  LED driving and controlling methods AuthorGiorgos LazaridisFebruary 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 VBof the transistor depends on the values of the resistors R1 and R2 which perform a voltage divider.

VB = VDD x R2 / ( R1 + R2 )

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

VE = VB - VBE

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

IE = VE / RE

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

IC = IE = VE / RE = ( VB - VBE ) / RE

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

PRE = I2C x RE

Small resistor values should be selected, between 1 and 220 Ohms. But this brings up another problem: Transistor are temperature-sensitive devices. The current IC increases with temperature. RE 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 VCE voltage:

VCE = VDD - VF_LED - VE

Now we can calculate the power dissipation on the transistor:

PT = VCE x IC

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 IC current), we can calculate the minimum IB:

IB_Min = IC / hfe_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:

IVD = Vdd / (R1+R2)

So the condition is that IB_Min < 0.01 x IVD (or 0.1 for firm). If this condition is true, then the circuit will work properly.

An Example

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 RE with an educative guess. From experience i choose RE = 47 Ohm. The power that will be dissipated on this resistor is:

PRE = I2C x RE => PRE = 0.022 x 47 = 18.8mW

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

VB = IC x RE => VB = 20mA x 47 Ohm => VB = 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 hfe of 100. The minimum IB must be 20mA / 100 = 0.2mA. The current through the voltage divider is:

IVD = 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
IVD = 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 VCE:

VCE = VDD - Vf_LED - VE = 12 - 3.2 - 0.94 = 7.86 Volts

PT = VCE x IC = 7.86 x 0.0198 = 155.6 mW

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

PRE = I2C x RE = 0.01982 x 47 = 18.4mW

• 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

• 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.

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