In many applications, a constant voltage source is required. However, there are more applications of constant current sources than constant voltage sources. Most power sources are voltage sources, including batteries and bench power.
A constant current source can be used anywhere a fixed current is needed. A resistor can also be used, but the current varies depending on the voltage. This would lead to noticeable variations in the output current.
Current sources are never truly constant as they vary with temperature. But they can still be used as a decent source of constant current.
The current source and load limit
In a constant current source, a load is connected to its output terminal and supplies a constant current to the load.
But what happens to the output voltage in this situation?
The output voltage plays a critical role. If you exceed the current source input limit, the source will no longer provide a constant current. To avoid this situation it is necessary to choose a maximum load limit.
There are two methods for generating a constant current source:
1. Combining transistors and operational amplifiers (op amps)
2. Using a Buck Regulator
Creating a constant current with LM350
In this experiment, we will design a constant current source using an LM350 linear buck converter. It is a step-down converter, which reduces (or steps down) the input voltage to a desired value .
The LM350 specifications:
- Maximum output current – 3A
- Input voltage range – 1.2 to 32V.
Basic design
Block diagram of a constant current source using LM350.
Operating principles
An adjustable linear regulator is required to design the current source.
This type of regulator consists of input pins, output pins, and a feedback pin (the adjustable pin), which powers the output. When the input or load changes, the feedback circuit regulates the output voltage.
Internally, the integrated circuit (IC) has an error amplifier (based on the operational amplifier), a transistor and a reference voltage source. The error amplifier monitors the output voltage, comparing it with the regulator reference voltage. The amplifier also tries to keep the reference voltage fixed to maintain a constant current flow through the feedback network.
Therefore, a constant current flows through the output even with different load values.
Necessary parts
Circuit Diagram
Circuit diagram of a constant current source using an LM350.
LM350 Features
1. Output short circuit protection
2. Thermal overload protection
3. Safe area protection
Building the current source
Here are the steps to design the current font.
1. Choose the output current
The LM350 provides a continuous current of 3A, so the project will have a maximum constant current source of 3A. We will make a 2.5A constant current source using IC LM350.
2. Select a feedback network
The feedback network is simply a network of feedback resistors. This resistor will decide the output current value by maintaining a fixed voltage on the feedback pin of the IC.
Below is the equation to use to calculate the feedback resistor.
Applying the famous Ohm's law (V = IR) We have…
V reference = regulator voltage reference
LM350 regulator, reference V =1.25V
Output current, I out = 2.5A
Feedback resistor, R Facebook =V reference / EU out
R Facebook = 1.25/2.5
R Facebook = 0.5
1. The rated power of the feedback resistor
The high current at the output causes a large power dissipation through the feedback resistor. As a result, it is necessary to calculate the exact power of the resistor, otherwise the resistor will begin to overheat and burn out.
The equation to calculate the nominal power of the resistor:
P R = me 2 *R Facebook
P R = (2.5*2.5)*0.5
P R = 3.13 W
2. The input capacitor
The capacitor will ensure that any spike at the input does not harm the circuit. It grounds all ripples and spikes from the input source.
Furthermore, to reduce the total ESR of the capacitor, we can connect a ceramic capacitor and an electrolyte.
3. Thermal management
The heatsink must be assembled with IC LM350 because we are using a large amount of current (about 2.5A). This will produce a lot of heat on the surface of the IC and deteriorate it. To reduce the temperature, we can use a fan or heatsink with IC thermal paste.
How the circuit works
The circuit works on the principle of a feedback loop, as already discussed. The feedback resistor continuously feeds output voltage to the feedback pin, maintaining a constant voltage across the resistor.
According to Ohm's law (V = IR), we know that a constant current will flow making the voltage and resistor constant. ET This is exactly what the linear regulator creates here.
In our constant current source circuit, the feedback circuit maintains a fixed output voltage across the feedback resistor – and the value of the feedback resistor is also constant. This allows a constant current flow at the output, regardless of the load.
Therefore, we have successfully designed a constant current source using just a resistor and a linear regulator IC.
Connecting the load to the output
When a load is connected to its output, a constant current will flow through it, regardless of its value.
In this situation, the load “decides” the output voltage. A voltage drop will be required according to its impedance value. Therefore, before connecting any load to the output, we must consider its impedance.
Calculating the maximum load limit
The law of conservation of energy states that energy can neither be created nor destroyed. Linear regulators are not power converters. Instead, the input power of the linear regulator is always less than the output power. Any remaining energy is dissipated into heat.
Ultimately, they provide a lower voltage than the input voltage.
Consider the following example to understand this concept of a linear regulator.
V out =V in –V d
V out = 12 – 2
V out = 10V…………………………Eq.2
V R = 2.5*10
V R = 25 V
However, according to the calculation above, the regulator cannot supply 25V at the output. We must first calculate the output load limit using the following equation.
Maximum value of the resistive load, R i = 4E
Thermal management
For cooling purpose, we must connect a heatsink to the IC LM350, which helps to maintain the temperature of the IC. See this article “Power Supply Thermal Management” to learn how to select the optimal parameters of a heatsink and cooling fan.
Practical observation
- Input voltage, Vin = 12V
Constant current graph with 3 ohm load.
Forms
1. Driving transistors for amplifiers
2. Charging constant current-based batteries
3. Lighting systems
Precautions
- Make sure the power rating of the feedback resistor is as per the stated requirement.
- Always place a load below the limits of the current source.
- A capacitor must be connected between the input pin and ground to keep the DC input voltage regulated.
- The capacitor used in the circuit must have a higher voltage rating than the input supply voltage. Otherwise, the capacitor will start to leak current due to excess voltage on its plates. Eventually it will burst.
- Make sure the entire capacitor is discharged before working on a DC power supply.
- Avoid using an input voltage higher than the operating input voltage range.
- Do not short the output terminals – this will reverse the current flow in the IC and cause a failure.
- Do not short the input terminals – this will generate a large current in the circuit and the components will fail.
- Always connect a heat sink or fan for heat dissipation around the IC.
Circuit configuration
The circuit
Real-time prototype image of the circuit.