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Help needed in modifying circuit.

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srsdesign

New Member
Hi all,

Newbie here as i'm just getting back into electronics and looking for some guidance in modifying an existing off the shelf circuit.

The circuit is an Electronic wheel of Fortune link: https://www.kemo-electronic.de/en/Hobby-School/B239-Electronic-wheel-of-fortune.php

IMG_2944.jpgIMG_2945.jpg

What i'm trying to achieve is to upscale the circuit to drive lengths of LED tape (12v 4.8W 6000k White 60 x 3528 / metre LED tape IP20 rated at 4.8w p/M. Using a 12V 3Amp power supply which are then fixed into a segmented unit which will have a printed acrylic face on the front.

IMG_2948.jpgIMG_2949.jpg

Currently the circuit works illuminating all of the LEDs with one unlit one spinning round when the button is depressed and released which then gradually comes to a halt. Obviously this is quite a draw on the power driving 10 x 720mm lengths of LED tape (see attached picture of the wheel) so I would like the LEDs to be brighter than they currently are.. Therefore, what I would like it to do is only have one LED/tape strip lit / spinning and coming to a halt. Also the speed at which it spins I would like to slow down, would I be correct in that this is controlled via a capacitor / transistor connected to pin 14 of the HCF4017 Decade Counter if so how can I slow this spinning down as from searching on this IC it is usually placed with a timer chip / variable resistor to control the speed of the counter?

I have attached a couple of photos of what I have done so far and schematic.

Many thanks in advance for any guidance on this.

Cheers

Stu

Wheel of Fortune.jpeg
 

sagor1

Active Member
Be careful in trying to drive a string of LEDs with that one chip. The 4017 counter is rated for a maximum 10mA of current sink, typically less than that in actual circuits. Driving one LED is possible for each pin, but not more than that.
You will need some form of buffer/driver between the IC and the LED strings. Also, depending on the buffer circuit, you may have to invert the logic of the pins so they are active HIGH instead of LOW A direct driver for low logic could be something like the 8 channel sink driver ULN2804. Those can handle up to 500mA and 50V, and is designed to interface with CMOS ICs. However, the logic is inverted, high turning it on low (current sink enabled). You would need an inverter IC between the 4017 and the buffer.
You could also make a double/inverter logic with two 2N7000. With high on the first gate, it conducts and Drain is connected to second 2N7000 gate which is now low. Second 2N7000 is off. When first gate goes low, first 2N7000 stops conducting, allowing second to go high via pull-up to Vcc, conducting. Each 2N7000 is rated for about 200mA and up to 60V. For higher currents, the second MOSFET could be a higher power one like IRF540. See diagram below. R2 and R4 are optional and used only if the Vcc is over 20V, as to not damage the devices by exceeding maximum gate voltage. Q2 can be 2N7000 for up to 200mA, or IRF540 for high currents.

2 MOSFET buffer.jpg

In your circuit, C1 charges instantly when the button is closed. With button closed, C2 also charges slowly via R1. When the button is released, C1 keeps supplying a voltage for a short while, and C2 discharges slowly via T1 and T2. In effect, C1 and C2 both supply the discharge current when the switch is open.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
You need to add a suitable transistor or FET on each output of the 4017.

To switch that size load, almost half an amp, I'd use a fairly high rated device so you do not have problems with overheating after a time.

This is the basic principle.
(All your LEDs are in a single unit with just two wires, connected between the FET drain (LED -) and +12V (LED +), while the drawing shows several separate LEDs; ignore that).

Use one FET per output / light bar.
The FET source pins are all connected to the supply & PCB negative.

Connected like this, only one bar will be on at a time.
https://www.circuitlab.com/editor/#?id=69un99

The FETS could be IRF540 or IRF740 etc. - they see readily available and quite cheap on ebay.

Those are rated at tens of amps each if attached to a suitable heatsink, but should be OK at half an amp without needing heatsinking as long as they are in open air.

Edit; clarifying the description.
 
Last edited:

sagor1

Active Member
As a side note, the final MOSFET for high power can be just about any IRF520, 530,540 series, whatever is most cost effective. 2N7000 can be bought real cheap on places like Ebay, like 100 of them for a few dollars. Of course, I test each one before using in any circuit, Chinese products on Ebay have a tendency to have a few defective ones at times...
 

sagor1

Active Member
Kingsb, you may be right. I was assuming, from the schematic, that only one LED would be on by current sinking. However, looking at the datasheet, the pin goes high as it counts. That means one LED goes out, the rest stay on.
So, it now depends on what the asker of the question wants to do. Using just one MOSFET will do if he wants the one string to be on, the rest off. If he wants to invert the logic, he could use my diagram. Both would work.
Thank goodness for those N Channel MOSFETs....
 

alec_t

Well-Known Member
Most Helpful Member
In your circuit, increasing the value of R1 and/or C2 will slow down the LED cycling. As noted above, N-FETs could drive the LED strings one at a time.
 
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srsdesign

New Member
Hi all,

Many thanks for your quick responses!

So if I have it correct (please excuse me as I am a newbie to this) can I spur out from the +12v power to all 10 strips of LED tape and then take the control for each LED pin on the circuit board into a IRF540 or IRF740 then onto the - to each LED strip?

Please excuse my schematic and diagram attached.

Also, to reduce the speed say by 50% what values of R1 and C1 would I be looking for?

Once again many thanks for all your help, have been thrown in the deep end so to say with this as I am making the main unit to house this for a charity event and now seem to be trying to work out the electronics.....enjoying it though!!

Stu
Schematic_Wheel-of-Fortune_Sheet-1_20190113182707.pngCIRCUIT-01.png
 

alec_t

Well-Known Member
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JimB

Super Moderator
Most Helpful Member
You will need more than 12v to power all those LEDs in series.
The voltage you will need depends on the colour of the LEDs.

Also, there is no current limiting resistor for the LEDs.
Yes, I can see that the original circuit had no current limiting resistors either, but it was a simple (poor ?) design, relying on parasitic resistance in the integrated circuit.

JimB
 

rjenkinsgb

Well-Known Member
Most Helpful Member
So if I have it correct (please excuse me as I am a newbie to this) can I spur out from the +12v power to all 10 strips of LED tape and then take the control for each LED pin on the circuit board into a IRF540 or IRF740 then onto the - to each LED strip?
Spot on, that should work fine


Just for info the LED tape does not actually have all the LEDs in series; each LED (or possible 2 -3 LEDs) will be connected to the 12V circuit via a resistor, like a ladder rung arrangement.

Your diagram is fine functionally, but that's what JimB is pointing out, not realising it's a mass produced 12V rated tape setup.
 

gophert

Well-Known Member
Most Helpful Member
Alternatively, you can just couple the light from your existing circuit to control your new light strips...

DCA1C4CE-18C7-4346-878A-81C1A7326FF3.jpeg
 

JimB

Super Moderator
Most Helpful Member
Your diagram is fine functionally, but that's what JimB is pointing out, not realising it's a mass produced 12V rated tape setup.
Well spotted.

I have used that "LED tape" myself a few times, I am quite familiar with it.
I was lead astray by the schematic.

JimB
 

srsdesign

New Member
Hi again,

Many thanks for all your help.

Just a few more questions to clarify.

Using this circuit:

Schematic_Wheel-of-Fortune_Sheet-1_20190113182707.png
I understand there will only be one strip of LED's lit at a time?

By spuring off the power from the 12v supply to the board and LED strip will be ok? As in the board will draw what it needs and the LED's will do the same and I wont overload the board still?

Currently the circuit works by pressing the button which starts the wheel spinning and when released then it starts to slow down to rest on the chosen segment. What I am trying to achieve is that this circuit / wheel is housed in a standing scale (see attached). When no one is standing on them then the last segment that it finished on is lit, when someone stands on the scales then it starts spinning then slows and stops on the random segment. How easy would this to be achieve?

SCALES v21.png

Once again apologies for all the questions.

Thank you all for your help.

Stu
 

rjenkinsgb

Well-Known Member
Most Helpful Member
You could disconnect C1 from R1, then use a changeover switch with the common to C1, normally closed to +12V (preferably with a low value resistor is series, eg. 100 ohms) and normally open to R1.

That means the cap is always charged ready to start the circuit, but it only connects when someone stands on the platform and the switch clicks across. The circuit will then start cycling and slow down as C1 discharges.

That's the simplest way, the only downside is the "spin" would stop instantly if they got off the platform too soon.

I'd use something like a generic V3 microswitch with a lever actuator as the changeover switch.
eg. https://www.ebay.co.uk/itm/Microswi...h=item3d66e8b79b:g:OycAAOSwMZFbl8Me:rk:8:pf:0
 

alec_t

Well-Known Member
Most Helpful Member

srsdesign

New Member
You could disconnect C1 from R1, then use a changeover switch with the common to C1, normally closed to +12V (preferably with a low value resistor is series, eg. 100 ohms) and normally open to R1.

That means the cap is always charged ready to start the circuit, but it only connects when someone stands on the platform and the switch clicks across. The circuit will then start cycling and slow down as C1 discharges.

That's the simplest way, the only downside is the "spin" would stop instantly if they got off the platform too soon.

I'd use something like a generic V3 microswitch with a lever actuator as the changeover switch.
eg. https://www.ebay.co.uk/itm/Microswi...h=item3d66e8b79b:g:OycAAOSwMZFbl8Me:rk:8:pf:0
That's great will give that a go. Will the length of cable from the switch to C1 to R1 have an effect? I am looking to house the circuit board behind the wheel face and the switch will be in the base plate about 1.4m distance using Equipment wire 7/0.2mm tinned copper?

Many thanks

Stu
 

srsdesign

New Member
Hi All,

Sorry not having much fun here today with this.

Ok have the IRF540N's and have wired them up as below pic to the LED tape strips.


Wheel FETs-01.jpg

IMG_2953.JPG

Will be connecting up to the PCB as below



Is there any easy way to test the FETs without connecting to the PCB as I think I may of blown the HCF4017 and awaiting replacement tomorrow but have to finish this tomorrow too.

On powering up the LED tape is lighting up randomly as below some are lit some just glowing. The panel that all of this is mounted on is ACP panel, would that effect it?

IMG_2954.JPG


Sorry starting to panic now to get this done.

Many thanks for any insight.

Stu
 

sagor1

Active Member
The fact many LEDs are on mans the 4017 has lots of high outputs at the same time, something that should not happen. If you toggle RESET (pin 15) to Vdd (+12V), only the "0" pin should be high.
Also, you should make sure each FET gate has a pull-down resistor to ground, to keep the gate from floating up. Even a 100k resistor will do. You can put those at the circuit board itself. I would test with the 100k pull-down resistors first, before changing the 4017 chip... If you don't have 100k, anything from 10k upwards will work.
 
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