PAGE 4 of 6 - Transistor - MOSFET Constant Current Driver
Transistor - MOSFET Constant Current Driver
Like the single transistor current regulators, this type of LED driver is also a linear supply. Some additional advantages have to do mainly with the circuit's efficiency, which many times is not a balancing factor at all. Here is a typical circuit with two transistor, one being a MOSFET:
If you've learned how the single transistor driver works, then this circuit is very simple to understand. When power is applied, the gate resistor RG turns on the MOSFET. This allows current to run through the LED, the MOSFET and the sensing resistor RS. As current increases, the voltage drop across RS is increased as well. When this voltage drop reaches the base-emitter voltage of the transistor (VBE=0.7V), the transistor is turned on. This will pull the MOSFET's gate to the ground turning it off. Therefore, the current through the LED is regulated to the one defined by the resistor RS.
The calculations for this circuit are very simple. To choose an RS resistor you only need to solve this formula:
RS = 0.7 / If_LED
The Voltage portion of the previous formula is always 0.7 Volts. This value is derived from the transistor's VBE contact.
Like before, the power dissipation on the MOSFET is calculate by the voltage across the Drain-Source multiplied with the current. The voltage across the MOSFET is calculated with this formula:
VM = VDD - Vf_LED - VRS
The voltage drop across the sensing resistor is always 0.7, because the BJT transistor will not let the voltage drop to go higher than the VBE. Therefore, we can simplify the above formula to:
VM = VDD - Vf_LED - 0.7
The power dissipation on the MOSFET is:
PT = VM x If_LED
A few words about the Gate resistor RG. Since MOSFETS are voltage components (unlike BJT transistors that are current components), this resistor is not very important to have an exact value. Make sure that it is big enough not to damage the transistor, but it is not too big. Roughly, the resistor can be calculated to provide about 0.8 to 1 mA of current.
Let's see a simple example. We will drive an LED with 30mA. The supply voltage is 12 volts. We start with the sense resistor. Since we need 30mA, we simply apply the numbers to the formula:
RS = 0.7 / If_LEF = 0.7 / 0.03 => RS = 23.3 Ohms
As for the Gate resistor, i will use a 10K resistor. The current will be 12/10000=1.2mA, but as i said this value is not critical. Just keep it low. For the power dissipation, we need to know the voltage drop of the LED for the specific current. Since i do not have the manufacturer's datasheet for this LED, i can only measure it with the voltmeter. It is 3.2 Volts. So we can now calculate the voltage across the MOSFET:
VM = VDD - Vf_LED - 0.7 = 12-3.2 - 0.7 => VM = 8.1 Volts
The power dissipation is:
PT = VM x If_LED = 8.1 x 0.03 = 243 mW
Here is the circuit mounted on a breadboard for test:
With the same methodology, we can adjust the circuit to provide about 300mA for a high brightness LED. The sense resistor is set to 2.5 Ohms (combination of 1 Ohm resistors)
Here are both circuits with probes. The orange (big) multimeter measures the current (left image is set to 200mA scale, right image is set to 10A scale). The black multimeter in the middle measures the voltage across the LED. The yellow multimeter on the right side measures the voltage across the MOSFET:
The sense resistor dissipates less power due to the low voltage (0.7V)
The current is kept stable regardless of the voltage supply
The MOSFET can deliver more current with less I2R losses
The system has much higher efficiency since the gate resistor needs to deliver only a small current
More stable in temperature changes
Not very efficient since a lot of power is lost across the MOSFET
|At 22 March 2014, 9:13:13 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]|
@Leon Not quite sure, sorry
At 18 March 2014, 12:55:22 user Leon wrote: [reply @ Leon]
can we apply this to operate a laser diode?
At 2 February 2014, 1:34:02 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, 1:28:06 user karthikeyan wrote: [reply @ karthikeyan]
dear sir can you send to my email about ampere and voltage of one led
At 26 December 2013, 21:21:45 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, 15:58:32 user fabelizer wrote: [reply @ fabelizer]
Very helpful! Thanks for all your work!
At 14 October 2013, 12:14:30 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, 6:44:32 user yony wrote: [reply @ yony]
Can you please show how can these circuits be incorporated with a microcontroller?
I'm in a search for an efficient way to drive a 1 watt LED via Arduino (right now I'm using a non-efficient way to do this with a TIP120 transistor and 2 resistors which one of them needs to be at least 1 watt resistor).
At 16 June 2013, 1:03:14 user jai ochani wrote: [reply @ jai ochani]
please advice me the circuit diagrm for 12v out put, 2Amp led driver.
At 14 February 2013, 9:14:10 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]
At 11 February 2013, 3:46:22 user shamim wrote: [reply @ shamim]
we wanted to drive a LED using Mosfet. what Mosfet to use???
At 16 January 2013, 7:39:39 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, 5:22:32 user Ian wrote: [reply @ Ian]
I'm wishing to use currents between 10mA and 1A depending upon the type of LED connected. I'd like the efficiency to be 90% plus if possible. I've seen Rds(on) values as low as 0.021 ohm, which I'd guess would translate into less power losses?
I'm guessing that from a 5v PIC micro that a logic level mosfet is the only way to go as opposed to a normal mosfet?
Thank you, Ian
At 14 January 2013, 1:17:48 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, 9:49:19 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 6 January 2013, 21:34:43 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, 14:01:19 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, 0:41:11 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, 8:17:29 user Ajay wrote: [reply @ Ajay]
Pls. give detail related to LED Driver Design.
At 28 September 2012, 11:38:22 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, 0:44:27 user kwstas wrote: [reply @ kwstas]
@Giorgos Lazaridis Thank you very much but i'm asking about the zener diodes..
At 26 September 2012, 10:48:15 user Giorgos Lazaridis wrote: [reply @ Giorgos Lazaridis]
@kwstas same diodes (1n4001 if i remember well or similar)
At 26 September 2012, 10:15:07 user kwstas wrote: [reply @ kwstas]
I would like to tell me what diodes did you use for the 1w Led.
At 9 March 2012, 0:53:16 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, 0:46:17 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 8 March 2012, 23:46:04 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, 14:13:28 user Alvie wrote: [reply @ Alvie]
Regarding the BJT MosFET circuit:
Can you prove it's stability ? Just by looking at the diagram I think it might oscillate, eventually at high frequencies, and eventually amplify the thermal noise.
At 7 March 2012, 12:10:23 user Daniel wrote: [reply @ Daniel]
hey you should take a look at how leds could be used as sensors(ldrs)
At 25 February 2012, 22:32:30 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, 14:54:05 user Lupin wrote: [reply @ Lupin]
Just wanted to mention, tha in the example it seems you forgot to subtract V_Z in the calculation of I_RB_MAX.
On the other hand, I got an issue on how you calculated P_RB in the video (around 5:50). You added the base current to I_RB_MAX. Isn't that already included?
Still many thanks for all the explanations and videos! I'm teaching a beginners electronics course (diodes, transistors, opamps) next semester at uni. Teaching applications, when you got to do theory first always falls short. I guess I'll refer some students to your page if they want to know more.
HOT in heaven!