Thanks for all the suggestions and advice. I have tried various ways of getting the job done and come to the conclusion that the best way may be the 2576 switcher.
I have just ordered 3 of them, only cost £3.50 on fleabay too!
Anyway as soon as they come I will post the outcome.
You said the forward voltage of the LEDs is 3.6V to 4.2V and the ULN2803 has a 1.4V drop. Then a 5V supply is too low for 4.2V + 1.4V= 5.6V and additional voltage is needed to limit the current with resistors.
But you don't have anything to limit the current.
Just realised that I don't know if the inductors I have will be big enough to suit the project?
Can someone advise please also on how vital the value of the 100uH is? The ones shown which I have say 101 on them but measure 95 to 98 on my meter. Is that close enough?
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Here's 2 suggestions for inductors, I browsed down the spec sheet and it looks like around 40 - 60uh is max and I like 33uh, and if you need to, you can series or parallel them.
did you order fixed voltage or adjustable voltage smps
Kinarfi
PS your 101 inductor do look small, but if you have several, you could try paralleling 2 or 3 of them, since I think you only want 33uh any way.
The problem is that the forward voltage of LEDs, for a given current, varies from one to another as the result of manufacturing tolerances and as a function of temperature, so you don't know (until it's too late!) what the critical voltage is. Therefore it's safest to use a limiting resistor.
Buck converters are the simplest switchers and with the single chip converters available it makes it even simpler. Just make sure to get the right inductor and check all the wiring/connections before starting it up for the first time. A circuit with an inductor in it can blow out as soon as it is turned on if the catch diode is not connected right. Use an ohm meter etc. Purchase more than one regulator chip just in case. You'll get it working one way or another.
A circuit with an inductor in it can blow out as soon as it is turned on if the catch diode is not connected right. Use an ohm meter etc. Purchase more than one regulator chip just in case. You'll get it working one way or another.
It looks like you will use a LM2576 to drop the 12 to 14V to 6V.
Now the resistors will have only 2-3 volts across them not the 8 to 10 volts.
The inductance value is not real critical. A 100uH inductor (measured at 0 current) will probably be 60 to 70uH at full current. If you get +/-10% that is real good.
Just built the circuit with the LM2576 and it does drop the output to 5v. But at a load of only 950mA the voltage drops to 3.7v
Any ideas as to where I am going wrong? Could the inductor be wrong in some way. I built it on a ferrite core of 23mm OD and 17mm ID and it has 16 turns of 1.5mm wire.
Cut the inductance value to 47uh or try paralleling two, then 3 of your 101s, I, myself do not have faith in self wounds yet, I made a few, but the core material was wrong on some and I don't know what the problem was with some of the other winds, tried one with telephone wire from a cat 5 cable and I don't remember it being a success. My opinion is that the 100uh is quite a bit too much. You can also try https://www.ti.com switcher calculator, it may or may not give you the same ic number, but the inductor value is a good guide.
From my experiences, I say the higher the inductance, the less the current that is available, and if the circuit can supply 5 and you only need 3, it will only supply 3, I'm thinking you end up with close to 33uh, IMHO
Be aware that the lower the inductance the greater will be the ripple current in the output smoothing cap. So the cap value/type may be critical to prevent overheating.