED's. The leaves about 2.5V across the 160 ohm resistors, and

IC = 2.5/160 = 16mA

Then, 10*20/16mA = 12.5K for the PNP's.

And ib = .16mA for the NPN's driving the PHP's.

RB(NPN) ~= 10*11/1.6mA~= 60K.

considering the above, I'd try 56k for the NPN and 10k for the PNP.

Edit: corrected math errors

vne147 · Jan 24, 2010

BrownOut said:
All of them. I good method would be to start with the PNP's, using the modification of ponst #53, and work back to the NPN resisotrs. If you get really, really stuck, then re-post the current requirement for the LED's and I'll try to run the numbers for you.

The current requirement according to the OP was 20 mA but let me take a stab at it before you run the numbers. I'll post back when I'm done. Just to make sure I'm understanding things, your equation:

Ib = Ic/10

implies a DC current gain or hfe of 10, correct? It that is correct, why use 10 when the data sheet has values ranging from 100 - 800? Thanks.

BrownOut · Jan 24, 2010

Sorry, I got carried away in the previous post

The reason for using Hfe=10 is because your transistors will be operating in saturation mode, and hfe will be less than active mode. 10 is a generic number and is safe for most GP transistors.

PS: if you already previous post (#55), it containd errors, which have been corrected.

vne147 · Jan 24, 2010

BrownOut said:
Sorry, I got carried away in the previous post

The reason for using Hfe=10 is because your transistors will be operating in saturation mode, and hfe will be less than active mode. 10 is a generic number and is safe for most GP transistors.

OK, understand. Thanks.

BrownOut · Jan 24, 2010

vne, I had to correct my math in my last couple posts. Please read them over.

vne147 · Jan 24, 2010

Will do. I'm also going to run the numbers on my own first as an exercise and then compare them to yours for a double check. Thanks.

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