  LED driving and controlling methods AuthorGiorgos LazaridisFebruary 5, 2012

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.

An Example
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

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