# Circuit for LED flashlight

#### PsySc0rpi0n

Joined Mar 4, 2014
1,776
Hello. Fowllowing the discussion here and while I'm waiting for my UV LEDs, I'm using the time, to build at least 2 white LED light sources to use with my small USB microscope, because the LEDs that comes with the microscope, reflect a lot in the components under observation. So, using external light source, might improve the experience with reduced reflection from the LEDs, by controlling the angle of the LED light external sources.

We spoke about using a constant current source to provide constant current to the LEDs and have the best match regarding the light each LED will provide.

I want to ask a couple of things regarding the circuit I want to use to feed the LEDs.

I came up with the following circuit (file attached):

I started by setting the LEDs current to 60mA, 20mA to each one, as they are rated at 30mA.
But I think I'm not doing this correctly and I want to know where I am going wrong.
So, knowing that I'm using a 12V PSU and:
$$\displaystyle{V_{LED}=5V}$$
I'm using a BC548B with the following characteristics given by those Chinese transistor analysers:
hFE = 311
Vbe = 0.666V

9V Zener reverse bias Voltage

I would start by writing that:
$$\displaystyle{I_{E} = \left (\beta + 1\right )\cdot I_{b}}$$
$$\displaystyle{I_{Load} = I_{C} = \left (\beta + 1\right )\cdot I_{b}}$$
$$\displaystyle{I_{Load} = \beta \cdot \frac{V_{E}}{\left (\beta + 1 \right )\cdot R_{E}} }$$
$$\displaystyle{I_{Load} = \frac{V_{b} - 0.666}{R_{E}}}$$

So, the Emitter resistance would be:
$$\displaystyle{R_{E} = \frac{V_{b} - 0.666}{I_{Load}}}$$

Emitter resistance would be
$$\displaystyle{R_{E} = \frac{9V - 0.666}{60mA} = 140.56Ω}$$
for a 9V zener.

I'll keep going with the rest of the math, while I wait for confirmation!

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#### ericgibbs

Joined Jan 29, 2010
18,997
Hi Psy,
What is the forward voltage drop of your White LED, typically they are ~3.2V.?
E

#### JohnnyC1951

Joined Jun 27, 2020
22
As a side point, maybe you could consider feeding each individual LED from it's own resistor? Due to small variations between LEDs, it is possible that one or more will 'hog' the current and giving uneven illumination.

#### ericgibbs

Joined Jan 29, 2010
18,997
Hi Psy,
With your posted circuit, this is what LTspice shows.
ie: 15mA divided into 3 LED's ~5mA/LED
E

#### djsfantasi

Joined Apr 11, 2010
9,183
Hi Psy,
What is the forward voltage drop of your White LED, typically they are ~3.2V.?
E
The TS mentioned that Vled was 5V. Which is an unusual value, unless the LEDs chosen were made for 5V and have an internal resistor.

#### ericgibbs

Joined Jan 29, 2010
18,997
hi Psy,
IF the LED's are 3.2Vfwd, this is a simple option,
E

Updated:
Uses two diodes

Last edited:

#### PsySc0rpi0n

Joined Mar 4, 2014
1,776
Hi Psy,
What is the forward voltage drop of your White LED, typically they are ~3.2V.?
E
Aparently I always find ways of screwing things up

See, I found this as one of the first links I got from a web search

So, I assumed these LEDs would drop 5V.

As for your LTSpice circuit, I'm not using LTSpice anymore because I don't want to have to install Wine in my Debian.
That's why I'm using falstad simulator. I entered some values manually in falstad, such as transistor Vbe and hFE.

The TS mentioned that Vled was 5V. Which is an unusual value, unless the LEDs chosen were made for 5V and have an internal resistor.
Well, seems that I got it wrong from some datasheet. See my quote above to @ericgibbs

#### PsySc0rpi0n

Joined Mar 4, 2014
1,776
hi Psy,
IF the LED's are 3.2Vfwd, this is a simple option,
E
I don't have 2V5 zeners at hand. Only 4V7 (which I think they are regulating at 4v1 instead) and 9V1 zeners!

@ericgibbs I measured the votlage drop across the LEDs, and I got ~2.8V
However, I can't get the 9V1 at the zener. I measure around 7V. No idea why!
And I wanted to follow up with the math! Because if I don't follow the math, I don't understand how to chose the values for resistors or how to get the equations out of the circuit!

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#### bertus

Joined Apr 5, 2008
22,289
Hello,

The line you show in the datasheet is the REVERSE voltage.
You need the FORWARD voltage for the calculations.

Bertus

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#### ericgibbs

Joined Jan 29, 2010
18,997
hi Psy,
Relook at post #6, using diodes not a 2.5Vref.
E

#### PsySc0rpi0n

Joined Mar 4, 2014
1,776
Hello,

The line you show in the datasheet is the REVERSE voltage.
You need the FORWARD voltage for the calculations.
View attachment 258437
Bertus
I overlooked the following pages on that link. You are absolutely right!

hi Psy,
Relook at post #6, using diodes not a 2.5Vref.
E
Yes, but that also means a 3 * V_Led voltage drop! Is it still a better solution than using LEDs in parallel? Another thing is that Re will take some power to dissipate!

#### Audioguru again

Joined Oct 21, 2019
6,758
You might have the collector and emitter of the transistor reversed then its base current is way too high which reduces the base voltage. The pins of a European BC548 are CBE but the pins of American 2Nxxxx transistors is EBC.

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#### ericgibbs

Joined Jan 29, 2010
18,997
hi Psy
So 3 *3.2V = 9.6V, leaving 12V-9.6V = 2.4V across the transistor and emitter resistor.

Power dissipation in the 33R is ~ 13mWatt!

The series circuit, more efficient than the parallel version
E

Update:
You could use a 100R pot to control the brightness as it's going to be used on a microscope.

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#### PsySc0rpi0n

Joined Mar 4, 2014
1,776
hi Psy
So 3 *3.2V = 9.6V, leaving 12V-9.6V = 2.4V across the transistor and emitter resistor.

Power dissipation in the 33R is ~ 13mWatt!

The series circuit, more efficient than the parallel version
E

Update:
You could use a 100R pot to control the brightness as it's going to be used on a microscope.
Instead of those diodes, aren't Zeners better as they will make the base voltage constant? I don't have any other diodes other than 1N4007.

#### ericgibbs

Joined Jan 29, 2010
18,997
Hi Psy,
We could analyse your original circuit if you would like to do so.

As a first step, disconnect the Zener and measure the Base voltage.??

E

#### PsySc0rpi0n

Joined Mar 4, 2014
1,776
Hi Psy,
We could analyse your original circuit if you would like to do so.

As a first step, disconnect the Zener and measure the Base voltage.??

E
I tell you what, if you allow me!
Let's discuss your version first, because I like the rpot thing. It's actually handy the rpot.
And after we agree with the circuit components I have available, if possible, then we deal with the math.
Is it ok?

So, for the diodes. 2 Questions:
1 - Are the diodes a better solution than a Zener? Why?
2 - If they are better than the Zener, could I get away with the 1N4007 as these are the only ones I have with me?

#### ericgibbs

Joined Jan 29, 2010
18,997
hi Psy,

1 - Are the diodes a better solution than a Zener? Why?
We only require a Base voltage in the order of 1.4V, so two diodes at 0.7V each give us that voltage.

2 - If they are better than the Zener, could I get away with the 1N4007 as these are the only ones I have with me?
Yes, you can use those 1N4007.

E

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#### BobTPH

Joined Jun 5, 2013
9,148
You are way overthinking this. Put them in series with a 120Ω resistor. For this low power, anything else is silly.

Bob

#### JohnnyC1951

Joined Jun 27, 2020
22
>> For this low power, anything else is silly.
Especially if this is an alkaline or lithium battery which will be pretty self regulated anyway..

#### PsySc0rpi0n

Joined Mar 4, 2014
1,776
hi Psy,

1 - Are the diodes a better solution than a Zener? Why?
We only require a Base voltage in the order of 1.4V, so two diodes at 0.7V each give us that voltage.

2 - If they are better than the Zener, could I get away with the 1N4007 as these are the only ones I have with me?
Yes, you can use those 1N4007.

E
Ok, I see you changed Re = 33Ω. Any reason for this?

You are way overthinking this. Put them in series with a 120Ω resistor. For this low power, anything else is silly.

Bob
Nevermind. It's a way of learning beyond academic stuff!