(MicroLumina) current limiters?

[deemery]

I can't find my MicroLumina Lighting Tips that talks about how to use MicroLumina current limiters.  Does anyone have a quick tutorial on those?  I do remember that each LED takes 3v and the current limiter takes something like 2v, so you can light three 3v LED in a circuit with a current limiter and 12v input.  But I don't quite remember how those are wired, the 3 LEDs in series with the current limiter?  Or is the current limiter in parallel?   And if you have to align the current limiter a certain way, which pin gets + vs -?

Thanks in advance!

dave

[jbvb]

Any LEDs used with a current limiter must be wired in series with the current limiter. If you need to know the polarity of everything, ask: I have both MicroLumina instructions and those that came with some I got from Digi-Key.

[deemery]

So that's simple:  CL and 3 LEDs in series on a single 12v circuit...  

dave

[jbvb]

With incandescent bulbs, there's a valid argument against putting lamps in series: requires trial & error to identify the one that burned out. But much less important with long-lived LEDs.

[deemery]

A question on current limiting resistors:  If I have 4 LEDs in series, does each need its own limiting resistor?  Or, if I have 3 that come pre-wired with resistors and add a 4th one that doesn't have a resistor, will that work?

dave

[jbvb]

If LEDs are in series (anode on 1st connected to cathode on 2nd etc.) then only one resistor is needed. But it won't be the same value as a resistor for a single LED. Each LED drops between ~2 (red, green & most colored LEDs) and ~3 volts (white and blue LEDs). The resistor only needs to limit current for the voltage left over after the LED voltage drops are added up.

This is one reason I bought the lab power supply: It lets me adjust both current and voltage accurately, so I can figure out things like this without worrying about letting the magic smoke out.  Mine is likely to be idle into mid-September if you want to borrow it.

[deemery]

If LEDs are in series (anode on 1st connected to cathode on 2nd etc.) then only one resistor is needed. But it won't be the same value as a resistor for a single LED. Each LED drops between ~2 (red, green & most colored LEDs) and ~3 volts (white and blue LEDs). The resistor only needs to limit current for the voltage left over after the LED voltage drops are added up.

This is one reason I bought the lab power supply: It lets me adjust both current and voltage accurately, so I can figure out things like this without worrying about letting the magic smoke out.  Mine is likely to be idle into mid-September if you want to borrow it.

I have a lab power supply, I just don't know all the ways to use it, or more importantly how to calculate the resistor value to prevent the magic smoke from escaping.  What I'm thinking right now for the machine shop is 3 pre-resistored larger LEDs over the shop area and a single smaller size LED without a resistor for the office.  Those 4 LEDs should work with my 12v aux power bus, as long as I wire them correctly.  I'll probably do a similar set-up for the 2nd floor space, too.

dave

[ACL1504]

Is this what you were looking for?

Tom

[deemery]

Is this what you were looking for?

Tom

Yeah, that works for current limiters, and a CL "eats" about 3 volts, apparently.  But if I put 3 LEDs on that 9V battery, with no current limiter but with the correct sized resistor, I should be able to drive all 3 LEDs that way.  

dave

[Bernd]

Here's how to figure out what resistor you need.

Ohm's Law

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage. This is true for many materials, over a wide range of voltages and currents, and the resistance and conductance of electronic components made from these materials remain constant. Ohm's Law is true for circuits that contain only resistive elements (no capacitors or inductors), regardless of whether the driving voltage or current is constant (DC) or time-varying (AC). It can be expressed using a number of equations, usually all three together, as shown below.

V = I × R

R =V/I

I =V/R

Where:

V is voltage in Volts
R is resistance in Ohms
I is current in Amperes

So if your LED draws .05 amps you dive the voltage (12volts) by how much current the LED draws and you come up with (240 ohms) the correct resistor to use. Always use the next size up. If you come up with an odd number like 975 ohms, use a 1K ohm resistor.

Now let's see how many LED's you can blow up.

Bernd

[deemery]

Here's how to figure out what resistor you need.

Ohm's Law

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage. This is true for many materials, over a wide range of voltages and currents, and the resistance and conductance of electronic components made from these materials remain constant. Ohm's Law is true for circuits that contain only resistive elements (no capacitors or inductors), regardless of whether the driving voltage or current is constant (DC) or time-varying (AC). It can be expressed using a number of equations, usually all three together, as shown below.

V = I × R

R =V/I

I =V/R

Where:

V is voltage in Volts
R is resistance in Ohms
I is current in Amperes

So if your LED draws .05 amps you dive the voltage (12volts) by how much current the LED draws and you come up with (240 ohms) the correct resistor to use. Always use the next size up. If you come up with an odd number like 975 ohms, use a 1K ohm resistor.

Now let's see how many LED's you can blow up.

Bernd

I guess the current draw for the LED should be on the LED package?  For 2 of the LED models I have, the spec sheet https://www.mouser.com/datasheet/2/143/17-21-BHC-AP1Q2-3T_datasheet-51933.pdf says "forward current 25ma, peak forward current 100ma"    For 4 LEDs, each at 25ma, that's 100ma or .1a.  R= V/I, R = 12v/.1A or 120 ohms?  

dave

[jbvb]

Usually, you want the steady-state current.  Peak inrush current is what won't *quite* let the magic smoke out.

If the LEDs are in parallel, the currents add up, but the voltage drop is that of a single LED.

If the LEDs are in series, the voltage required to light them all is the sum of the voltage drops of each LED. The current is that of a single LED.

[deemery]

Usually, you want the steady-state current.  Peak inrush current is what won't *quite* let the magic smoke out.

If the LEDs are in parallel, the currents add up, but the voltage drop is that of a single LED.

If the LEDs are in series, the voltage required to light them all is the sum of the voltage drops of each LED. The current is that of a single LED.

So if I have 4 pre-resistored LEDs in series, I can cut the resistor off of 3 of them and everything will work OK?

dave

[jbvb]

Usually, you want the steady-state current.  Peak inrush current is what won't *quite* let the magic smoke out.

If the LEDs are in parallel, the currents add up, but the voltage drop is that of a single LED.

If the LEDs are in series, the voltage required to light them all is the sum of the voltage drops of each LED. The current is that of a single LED.

So if I have 4 pre-resistored LEDs in series, I can cut the resistor off of 3 of them and everything will work OK?

dave

What I'd expect is I'd need to remove all 4 resistors, then add a different resistor in series with the 4 LEDs. Resistance calculated to pass the proper current after the voltage drop of the 4 LEDs is subtracted from the supply voltage.

[deemery]

Usually, you want the steady-state current.  Peak inrush current is what won't *quite* let the magic smoke out.

If the LEDs are in parallel, the currents add up, but the voltage drop is that of a single LED.

If the LEDs are in series, the voltage required to light them all is the sum of the voltage drops of each LED. The current is that of a single LED.

So if I have 4 pre-resistored LEDs in series, I can cut the resistor off of 3 of them and everything will work OK?

dave


What I'd expect is I'd need to remove all 4 resistors, then add a different resistor in series with the 4 LEDs. Resistance calculated to pass the proper current after the voltage drop of the 4 LEDs is subtracted from the supply voltage.

Now I'm back to being confused.  I thought each LED eats 3v, so 4 of them would totally consume 12v, so no resistor needed?  But how do I determine or measure the voltage drop for a single LED?

dave