30ma SMD LED Resistor

I did not know how to word the title but here is the issue. A member made this for me and said it will allow only 10ma to power the led regardless of the voltage. My project was a Buck converter with an input of 4.5-24 volts so that's the voltage range. Well I tested this on a breadboard with a led that had a 1.9fv. the mA must have been around 10-15 because it worked. My finished project is all SMD. The SMD LED is much higher 3.0v @ 30mA and it does not work in the circuit. I am pretty sure, OK absolutely sure, that the R2 needs to be lower than 62R. I just don't know what the formula is to figure this out. Thats what I need help with.

I also want the group to know that it has been 3 years since I asked for help with this BUCK converter project. I did have the board made and I soldered one up. My soldering is embarrassing, I forgot to order the correct pot, so I will do another after I solve the LED issue. Oh and the picture sucks too.

What works is when you have a problem to report all the node voltages clearly with labels or Ref Des. Even if you pencil on a schematic is OK.

Your LED may also be burnt out.

This is an active current limiter cct. Ic~0.62/Re= ~ 10mA depending on temperature and Q2 Vbe & base current.

The setpoint is Ic= Vbe(Q2)/R1 but Q1 needs to have enough current gain to work. When Q1 Vce = 0.7V is in the saturation region with low Vcc, hFE drops to about 10% of the max linear hFE.

Check that Vbe~ 0.6 with Ib=400 uA thus record Q1 Vc,Ve and Vf of diode.

Tony Stewart said:

What works is when you have a problem to report all the node voltages clearly with labels or Ref Des. Even if you pencil on a schematic is OK.

Your LED may also be burnt out.

This is an active current limiter cct. Ic~0.62/Re= ~ 10mA depending on temperature and Q2 Vbe & base current.

The setpoint is Ic= Vbe(Q2)/R1 but Q1 needs to have enough current gain to work. When Q1 Vce = 0.7V is in the saturation region with low Vcc, hFE drops to about 10% of the max linear hFE.

Check that Vbe~ 0.6 with Ib=400 uA thus record Q1 Vc,Ve and Vf of diode.

Click to expand...

Thank you. I understand most of what you replied and I will work on the node voltages. To clear up a few things. I believe the LED does work. But if you think the LED may be burnt then are you also saying that the 3.3fv 30mA LED should be working in this circuit? If that is accurate I could just change the LED however while it is difficult for me to remove the LED I did probe it with the MM on the board in diode mode. It read 3.0 and the LED glowed. So that led me to believe the LED is not burnt.

I took a white through hole led with a fv of 1.9 and just held it in place on the board over the SMD led and the white TH worked. It worked as expected as I adjusted the voltage up and down.

Here are the voltage readings taken at every node. I apologize if I missed any.

I think there is no connection between the collector of Q1 and the LED, or no connection between the Q1 and the positive supply, or the LED is connected the wrong way round. The circuit diagram you have posted will work if everything is connected as shown and all components work.

You haven't said what the voltage on the collector of Q1 is.

From the voltages on your circuit, there is 1.05 mA flowing through R1 and through R2. Q2 is not taking any current because there is not enough voltage between base and emitter to make it conduct. So Q2 is doing the right thing as the current is so much less than the design current.

The current flowing in R1 and R2 should therefore be flowing through the base of Q1. Q1 has a gain of about 100 so 1.05 mA in the base would allow 100 mA in the collector, but there fact that the base and emitter currents are the same means that there is no collector current. With no collector current and 1 mA base current, Q1 is saturated (turned fully on) and the collector voltage should be only a tiny bit higher than the base voltage.

That gives over 11 V across the LED and no current through the LED, which is what leads me to think that there is a problem in that area.

Diver300 said:

You haven't said what the voltage on the collector of Q1 is.

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Thank you so much. The reading at Q1's collector is .076 v

ThomsCircuit said:

Thank you so much. The reading at Q1's collector is .076 v

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That shows that the transistors and resistors part is doing what it should. That leaves just over 11 V across the LED, no current flowing and no light.

You said that the LED glows at 3.0 V, so it must be not connected or the wrong way round.

Oh my gosh! The LED was backwards.

Diver300 said:

That shows that the transistors and resistors part is doing what it should. That leaves just over 11 V across the LED, no current flowing and no light.

You said that the LED glows at 3.0 V, so it must be not connected or the wrong way round.

Click to expand...

You are the smartest! I was misinterpreting the arrow on the bottom of the component.

For future reference there are constant current diodes out there,
simple, that are simple to use. See attached. I think its more accurate
over T as well. Note part bottoms out as 15 mA lowest available
current version.

Your design, over 20 - 50 Centigrade Temperature, with a 3.4V LED :

danadak said:

Your design, over 20 - 50 Centigrade

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Yep. The light is way to bright. Its more like a flashlight.

ThomsCircuit said:

Yep. The light is way to bright. Its more like a flashlight.

Click to expand...

If it's too bright, increase the value of R2, the circuit is a simple (and common) constant current source, with the current set by R2 - 62 ohm should give about 10-12mA.

"it is difficult for me to remove the LED"
Did you succeed? in reversing the wires?

LED's have a weakness to failure when the voltage is reversed more than 5V, like trying to reverse a finger.
The datasheet on most LEDs dictate the absolute maximum is -5V and this is also true of most transistor base-emitter voltages. That means it may or may not break with the same current in reverse.

White LEDs are tested to leak 10 uA max at -5V, but beyond this voltage accelerates rapidly and that is bad news.


When Q2-b = Q1-e is 0.65 , then Q1 is in active current limiting mode limited by Q2 which shunts Q2's base current by this voltage.

We can use Ohm's Law to see if the transistor and LED are working by looking at the voltage drop on each part.

Since Re emitter is also in series, then we can compute the current within some tolerances.

From your info for Q1 using Ohm's Law;
Q1-c = 0.76 V
Q1-e = 0.65 V thus C-E or Vce = 0.11 V or 110 mV.
Ie=Ve/Re = 0.65 V / 62 Ohms or 650 mV/62 is about 10 mA
This is alot more than the estimate base current of 11V/10k = 1mA.
The good news is a good switch design uses Ic/Ib= 10 as this does 10mA/1mA.
It also means Q1 working as a switch Vce(sat)/Ic = Rce = 110 mV / 10 mA = 11 Ohms

But the LED drop must never exceed 5V and now had -11V across the LED. Oh oh.


So hopefully you surgically reversed the wires to the LED and it still works, I think you will be lucky this time.
But there is a risk it can be partly damaged inside.

In future verify your installation with the Diode test mode or the datasheet mechanical diagram.

Tony Stewart said:

"it is difficult for me to remove the LED"
Did you succeed? in reversing the wires?

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Yes, I sucked up the little solder there was then carefully disconnected the LED. It took me about 10 minutes. Most of that time was trying to align and solder on the new one.