Another Scout Project

[JohnW2]

...If you are using multimeters arround mains on anything like a regular basis and/or you are using multimeters at all arround systems with higher fault currents than a normal domestic system ....

Do you think that a significant number of people do either of those things? For any who do, for whatever reason, then I would certainly agree with...

... then I would STRONGLY reccomend both a good quality multimeter and a set of approved fused test leads.

... but that's only really in relation to safety. Most people do not need the degree of functionality that Monkeh appears to require.

Kind Regards, John

 

[333rocky333]

I once had a 200 pound clampmeter and never used the plug in leads and voltage setting.
Then one day i broke my voltage tester and for some stupid reason i used it to test the output of an ignitor.
That was the end of that

 

[ban-all-sheds]

[net]A photo of a cheapo meter which is identical to the cheapo one I have knocking about in a toolbox[/net]

It and the leads fit neatly into an old Philishave case

which protects it from the depredations of hammers and saws etc.

 

[securespark]

Righto.

Time for an update.

I started the LED card project last Thursday.

Few problems with the white LED's.....

Put one with a red LED and the white ones go out.

Came to the conclusion that maybe they are drawing too much.

So I put two button cells together, still no joy.

Any ideas?

 

[bernardgreen]

If the LEDs are in parallel across the battery then the red LED will be pulling the battery voltage below the voltage needed for the white LED to start conducting ( and thus lighting up ).

Either connect in series with a couple of batteries in series ( brightnesses may be very different ) or use two resistors, one in series with each LED so the battery voltage is not pulled down by the red LED.

 

[JohnW2]

If the LEDs are in parallel across the battery then the red LED will be pulling the battery voltage below the voltage needed for the white LED to start conducting ( and thus lighting up ).... Either connect in series with a couple of batteries in series ( brightnesses may be very different ) or use two resistors, one in series with each LED so the battery voltage is not pulled down by the red LED.

...or, even, just one resistor, in series with the red LED, may well be enough.

Kind Regards, John

 

[securespark]

Thanks, guys.

I just don't understand why this happens.

The other LED's in parallel don't have an issue working together.

 

[GarethH]

Semiconductors including LEDs have a relatively fixed voltage drop across them when conducting - called Vf in the case of an LED.

The LEDs you have will have different Vf values but, by connecting the anodes and cathodes together, you have ensured that both anodes are at the same potential and both cathodes are at the same potential. i.e. the LEDs have the same voltage drop across them.

The LED with the lowest Vf (let's call it Vf1) is conducting and dropping Vf1 volts across it. As they're tied together, the other LED - which has a higher Vf - does not conduct (significantly, at least) because the voltage across it is less than its Vf; As the LEDs are connected, they'll both have the same voltage across them and that voltage will be the lower forward voltage.

Semiconductors in parallel will behave very differently to resistors in parallel - if the voltage across one of them is less than the forward voltage then very little current will flow through that device. This is what you're seeing.

Put a series resistor in for each LED. For each LED, look up Vf and the forward current. Look up the maximum cell voltage - which may be more than the rated voltage for a lithium cell - and subtract Vf for the LED. This gives a value that approximates the required voltage drop across the series resistance. Use Ohm's law to calculate the value of resistance that gives the LED's rated forward current at this voltage and put that resistance in series with the LED. Repeat for each LED and then connect the resistor-led pairs in parallel.

You can increase the resistor values if the LED is too bright or you want the battery to last longer.

 

[securespark]

Thanks Gareth.

As it's a relatively simple LED Christmas card (and I don't have time to source some resistors!!), I'll try putting another cell in series and reconnecting the LED's in series.

 

[GarethH]

If the sum of the forward voltages of the LEDs is close to the total cell voltage that may be fine but you would probably be exceeding the LEDs' rated current.

Depending on the difference, this may or may not be OK. If the difference is too great, the LEDs may fail and the battery and LEDs could get hot. I'm not sure whether this would pose a risk or not...

Personally, I'd be more comfortable with a series resistance in place.

 

[securespark]

Right. I'm obviously not cut out to be a spark, as I am not understanding this at all.

The forward voltages of the various LED's all vary between 2.8- 3.6V.

The currents all vary between 20 and 30mA.

Red is 30mA and white is 20mA (which is where we had a problem).

But the others were mismatched too, yet they worked together OK.

 

[GarethH]

I expect that's due to the voltage current characteristics for the specific LEDs. An "ideal" diode doesn't pass any current until its Vf is exceeded at which point it starts conducting. When conducting, it always drops Vf volts across it regardless of the current.

That's just an idealised model though, in practice diodes including LEDs start conducting before their Vf is reached but the amount of current they pass (and hence light emitted for a LED) increases sharply as Vin approaches Vf.

Exactly what happens varies from LED to LED depending on the technology and physics of the device though. Some different colour LEDs will work in parallel for this reason but it's a bit of a hit-and-miss thing to rely on in terms of electronic circuit design.

Searching for "current voltage characteristics led" or similar should bring up some graphs that shows this for different LEDs if you're interested.

 

[JohnW2]

Right. I'm obviously not cut out to be a spark, as I am not understanding this at all. ... The forward voltages of the various LED's all vary between 2.8- 3.6V.

Are you talking about measured Vf figures or 'what it says on the tin?

I think the problem is that, since you aren't using explicit series resistors, the 'series resistance' limiting the current is the internal resistance of the battery. If you put an LED with a Vf of 2.8V directly across a battery, the voltage between the battery terminals (and across the LED) will fall to roughly 2.8V (the 'drop' being across the internal resistance of the battery). If you then put an LED with a Vf appreciably above 2.8V in parallel with the first one, the 2.8V available there might not be enough for the second one to conduct and light up. If you put a resistor in series with the 2.8V one, most of the voltage drop would then be across that resistance (rather than the battery's internal resistance), and there would then be >2.8V across the battery terminals,possibly now enough to work one of higher Vf connected across the battery terminals (with no resistor).

Kind Regards, John

 

[GarethH]

My interpretation of securespark's last post was that some combinations of LEDs with differing / mismatched Vf values light when put in parallel and some other combinations do not.

I believe this is because the LEDs in question won't have the "ideal" brick wall V/I characteristic - a real LED will start conducting before its forward voltage is met and the voltage across it will increase with increasing forward current whereas A perfect / model LED would not conduct at all until Vf is exceeded and would always drop Vf volts when conducting.

I agree on the benefits of a series resistance but advocate adding a current limiting resistor to both LEDs rather than one. If a resistor is put in series with one LED only, the current through the LED without a resistor is only limited by the battery's internal resistance and the slight increase of that LED's Vf when forward current increases. This is will cause that LED to burn out as its rated forward current is exceeded which, at best, will shorten its life (maybe not much of a concern for a Christmas card ).

The excess power is dissipated in the LED and battery which may end up getting warm / hot.

 

[JohnW2]

I agree on the benefits of a series resistance but advocate adding a current limiting resistor to both LEDs rather than one. If a resistor is put in series with one LED only, the current through the LED without a resistor is only limited by the battery's internal resistance and the slight increase of that LED's Vf when forward current increases. This is will cause that LED to burn out as its rated forward current is exceeded which, at best, will shorten its life (maybe not much of a concern for a Christmas card ). The excess power is dissipated in the LED and battery which may end up getting warm / hot.

Indeed - but we went through all that many pages back in the thread, and securespark was keen to keep the component count to the minimum, which is why he ended up using LEDs without series resistors - with all the potential problems you mention.

Even if one used a pair of LEDs of nominally identical Vf, if one put them in parallel straight across a battery with no resistors at all, it would not surprise me if, at least with some pairings of those 'identical' components one lit up whilst the other didn't (much or at all).

Kind Regards, John