The Outputs of the CD4017 go to the positive supply and to ground when they feed a CD4xxx operating from the same supply voltage. Therefore the inputs of the CD4069 work perfectly when fed from the outputs of a CD4017.

What is the part number of the Mosfets and the banks of LEDs you will use?

 

looks like your on the right track, inverters going to P Channel MOSFET is a good solution

 

The Outputs of the CD4017 go to the positive supply and to ground when they feed a CD4xxx operating from the same supply voltage. Therefore the inputs of the CD4069 work perfectly when fed from the outputs of a CD4017.

What is the part number of the Mosfets and the banks of LEDs you will use?

I will most likely take your advice in post#8 and use the circuit as is minus the base resistors, it seems the easiest/cheapest option.

However, I am interested in exploring an alternative solution and learning something along the way. I don't understand what you mean when you say, the outputs of the CD4017 go to the positive supply and to ground. Looking at the datasheets I would have connected pin3 (CD4017) to pin1 (CD4069) and pin2 (CD4069) to the mosfet gate?

I haven't looked at which mosfet I would use but I am open to suggestions. The leds are of unknown origin, they were given to me by a friend. I have measured current draw on each bank and it is 31mA @ 12V.

 

When it its time, each output of the CD4017 goes to a logic high which is the positive supply voltage when it drives the input of another CD4xxx logic IC powered from the same positive voltage. Other than its time, each output of the CD4017 goes to a logic low which is 0V.

Why do you say each bank of LEDs must have their cathode connected to 0V? Please sketch how a bank of LEDs is wired.
Are the five banks of LEDs connected together somehow?

 

That makes sense now, I misinterpreted what you said previously. When you said the outputs of the CD4017 go to the positive supply and to ground I thought you were saying the outputs are physically connected to the positive supply and ground. I see now you were referring to the outputs going logic high and logic low.

I know each output pin on the CD4017 is low by default and goes high when triggered by the clock pulse from the 555 and the CD4069 will invert this logic high to low on its own output. What I am asking about is the physical connection between the CD4017 & CD4069, ie pin3 connects to pin1 etc and the resulting voltage & current from the CD4069 (and its ability to switch a mosfet/pnp).

The leds's are encased in a single sealed plastic block, the block is divided into 5 rows of leds. There are 6 wires exiting the sealed block, one ground and five positive feeds. If 12V is applied to each of the five positive feeds a separate row of leds illuminates.

 

An LED is a diode that has a forward voltage depending on its color. A red LED will be about 1.8V to 2.2V and a blue or white LED will be about 3.0V to 3.6V. LEDs without current limiting will blow up if a voltage slightly too high is fed to them and need to have their current limited by a circuit or by a resistor. If the voltage fed to the rows of LEDs without current limiting is slightly low then it will cause no light.
Your rows of LEDs might or might not have internal current limiting.

I do not know any and have never used little P-channel Mosfets.

 

For a small P ch. FET, BS250FTA, 90 mA, 40 V, SMT SOT23, G to S about 2 mm, US $ .45, JAMECO.
I would use a 4049 inverter , 2N3906 PNP, 2.7k base R. If it had to be FET, then 4049, FDS9435A
( because I have some )LL, P ch., SMT, SOT23. Would require a V divider to reduce gate to 10 V.
Where do you buy your parts ?

 

Thanks Bernard, I am going to redo the circuit using a hex inverter and pnp transistor with base resistor. I will go with your recommendations regarding the parts. May I ask why you would use the 4049 over 4069? To check my calcs, a base resistor of 2.7k should fully saturate the 2N3906 (hfe 50) allowing for its max collector current of 200mA , does this sound right (11.4-0.7V)/(0.2A/50) = 2675ohms?

I live in Australia and have two local stores where I can buy parts, Jaycar and Altronics. If I have to I order online I use Mouser, Digikey or RS Components. My location is also why my responses are delayed.

 

Check the datasheets 4049 sinks much more current if you’re using BJTs.

 

Thanks Wolframore, just to be certain I am looking at the correct figures is the sinking current IOL on the datasheet?

 

To check my calcs, a base resistor of 2.7k should fully saturate the 2N3906 (hfe 50) allowing for its max collector current of 200mA , does this sound right (11.4-0.7V)/(0.2A/50) = 2675ohms?

No. The datasheet shows that it saturates well when its base current is 1/10th its collector current at reasonable currents, not at its max allowed current of 200mA.

hFE (beta) is used when the transistor is used with lots of collector to emitter voltage like in an amplifier where it is never saturated.
The curves in a datasheet are for a transistor with typical specs but some are better and some are worse. You should design so that a transistor with the worst written (minimum and maximum) specs works properly then all worst, typical and better transistors will work properly.

If you want 200mA then use a 2N4403 PNP transistor with a max allowed current of 600mA.

 

Thanks again and please excuse my ignorance, I did say I was a newbie. Ok, so is it reasonable to say that when selecting a transistor as a switch I should look at the required collector current and build in headroom so the transistor isn't maxed out. Then calculate the base resistor value using the required collector current.

Can you please tell me where in the attached datasheet to find the 1/10th figure?

 

When a transistor is used as a switch you want it to turn on hard (saturation) with a low Vce.

 

So you actually prefer to use 1 inverter chip, 5 resistors and 5 transistors, over the one chip solution?

Now don't get me wrong, I'm not trying to convince you of anything, especially since this board seems more like a competition then anything else sometimes.

I'm just curious.

 

First pass at the circuit I described in post #12, an all-MOS solution. The inverters can be any CMOS inverting device with at least 5 sections, such as the CD4069.

This is an edit of something from another thread, so ignore the LED specifics, MOSFET part number, etc. and use what you have / can get / whatever.

The U2F input can be tied to anything, just not left floating. Also, the circuit needs two power supply decoupling capacitors.

ak

 

So you actually prefer to use 1 inverter chip, 5 resistors and 5 transistors, over the one chip solution?

2981-type devices are darlington emitter followers, not saturating switches, and would reduce the voltage to the LEDs by almost 15%. The single-chip thing is very convenient, but comes at a price.

ak

 

At 40 mA I seriously doubt that, but yea, there is a trade off on each of the circuits.

I doubt very much if the OP would notice the brightness difference, but on the other hand I'm sure they would notice the build time difference.

 

Can you please tell me where in the attached datasheet to find the 1/10th figure?

It is not stated explicitly, it is a test conditiion based on a rule-of-thumb. My problem with it is that it is a leftover from the 1950's and needs an update. Since by definition we are talking in broad generalities, today's transistors have both better gain, lower saturation voltages, and a sharper saturation knee. For a 2N3904 / 4401 -type of small signal, general purpose transistor, the difference in saturation voltage for collector-base current ratios of 10:1 vs 30:1 is a few millivolts. When you're talking about driving bipolar parts with old CMOS parts, a 3:1 difference in output current can be important.

ak

 

Re: Post # 32, check Figure 14 & 15.

 

It is not stated explicitly, it is a test condition based on a rule-of-thumb. My problem with it is that it is a leftover from the 1950's and needs an update. Since by definition we are talking in broad generalities, today's transistors have both better gain, lower saturation voltages, and a sharper saturation knee. For a 2N3904 / 4401 -type of small signal, general purpose transistor, the difference in saturation voltage for collector-base current ratios of 10:1 vs 30:1 is a few millivolts.{/QUOTE]
I think a 2N3904 made today has the same spec's as one made many years ago as per the datasheet. If you want better spec's then pick a transistor with better spec's which might be a new part number.

The first transistor I ever used was an old BC107 in a metal case 60 years ago. The BC547 made today (plastic case) has the same spec's. it is European and has poorer saturation when the base current is 1/20th the collector current.
The spec's guarantee that all passing transistors meet the spec's when the base current is 1/10th for American transistors or 1/20th for European transistors. The American spec has more base current so its saturation voltage is lower than the European spec.