Note that the 5% guideline for BC547 doesn't apply to all transistors. General purpose 2N transistors use 10%. The appropriate number can be found in the datasheet.
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Completely understandable that you wouldn't have expertise in everything your employer did.

When I worked at HP Labs, we reported to Bill Hewlett. I doubt that he had detailed knowledge of everything the labs worked on.

Absolutely understood... Thanks for the datasheet! The one I have I think is incorrect.

Note that the 5% guideline for BC547 doesn't apply to all transistors. General purpose 2N transistors use 10%. The appropriate number can be found in the datasheet.
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Completely understandable that you wouldn't have expertise in everything your employer did.

When I worked at HP Labs, we reported to Bill Hewlett. I doubt that he had detailed knowledge of everything the labs worked on.

The BC547B's worked like a charm - perfect brightness now! Thank you for the suggestion.

Well, at least it was perfect for the first 30 seconds.

It ran perfectly, then just froze with only one channel lit. So I swapped out the 547's just in case. Same thing.

I ended up swapping out ALL the components. Same thing.

I have ZERO clue as to why this is happening.

Is there something else needed connected to the 547's maybe? I am guessing no.

Without the transistors, it runs perfectly 'forever'.

I have 3 CR2032's in series powering this.

Thanks again...

Mike








I would think if the 4017 had failed, it wouldn't work at all.

 

Can you provide a schematic? It’s hard to debug without seeing the circuit.

I immediately have a few questions.

• What voltage do each of the three batteries have after it freezes?
• Do you have bypass capacitors on your CMOS chips?
• Are you using a CMOS 555?
• Are ALL inputs to the 4017 connected to something? Particularly Enable… as clock and reset are used in the circuit.

 

I ended up swapping out ALL the components. Same thing.

I have ZERO clue as to why this is happening.

Is there something else needed connected to the 547's maybe? I am guessing no.

Post a schematic so we can suggest some more informed troubleshooting steps.

 

Can you provide a schematic? It’s hard to debug without seeing the circuit.

I immediately have a few questions.

• What voltage do each of the three batteries have after it freezes?
• Do you have bypass capacitors on your CMOS chips?
• Are you using a CMOS 555?
• Are ALL inputs to the 4017 connected to something? Particularly Enable… as clock and reset are used in the circuit.

Thank you for you awesome replies and questions...

I am sorry my troubleshooting skills have been lacking the past 20 years.

One thing I noticed I forgot to put resistors before the Base of each transistor from the outputs. I will do that right now. Please stand by (or sit, if you'd prefer.) Ok I'm back. That didn't help.

Before I answer your most recent questions, I think I did mention that this circuit worked perfectly, running for almost 3 days straight before the transistors were added.

Answers to your questions:

1. Battery voltages: After the most recent freeze, here's the voltages: 2.773, 2.794, 2.826. They were all fresh and "brandy new" this morning.
2. Bypass caps: I do NOT have bypass caps. Can you please recommend a value. (And placement please, I forget where they go.)
3. CMOS Timer: Yes, it is the NE555N
4. Inputs: Yes, all are tied. Reset to pin 7, Clock to 555, and the Clock Inhibit to Ground.

Thanks again for all your help!

Mike

 

Battery voltages: After the most recent freeze, here's the voltages: 2.773, 2.794, 2.826. They were all fresh and "brandy new" this morning.

Without a schematic, we can only guess. Do you have current limiting resistors for the LEDs?

Bypass caps: I do NOT have bypass caps. Can you please recommend a value. (And placement please, I forget where they go.)

0.1uf are commonly used.

CMOS Timer: Yes, it is the NE555N

Since the bipolar timer can sink/source 200mA, it can put spikes on the battery voltage. You should use an electrolytic cap to smooth the battery voltage.

 

Since the bipolar timer can sink/source 200mA, it can put spikes on the battery voltage. You should use an electrolytic cap to smooth the battery voltage.

I’d place a 47μf electrolytic capacitor across the battery supply for this purpose.

The 0.1μf bypass capacitors need to be placed as close as possible to the IC’s Vcc and Gnd pins. On each IC.

 

Without a schematic, we can only guess. Do you have current limiting resistors for the LEDs?
0.1uf are commonly used.
Since the bipolar timer can sink/source 200mA, it can put spikes on the battery voltage. You should use an electrolytic cap to smooth the battery voltage.

I never posted a schematic? Thought I did - sorry! It's attached now.

Feel free to mark it up as you wish!

Thank you for the other advice as well!

 

I never posted a schematic? Thought I did - sorry! It's attached now.

Feel free to mark it up as you wish!

Thank you for the other advice as well!View attachment 274529

Wow!

With your LEDs wired that way, you’ll need a resistor FOR EACH LED. Otherwise, it’s likely they will work for a while and then burn out. Or stop working. Plus, with all in parallel, each one is getting ~8.2V!!! Way to much. Surprised they worked at all.

Also, based on the current draw of one string, you’ll need to select a base resistor for each transistor. Ten LEDs in parallel where each is drawing 10ma results in the string needing 0.1A. Thus, the transistor base resistor should set a current of 10ma. At 8.2V that’s a 8k ohm resistor

You can also wire the LEDs in a series|parallel arrangement. This way reduces the current draw of the LEDs. For example, if you could use 9 instead of 10 LEDs in each string, you could wire the LEDs as 3 strings of 3 LEDs in series, and all three in parallel. It would reduce the current needed from 0.1A to 0.03A… in any case, unless you have a constant current supply you will need a resistor in front of the LEDs or string of LEDs in series.

 

Wow!

With your LEDs wired that way, you’ll need a resistor FOR EACH LED. Otherwise, it’s likely they will work for a while and then burn out. Or stop working. Plus, with all in parallel, each one is getting ~8.2V!!! Way to much. Surprised they worked at all.

Also, based on the current draw of one string, you’ll need to select a base resistor for each transistor. Ten LEDs in parallel where each is drawing 10ma results in the string needing 0.1A. Thus, the transistor base resistor should set a current of 10ma. At 8.2V that’s a 8k ohm resistor

You can also wire the LEDs in a series|parallel arrangement. This way reduces the current draw of the LEDs. For example, if you could use 9 instead of 10 LEDs in each string, you could wire the LEDs as 3 strings of 3 LEDs in series, and all three in parallel. It would reduce the current needed from 0.1A to 0.03A… in any case, unless you have a constant current supply you will need a resistor in front of the LEDs or string of LEDs in series.

You are definitely the digital guru!

The way the board is (will be) laid out, I do need to keep 10 per string. That's a great idea tho! Maybe 2 strings of 5?

You did say that it would burn out quick or not work at all. Why it worked continuously almost 3 days then (before the transistors) surprises me!

I will add the resistors as you recommended and see what happens.

So maybe I can assume that it's freezing because of the current draw?

Thanks for all your help!

 

The way the board is (will be) laid out, I do need to keep 10 per string. That's a great idea tho! Maybe 2 strings of 5?

What is the Vf specification for the LEDs? If you don’t know, what’s the color?

It’s unlikely that two strings of 5 would work. First, I want to remind you that we’re using a series | parallel circuit. That is, we put some LEDs in series with a ballast resistor to set the desired current. Then we wire these series strings in parallel. So, given a Vf of 2V (typical of red LEDs at one point in time. 5 @ 2V in series requires 10V to operate. You have 8.2V. See the problem? Three LEDs in series requires 6V. Which is less than 8.2V. All is good.

How about two sets of 3 and one set of 4? 4 LEDs requires 8V which is too close to 8.2V. You need at least 1V headroom.

But you could use 3 sets of 3 plus a single LED and resistor. (I might go 2 sets of 3 and 2 sets of 2.

But with all these patterns. You’ll need different resistors.

To figure out what resistor you’ll need is very simple. Use these values:

• Vftot - the sum of the individual Vf values for the LEDs in a series string
• Aled - the operating current required for an LED
• Vss - the supply voltage (8.2 in your case)

To calculate the series resistor, use this formula:

​

R = (Vss- Vftot) / Aled​

So, for a series string of 3 LEDs (with an operating current of 10ma and a Vf of 2V) the desired resistor would be

​

R = (8.2 - 6) / 0.01​

= 2.2 / 0.01​

= 220Ω​

I’m this case 220Ω is a standard value. If necessary, pick the closest standard to whatever you calculate.

 

Feel free to mark it up as you wish!

We prefer for the flow in schematics to be primarily left to right and top to bottom. The symbol your schematic uses for a 555 timer is terrible. Pin order symbols don't convey function or contribute to the preferred schematic flow.

You can't wire LEDs that haven't been matched for forward voltage in parallel. The only thing preventing bad things from happening is the fact that CR2032 can't source much current.

It was never clear to me that you were blinking 10 LEDs at the same time. Your circuit isn't going to run long on coin batteries. You got several days runtime before because current was being limited by the CD4017.

 

What is the Vf specification for the LEDs? If you don’t know, what’s the color?

It’s unlikely that two strings of 5 would work. First, I want to remind you that we’re using a series | parallel circuit. That is, we put some LEDs in series with a ballast resistor to set the desired current. Then we wire these series strings in parallel. So, given a Vf of 2V (typical of red LEDs at one point in time. 5 @ 2V in series requires 10V to operate. You have 8.2V. See the problem? Three LEDs in series requires 6V. Which is less than 8.2V. All is good.

How about two sets of 3 and one set of 4? 4 LEDs requires 8V which is too close to 8.2V. You need at least 1V headroom.

But you could use 3 sets of 3 plus a single LED and resistor. (I might go 2 sets of 3 and 2 sets of 2.

But with all these patterns. You’ll need different resistors.

To figure out what resistor you’ll need is very simple. Use these values:

• Vftot - the sum of the individual Vf values for the LEDs in a series string
• Aled - the operating current required for an LED
• Vss - the supply voltage (8.2 in your case)

To calculate the series resistor, use this formula:

​

R = (Vss- Vftot) / Aled​

So, for a series string of 3 LEDs (with an operating current of 10ma and a Vf of 2V) the desired resistor would be

​

R = (8.2 - 6) / 0.01​

= 2.2 / 0.01​

= 220Ω​

I’m this case 220Ω is a standard value. If necessary, pick the closest standard to whatever you calculate.

Thank you for all that outstanding info. You guys are all awesome for helping out an old dude with a bad memory!

I can always add another battery, but for now I will stick with the formulas you gave me.

 

I can always add another battery, but for now I will stick with the formulas you gave me.

Were you planning to add that battery in series with the other three? That won't improve the current capability.

Normally we don't recommend putting LEDs in parallel the way you've done, but when using a battery like CR2032, it's more acceptable. You can't draw 10's of mA from them and expect them to last very long. Capacity, down to 2.0V, is about 200mAh.

If you operated each group of 10 LEDs at 20mA (total), you don't need to worry about current hogging causing a cascading failure. Runtime would be in the 10 hour range.

Using 3 CR2032 in series and operating at 20mA, the current limiting resistor would need to be

\( \large R=\frac{V}{I}=\frac{V_{SUP}-V_{LED}}{I_{LED}}=\frac{9V-2V}{20mA}=350\Omega\)

(use 330Ω) the base resistors would need to be

\( \large R=\frac{V_{OH}-V_{BE}}{I_B}=\frac{9V-0.7V}{1mA}=8.3k\Omega\)

(use 8.2kΩ).

 

Were you planning to add that battery in series with the other three? That won't improve the current capability.

Normally we don't recommend putting LEDs in parallel the way you've done, but when using a battery like CR2032, it's more acceptable. You can't draw 10's of mA from them and expect them to last very long. Capacity, down to 2.0V, is about 200mAh.

If you operated each group of 10 LEDs at 20mA (total), you don't need to worry about current hogging causing a cascading failure. Runtime would be in the 10 hour range.

Using 3 CR2032 in series and operating at 20mA, the current limiting resistor would need to be

\( \large R=\frac{V}{I}=\frac{V_{SUP}-V_{LED}}{I_{LED}}=\frac{9V-2V}{20mA}=350\Omega\)

(use 330Ω) the base resistors would need to be

\( \large R=\frac{V_{OH}-V_{BE}}{I_B}=\frac{9V-0.7V}{1mA}=8.3k\Omega\)

(use 8.2kΩ).

Thank you sir. I actually tried the 8.2K's earlier today and it froze again. It worked maybe 2-3 minutes then froze, over the 30-40 second without the resistors.

I just wish I knew why it was freezing!

 

You have three low current and low capacity CR2032 batteries that produce a total voltage of 9V with no load. Then measure the 9V drop down to a very low voltage when the circuit is running.

Use 4 or 5 AA alkaline cells in series to make 6V or 7.5V instead to provide the current the circuit needs.
Also, do not use a little 9V battery that produces a low current for smoke alarms.

 

I actually tried the 8.2K's earlier today and it froze again.

By "froze", do you mean that the counter stopped incrementing? If that's the case, you should make sure the 555 timer is still oscillating. If the timer is still working, then it's a problem with the counter.

 

By "froze", do you mean that the counter stopped incrementing? If that's the case, you should make sure the 555 timer is still oscillating. If the timer is still working, then it's a problem with the counter.

Yes sir. That was the whole problem after adding the transistors. As mentioned, before I did that (when the lamps were a bit dimmer,) the circuit ran literally for almost 3 days.

I swapped out all the components thinking one of them decided to crap out.

The 555 still outputs a pulse consistently, even after it freezes. So that's when I changed the 4017. It still did the same thing. Even after all of you and the other folks' outstanding suggestions, it still freezes.

Nothing else was changed, every single jumper wire is intact and untouched.

Thank you!

Mike

 

Yes sir. That was the whole problem after adding the transistors. As mentioned, before I did that (when the lamps were a bit dimmer,) the circuit ran literally for almost 3 days.

I swapped out all the components thinking one of them decided to crap out.

The 555 still outputs a pulse consistently, even after it freezes. So that's when I changed the 4017. It still did the same thing. Even after all of you and the other folks' outstanding suggestions, it still freezes.

Nothing else was changed, every single jumper wire is intact and untouched.

Thank you!

Mike

Have you added the 0.1μf bypass capacitors across the ICs Vcc and Gnd pins and the 47μf filter capacitor across the batteries + and - ?

 

Have you added the 0.1μf bypass capacitors across the ICs Vcc and Gnd pins and the 47μf filter capacitor across the batteries + and - ?

Oh crap, thanks for reminding me. I meant to. (I take care of my elderly mom and I forgot!)

 

The 555 still outputs a pulse consistently, even after it freezes. So that's when I changed the 4017. It still did the same thing. Even after all of you and the other folks' outstanding suggestions, it still freezes.

The next time the circuit freezes, do some troubleshooting instead of replacing parts. Unless the CD4017 is defective, replacing it won't change anything.

My suspicion is that you're pulling so much current from the batteries that supply droop is causing the counter to misbehave.

If you don't have base resistors on the transistors, you're loading the outputs of the counter, but that doesn't affect count order.

From Harris: