# Current sources

The current sources are electronic circuits that provide a constant current. In this post, are shown some examples.

## Ideal and real current sources

Ideally, a current source provides constant current independent of load resistance or impedance. In practice, the current source has a parallel resistance that is very high, but not infinite. Therefore, the provided current will decrease with the increase of voltage drop on load, due to impedance increase.

## Why use current sources?

In many situations, it’s necessary to provide a constant current independent of voltage drop on load. Some application examples:

• Batteries are recharged with current sources.
• For resistive heating systems to keep a constant temperature, the current also must be constant.
• LEDs that need to keep the brightness use this type of source.

## The simplest current source

The simplest current source consists of a voltage source in series with a resistor.

This configuration is an economical solution to charge batteries. However, considerable power is wasted by the series resistor. In addition to this, the current changes when load impedance changes.

## Designing some current sources

### Current source with voltage regulator

The voltage regulator LM317T can be used as a constant current source.

Warning: to this circuit to work, you must use LM317T, not LM317. Otherwise, the output current will change if voltage input Vin changes.

The equation to calculate current output $I_{o}$.

$I_{o}=\frac{1.25}{R}$

The R value that I choose was 1 kΩ. Therefore, $I_{o}$ must be 1.25 mA.

### Current mirror

In this circuit, the current that passes through R resistor has the same value as current I through T2’s collector. In this post, are shown current mirrors which use BJT transistors, this component’s operation is in the following link.

To avoid burning transistors it’s necessary to put a resistor between ground and emitter terminal in each transistor.

• Both Re1 and Re2 are 68 Ω.
• The value Vcc from the source is 12 V.
•  T1 and T2 are TIP122. Because lower power transistors will heat up too much, could burn.
• I current must be 30 mA.

Calculating R using Kirchhoff’s voltage sum law. The base-emitter voltage $Vbe$ is 0.7 V.

$R+Re=\frac{Vcc-Vbe}{Io}$

$R+68=\frac{12-0.7}{30m}$

$R=308\Omega$

Was used the commercial value 330 Ω.

Designing a current mirror with pnp. This time the used model is TIP127.

• The supply voltage is 10 V.
• Resistors on emitter terminals have 150 Ω.
• For Rc, I choose the commercial value 470 Ω.

Calculating I2 current.

$V=I2\cdot Re1+Vbe+I2\cdot Rc$

$10=I2\cdot 150+0,7+I2\cdot 470$

$I2=15mA$