Voltage and current sources are two fundamental types of electrical sources used in electronic circuits. A voltage source, such as a battery, is commonly used in many devices. It’s a two-terminal device that maintains a constant voltage across its terminals, regardless of the current flowing through the source.

A current source is typically less discussed and more difficult and costly to design. It’s also a two-terminal device, but it maintains a constant current through its terminals, regardless of the voltage across the source. This means it’s used to ensure a steady current even if the load fluctuates.

In this experiment, we’ll design a constant current source application using a linear buck regulator.

Specifications

• The LM338 linear buck regulator
• Input voltage range – 1.2 to 32V
• Maximum output current – 5A

Block diagram 

Circuit diagram

Principle
When using any linear regulator as a current source, the circuit works on the principle of a feedback loop. 

The regulator consists of a feedback pin (the adjust pin) that feeds the output back to the regulator and maintains a fixed voltage at the feedback network. This network is a resistor, which determines the output current at the output. 

Circuit design
Here are the steps for the circuit design.

1. Input source – 24V DC
2. Linear buck converter – LM338 IC 

Features of the LM338 

• Overload protection – in case of overloading, the device will reduce its output current until the overload is removed
• Thermal regulation
• Output short-circuit protection

3. Output current set – the output current is based on the feedback network, which is the resistor

The equation for calculating the feedback resistor…
By applying famous Ohm’s law, we get:.  V_ref = voltage reference of the regulator

LM338 regulator, V_ref =1.25V

Desired constant output current, I_out  = 4A

Feedback resistor, R_fb  = V_ref / I_out

R_fb = 1.25/5

R_fb = 0.33E

4. Feedback resistor power rating

The equation for calculating power rating of the resistor…

P_R = I²*R_fb

P_R = (4*4)*0.33

P_R = 5.28W (minimum)

5. Filtering – the capacitor, which is at the input, grounds all of the ripples and spikes from the source. To reduce the total ESR, we can connect a ceramic capacitor along with an electrolyte.

6. Cooling system – the heat sink must be mounted with the regulator because we’re drawing a large amount of current. This will produce a lot of heat at the IC surface and can ignite it. So, it’s important to use a cooling fan or a heat sink to reduce its temperature.

How the circuit works
In our constant current source, the feedback loop maintains a fixed output voltage across the feedback resistor. As per Ohm’s law, with a fixed value of voltage and a resistor, a constant amount of current will flow. 

This will generate a constant current at the output, irrespective of the load.

Output current regulation
All current sources provide a regulated current but only for a specific load range. 

It’s possible to calculate the load limit for the linear regulator by using the following equation… 

V_out = 12 – 3

V_out = 9V…………………………Eq.2

V_R = 5*4.7

V_R = 23.5V

The regulator is unable to provide 23.5V at the output. So, we must first calculate the output load limit using the equation below…

The maximum value of resistive load, R_L = 2.25E

 

Testing results
Initially, we designed the current source for 5A since the IC is rated for 5A output current. However, the IC could not maintain a constant 5A output current and began dropping its value even with a heat sink and cooling fan. 

This might be because of the IC’s internal thermal regulation, which tries to maintain its temperature and drops the current if in a high-power dissipation state. 

So, instead, we tried the circuit using a 3.7A current, and it worked perfectly with the heat sink. 

Observation

The output current begins stabilizing at a load of 1 Ohm. The difference in the output current is due to the resistor’s tolerance values.

Circuit limitations

1. Input source – A high-voltage input source is needed to increase the maximum load limit
2. Cost – A high power rating feedback resistor increases the cost
3. Efficiency – There’s less efficiency because of high-power dissipation across the feedback resistor

Thermal management
For extra heat dissipation, heat is connected to the IC. It’s also possible to use a fan to disperse the hot air.

Applications

1. Battery charging
2. In transistor biasing
3. Lighting systems
4. Regulated supply

Precautions 

1. Ensure the feedback resistor’s power rating is per the product requirements. 
2. Always put a load under the limits of the current source. 
3. A capacitor should be connected between the input pin and ground to regulate the DC input voltage. 
4. The circuit’s capacitor must have a higher voltage rating than the input supply voltage. Otherwise, the capacitor will leak current due to the excess voltage at its plates and burst.
5. Ensure all the capacitors are discharged before working on a DC power supply.
6. Do not give a higher voltage at input than its operating input voltage range.
7. Always connect a heat sink or a cooling fan for heat dissipation around the IC.

Video

 

 

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Filed Under: Tutorials
Tagged With: buckconverter, constantcurrent, current, currentsource, lm338, powermanagement, voltage
 

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