LED driving and controlling methods

Home Contact Projects Experiments Circuits Theory BLOG PIC Tutorials Time for Science RSS Terms of services Privacy policy

Home

Projects

Experiments

Circuits

Theory

BLOG

PIC Tutorials

Time for Science

LED driving and controlling methods

Author
Giorgos Lazaridis
February 5, 2012

PAGE 2 of 6 - Single Transistor Constant Current Driver

Single Transistor Constant Current Driver

Before we begin let me inform you that i have upload an on-line calculator to speed-up your calculations for this circuit. You will find this calculator in the Dr.Calculus section of my site in this link.

Let's see some transistor theory... The following analysis is explained in details in my Transistor Theory page. You may wanna read this to have a better idea in the transistor operation. The circuit that we will use is a Common Emitter amplifier with voltage divider biasing and emitter feedback resistor.

From the theory we know that the base voltage V_Bof the transistor depends on the values of the resistors R1 and R2 which perform a voltage divider.

V_B = V_DD x R2 / ( R1 + R2 )

The voltage on the emitter is calculated by subtracting the V_BE voltage from the base voltage. A typical value for V_BE for silicon transistors is about 0.7 volts:

V_E = V_B - V_BE

The emitter current can be calculated with the Ohm's law:

I_E = V_E / R_E

Finally, from the transistor theory we know that I_C is almost equal to I_E. From this quick analysis we come to the conclusion that the current through the load (LED) is constant and it equals to:

I_C = I_E = V_E / R_E = ( V_B - V_BE ) / R_E

Circuit Considerations: The circuit works well for low current applications, but things get tougher as current increases. First of all, the resistor R_E must be properly selected to avoid excessive power dissipation. The amount of power that will be dissipated onto this resistor is calculated as follows:

P_RE = I²_C x R_E

Small resistor values should be selected, between 1 and 220 Ohms. But this brings up another problem: Transistor are temperature-sensitive devices. The current I_C increases with temperature. R_E provides a small feedback compensation, but since it is very small this compensation is usually inefficient. Another consideration is the power that the transistor is put to dissipate. From the Kirchhoff's law we can calculate the V_CE voltage:

V_CE = V_DD - V_{F_LED} - V_E

Now we can calculate the power dissipation on the transistor:

P_T = V_CE x I_C

Finally, one more consideration is the current amplification of the transistor. Since there is no AC voltage applied, we only need the DC equivalent to do the analysis. The current through the base must be sufficient to provide enough current for the LED. So, for the worst case scenario (lowest hfe for biggest I_C current), we can calculate the minimum I_B:

I_{B_Min} = IC / h_{fe_min}

The voltage divider must be either firm or stiff. This means that the base current must be 10 or 100 times smaller than the current through the voltage divider. The current through the voltage divider is:

I_VD = Vdd / (R1+R2)

So the condition is that I_{B_Min} < 0.01 x I_VD (or 0.1 for firm). If this condition is true, then the circuit will work properly.

An Example

This example made on a breadboard for test

Suppose that we want to control an LED that draws 20mA at 3.3 volts. We will use 12 Volts supply. First i will estimate the R_E with an educative guess. From experience i choose R_E = 47 Ohm. The power that will be dissipated on this resistor is:

P_RE = I²_C x R_E => P_RE = 0.02² x 47 = 18.8mW

A 1/4W resistor is more than enough for this application. To achieve I_C current of 20mA, the V_E must be:

V_B = I_C x R_E => V_B = 20mA x 47 Ohm => V_B = 940 mV

Therefore, the base voltage must be 0.94 + 0.7 = 1.64 Volts. Then, i use my on-line Voltage Divider Calculator to estimate a proper pair of R1-R2 resistors to get 1.64 Volts fro 12 Volts. After some trial-and-error, i decided to use R1=9.47 KOhms and R2 = 1 KOhms.

Let's test the current condition now for a firm voltage divider (x10). Suppose that we use the 2N2222 transistor which has minimum h_fe of 100. The minimum I_B must be 20mA / 100 = 0.2mA. The current through the voltage divider is:

I_VD = 12/(9470+1000) = 1.14 mA. The condition states that the base current must be 10 times smaller than this current, or else:

0.2 x 10 < 1.14 => 2 < 1.14 which is FALSE. So, we scale down both R1 and R2 by a factor of 10 and re-try:

R1 = 4735 Ohms
R2 = 750 Ohms
I_VD = 12 / 1150 = 2.188 mA

The previous condition is now TRUE, since 2 < 2.188. The closest values are R1=4.7K and R2=680 Ohm.

The circuit regulates the current of the LED at 19.8 mAmps. For this current, the voltage across the LED is 3.2 Volts. If another LED is placed instead, the same current will appear but the voltage may be different.

We need to make sure that the power dissipation on the transistor will not exceed the maximum rating (500mW). First we calculate the V_CE:

V_CE = V_DD - V_{f_LED} - V_E = 12 - 3.2 - 0.94 = 7.86 Volts

P_T = V_CE x I_C = 7.86 x 0.0198 = 155.6 mW

So, the circuit can operate normally without the smell of the toasting transistor. As for R_E:

P_RE = I²_C x R_E = 0.0198² x 47 = 18.4mW

ADVANTAGES:

It can handle higher power than a simple resistor - Depends on the transistor ratings, package and cooling.

Still it is cheap and simple

LED current is regulated regardless of the LED voltage

DISADVANTAGES:

Not very efficient since a lot of power is lost across the transistor

Unstable - A small change on the supply voltage will cause a large change on the LED current, since the base voltage from the voltage divider changes.

Any temperature change in the transistor package will result in output current change

	Continue reading. Click here to go to the next page >>>.
	OR click here to view the presentation.

Comments

At 19 October 2015, 14:13:48 user Jakoos wrote: [reply @ Jakoos]

At 19 October 2014, 16:40:28 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@Charles For 1W led better use a constant current driver, like the single transistor one.

At 18 October 2014, 2:28:07 user Charles wrote: [reply @ Charles]

I know most all of your suggestions talk about how the resister is a loss of power. If one was to use a buck converter to step the V down to 3.2, what option would be best to limit current to 1w LED's.

At 14 September 2014, 15:15:54 user Kothandapani wrote: [reply @ Kothandapani]

At 3 September 2014, 6:03:09 user duncan wrote: [reply @ duncan]

Just wanted to say thanks for sharing. Your presentations are very clear and informative.

At 2 August 2014, 2:56:09 user Ancel wrote: [reply @ Ancel]

@Giorgos Lazaridis Hi, I'd like to create a 7 segment constant current driver using these transistors. Is there a way to use a single resistor current reference or will I need one for each segment?

At 6 May 2014, 7:44:48 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@oliver You can connect the output of the arduino at an open-collector transistor circuit. Say a 1K resistor goes to the base of an NPN, the emitter goes to the ground and the collector goes to the input of this circuit. It will provide enough sink current.

At 20 April 2014, 16:23:39 user oliver wrote: [reply @ oliver]

At 18 April 2014, 10:51:40 user MaxiD wrote: [reply @ MaxiD]

I can put another transistor that controls the gnd bjt transistor for use with pwm?

At 22 March 2014, 16:13:13 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@Leon Not quite sure, sorry

At 18 March 2014, 19:55:22 user Leon wrote: [reply @ Leon]

can we apply this to operate a laser diode?

At 2 February 2014, 9:34:02 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@karthikeyan check out the datasheet of the LED, this is where you find the info

At 2 February 2014, 9:28:06 user karthikeyan wrote: [reply @ karthikeyan]

dear sir can you send to my email about ampere and voltage of one led

At 27 December 2013, 5:21:45 user thiru wrote: [reply @ thiru]

VR voltage 230vAc and LED voltage is 15vAc now resistor value? so please send the formula as soon as possible.

At 14 December 2013, 23:58:32 user fabelizer wrote: [reply @ fabelizer]

Very helpful! Thanks for all your work!

At 14 October 2013, 19:14:30 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@yony Simply provide PWM pulses to the proper PWM input of the circuits is the easiest way. Foe more sophisticated solutions you have to do the research yourself

At 14 October 2013, 13:44:32 user yony wrote: [reply @ yony]

At 16 June 2013, 8:03:14 user jai ochani wrote: [reply @ jai ochani]

please advice me the circuit diagrm for 12v out put, 2Amp led driver.

At 14 February 2013, 17:14:10 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@shamim get a n-channel

At 11 February 2013, 11:46:22 user shamim wrote: [reply @ shamim]

we wanted to drive a LED using Mosfet. what Mosfet to use???

At 16 January 2013, 15:39:39 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@Ian for 90% efficiency you will need other type of circuits, not a linear driver. An SMPS instead

At 14 January 2013, 13:22:32 user Ian wrote: [reply @ Ian]

At 14 January 2013, 9:17:48 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@Ian depends on the current and your efficiency requirements. Typically, on resistance of mosfets is very low. In many cases it is lower than 1Ohm (not high current mosfets). What current requirements you have? More than 1 amp?

At 11 January 2013, 17:49:19 user Ian wrote: [reply @ Ian]

Hi Giorgos, Thank you for your help with current measurement. Can you also tell me in the selection of a power MOSFET if Rds(on) should be as low as possible or doesn't it matter? I'm presuming the MOSFET is operating in saturation mode or would be if PWM was applied to the gate? Thank you, Ian

At 7 January 2013, 5:34:43 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@Ian I put the scale of the multimeter to amperes and connected it in series with the LED

At 6 January 2013, 22:01:19 user Ian wrote: [reply @ Ian]

Hello Giorgos, Could you tell me how you measured the current through the LEDs as shown in your photos please? Thank you, Ian

At 3 January 2013, 8:41:11 user adithyan wrote: [reply @ adithyan]

Circuits are plenty for pwm control of led brightness. It is agreed that there cannot be a linear control of led brightness. Assuming that the pwm controls applied and the led current exceeds the 20ma limit in case of 4mm and 5mm Leds and 350ma in one watt, how a feed back control either shuts off the led or keeps the led current at maximum is yet to be published

At 2 October 2012, 15:17:29 user Ajay wrote: [reply @ Ajay]

At 28 September 2012, 18:38:22 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@kwstas there is no "specific" number. I would not go above 3.3 volts though, just to maintain the power dissipation on the emitter resistor low.

At 27 September 2012, 7:44:27 user kwstas wrote: [reply @ kwstas]

@Giorgos Lazaridis Thank you very much but i'm asking about the zener diodes..

At 26 September 2012, 17:48:15 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@kwstas same diodes (1n4001 if i remember well or similar)

At 26 September 2012, 17:15:07 user kwstas wrote: [reply @ kwstas]

At 9 March 2012, 8:53:16 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@Ghlargh although i plan to use such drivers, i will not go into details.

At 9 March 2012, 8:46:17 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@Alvie it is the most stable of all. i tried it with 10w led. i prepare video for this.

At 9 March 2012, 7:46:04 user Ghlargh wrote: [reply @ Ghlargh]

You should, if possible, explain how multiple constant current drivers with a single reference resistor work. Such as the Allegro A6282

At 8 March 2012, 22:13:28 user Alvie wrote: [reply @ Alvie]

At 7 March 2012, 20:10:23 user Daniel wrote: [reply @ Daniel]

At 26 February 2012, 6:32:30 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]

@Lupin Thank you very much for pinpointing those mistakes. You are right for both. The first one is already corrected, as for the second i will recompile the video. Thanks for noticing it soon.

At 25 February 2012, 22:54:05 user Lupin wrote: [reply @ Lupin]