  LED driving and controlling methods AuthorGiorgos LazaridisFebruary 5, 2012

PAGE 3 of 6 - Single Transistor Constant Current Driver with voltage regulation      Single Transistor Constant Current Driver with voltage regulation

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.

The previous circuit works very well, but it has one serious disadvantage: any change at the supply voltage will cause a change at the LED current. That is because the base voltage will follow the supply voltage.

A small change can make the previous circuit more stable. The simplest and most logical is to add a zener diode to maintain the base voltage constant. Here is the circuit:

The only thing that changes here is the biasing method. Here is how it works: The Zener diode maintains a fixed voltage at the base of the transistor, regardless of the supply voltage. Of course the supply voltage must be higher than the Zener voltage (plus the voltage drop across RB). As a result, the voltage at the emitter will also be fixed, and equal with VB - VBE. Thus, the emitter current will be constant (fixed) since RE does not change.

This solution is also very simple to calculate the biasing components. The emitter base is calculated with the following formula:

VE = VZ - VBE

VBE is typical 0.7 for silicon and 0.3 for germanium transistors. The emitter current (and therefore the collector current) is defined by the resistor RE. Using the Ohm's law we define RE:

RE = VE / IE

That's it actually! RB is now easily defined as a result of the transistor's current amplification and the required collector current. Using the worst case scenario (lowest hfe for highest collector current), we can calculate the base current:

IB = IC / hfe_min

Do not forget to add some 1-2 mA for the Zener to achieve proper regulation. Usually, 1mA is enough. So with the Ohm's law we can approximate a value for RB:

RB = (VDD-VZ) / (IB + 1mA)

Now we only need to calculate the power dissipation on the components to see if we are out of limits. First, the dissipation on the transistor. To calculate this, we need to know the forward voltage across the LED for the specific current setup. ATTENTION!!! A 3.2 volts LED at 30mA will NOT have 3.2 volts drop if it is driven with less current, for example 20 mA!!! Here is how to calculate the power dissipation on the transistor. First we need to calculate VCE:

VCE = VDD - Vf_LED - VE

The power dissipation on the transistor is:

PT = VCE x IC

For the resistor RE we simply apply this formula:

PRE = I2C x RE

Now we must calculate the base resistor power dissipation. We will use the Ohm's law to calculate the maximum current through this resistor first:

IRB_MAX = (VDD_MAX - VZ) / RB

We use the maximum supply voltage possible for VDD_MAX. Now we can apply the I2R formula to calculate the power dissipation on the resistor. If you want to be academically correct, you need to add the base current to the IRB_MAX, but since this is usually very small you can omit it:

PRB = I2RB_MAX x RB

To calculate the power dissipation on the Zener diode, we only need to multiply the zener voltage by the maximum current through the base resistor.

PZ = VZ x IRB_MAX

An Example The circuit on a breadboard for test. Two LEDs are connected in series
This time we will drive 2 LEDs connected in series with 20mA. The LEDs are completely different, so we will expect to see different forward voltages for the same current. Instead of a voltage divider, we will use a Zener diode with voltage 5.1 Volts. The transistor is a 2N2222 with hfe=100. The supply voltage is 12 Volts.

Since the Zener regulates the base voltage at 5.1V, the emitter voltage will be 5.1-0,7 = 4.4 Volts. To regulate the LED current at 20mA (20 mA = 0.02 A), we need to calculate the resistor RE for this current:

RE = VE / IE => RE = 4.4 / 0.02 => RE = 220 Ohm

As for the base resistor, first we need to calculate the base current. I use the 2N2222 which has a typical hfe 100. So the base current is 20/100 = 0.2mA. I add 2mA more for the Zener diode, so:

RB = (12 - 5.1) / 2.2 = 3.1 KOhm

I will use a 2.2K resistor instead, to be on the safe side. Here are the results: The current through the transistor is 20.4mA. Notice that the voltage across the white LED is 3.08 Volts while on the green LED is 2.27 Volts.

Let's calculate now the power dissipation on the transistor. The VCE is:

VCE = VDD - Vf_LED1 - Vf_LED2 - VE = 12 - 3.08 - 2.27 - 4.4 = 2.25 Volts

PT = VCE x IC = 2.25 x 0.02 = 45 mW

Comparing with the previous circuit, it is obvious that this one has 3.5 times less power dissipation on the transistor. The reason is not the circuit itself, it is the second LED! This LED has an additional voltage drop, therefore the voltage drop across the transistor (VCE) is decreased. Let's see the power dissipation on RE:

PRE = I2C x RE = 0.022 x 220 = 88mW

Obviously the power that dissipates on the emitter resistor has radically increased, and that is due to the higher base voltage of the transistor. To avoid such problems it is recommended to use as small Zener voltage as possible.

We need now to calculate the base resistor power dissipation. First we calculate the maximum current through the resistor:

IRB_Max = (VDD_MAX - VZ) / RB = (12 - 5.1) / 2200 = 3.1 mA

The power dissipation on the resistor is:

PRB = I2RB_Max x RB = 3.12 x 2200 = 21.1 mWatts

For the Zener diode we have:

PZ = VZ x IRB_Max = 5.1 x 3.1 = 15.81 mWatts

An alternative to the Zener diode

I found an interesting alternative to the Zener diode here. That guy uses two diodes in series instead of a Zener diode to get a stable base voltage. But does it work? As a matter of fact, yes - almost! The theory behind this is simple: A diode has a typical 0.7 minimum forward voltage to turn on. This voltage appears as a voltage drop across its leads. Actually it is the same voltage as the transistor VBE voltage. By adding 2 diodes in series, the base voltage is regulated at 2x0.7 = 1.4 volts. This voltage is maintained stable regardless of the supply voltage.
You can connect as many diodes in series as you want. Each diode adds another 0.7 volts to the base voltage. Of course, one diode won't work, because the emitter voltage will then be 0. This method is perfect to maintain a low emitter voltage and therefore having low power dissipation on the emitter resistor. The theoretical analysis is the same as if a Zener diode was used, with Zener voltage equal to 0.7 multiplied by the number of diodes in series.
One big disadvantage that this circuit has is that it cannot maintain a very stable current if the supply voltage is changed. That is mainly because the emitter resistor is very small. As we know from the BJT transistor theory, the bigger the emitter resistor, the better the stability.

An Example
We will drive an LED with constant current 30mA powered from a 12V source. Instead of a Zener we will use two diodes connected in series to achieve 1.4 volts at the transistor base. This will result in 0.7 volts at the emitter of the transistor. Therefore, i need a 23.3 Ohms emitter resistor to achieve 30mA. Since i do not have such a resistor, i will use two 47 Ohm resistors in parallel with total resistance of 23.5 Ohms:  This is the schematic diagram of the circuit It works just fine!

The voltage across the LED is 3.3 volts (measured with a multimeter). So the VCE is 12-3.3-0.7=8 volts. So the power dissipation on the transistor is:

PT = 8 x 0.03 = 240 mW.

This is quite a high power for a small TO92 transistor to dissipate without heatsink. For ambient temperatures bellow 20o the circuit might operate normally, although the transistor will heat up.

A more extreme example: Driving a 1 Watt and a 3 Watts LED
All these 3 last circuits with the single transistor driver (including the one with the voltage divider) are capable to drive high power LEDs. You'll need to use a more powerful transistor though, probably some kind of a Darlington pair to provide all the current needed. For this reason, i used the mighty TIP142 power darlington pair transistor. Keep in mind that a darlington transistors have doubled VBE voltage. It is a typical of 1.4 instead of 0.7.

One more point to take into account is the power dissipation. A 1 Watt LED with 3.4 volts forward voltage will draw about 300mA, and a 3Watts LED will draw about 1 ampere. This means that if the transistor is called to drop 7 volts, it will dissipate 7 Watts of power! It will get really really HOT in seconds, so hot that it can melt the breadboard.

The best thing to solve this problem is to use lower voltage for the circuit. I used 7 volts instead of 12 that i used for the previous experiments. Additionally, a heatsink on the transistor is a good idea. Finally, for the 3Watts LED, i used 5Watts sense resistors. You can use 1 Watt resistors as well, but i did not have them in stock at the moment. To keep the dissipation on the resistor low, i used the circuit with the diodes in series. But this time i used 3 diodes in series to have a 2.1 volts at the base and therefore 0.7 volts at the emitter (2.4-1.7).   Setup for 1 Watt LED: VDD: 7 Volts Base voltage: 2.1 Volts with 3 diodes in series Emitter voltage: 0.7 Volts Sense resistor: 2 Ohms Measured LED current: 270mA Setup for 3 Watts LED: VDD: 7 Volts Base voltage: 2.1 Volts with 3 diodes in series Emitter voltage: 0.7 Volts Sense resistor: 0.7 Ohms (measured with ohm-meter) Measured LED current: 950mA The circuit under 24-hours test: Worked like charm.

• It can handle higher power than a simple resistor - Depends on the transistor ratings, package and cooling.
• The current is kept stable regardless of the voltage supply

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