Adjustable Constant Current Source

About Constant Current Source

A constant current source is used to provide a constant current regardless of the input voltage and output load. This type of circuit is also used as a current limiter. Every electronics device has some current rating above that rating the device may damage or it will not work properly. Therefore there is a big requirement of the constant current source in electronics. In this experiment, we are designing an adjustable constant current source which supplies 25mA constant current at the output regardless of load resistance and input voltage.

Overview

In this experiment, an adjustable current source is developed which can supply a constant current of 25mA at the output. The constant current of 25mA in this circuit can be changed as per requirement by just changing the value of a single resistance in the circuit. Hence we can say this circuit is an adjustable constant current circuit. The circuit can be adjusted to provide a constant current up to 1.5A.

Components Required

Components Required	Specifications	Quantity
TR1	12V-0-12V/2A	1
Voltage Regulator	LM 317	1
Resitance R	50ohm	1
Diode D1-D4	SR560	4
Fuse	1.5A	1

Basics of Power Supply

Every DC power supply needs some steps to follow for getting proper DC voltage at the output. Below diagram shows these basic steps by which we get a regulated DC power supply by AC.

Basics of power supply

BLOCK DIAGRAM

Working

• Step down the Mains supply by input transformer

The voltage of Mains (Electricity coming at our home from the government) is approximately 220V but as per the circuit requirement, only 12V is required at the output terminal. To reduce this 220V to 12V, a step-down transformer is used.

The best suitable step-down transformer that meets our voltage and current requirement is 12V-0-12V/2A. This transformer step downs the main line voltage to 12V, as shown in the below image.

Input transformer

• Rectification

The rectification is the process of converting AC to DC. There are two ways to convert an AC signal to DC. One is through half wave rectifier and another is by using a full wave rectifier. In this circuit, we are using a full wave bridge rectifier for converting the 12V AC to 12V DC. As full wave rectifier is more efficient than a half wave since it can provide complete use of both the negative and positive side or part of AC signal. In full wave bridge rectifier configuration, four diodes are connected in such a way that it generates a DC signal at the output, as shown in below image. During full wave rectification, at a time two diodes become forward biased and another two diodes are reversed biased. We have chosen the SR560 diode because they can allow 1.5A current through them when forward biased and in reverse biased condition, they can sustain reverse 12V supply. Due to this, we are using SR560 diodes in rectification purpose.

Rectification

• Voltage regulation by LM317

About LM317 IC

LM317 is one of the popular adjustable positive voltage regulators. The LM317 IC has various features like over voltage protection, internal current limiting, overload protection, low quiescent current (more stable output), safe area compensation (internal circuitry limit the maximum power dissipation so it does not self-destruct). This can provide output in the range of 1.25V to 37V with the input voltage up to 40V. It can provide a maximum current of 1.5A at the output.

Key Principle of 317

Internally an OPAM is there in the IC whose inverting input is connected with the adjustment pin. The non-inverting input is set by a band gap voltage reference whose voltage is independent of the temperature, power supply, and circuit loading. Thus LM317 gives a stable reference voltage of 1.25V under nominal conditions. The reference voltage can be from 1.2V to 1.3V.

• How 317 works as constant current source

A 12V DC supply is connected to the input pin of 317.For setting the constant current at the output a resistance R is connected from the output to adjustment pin of 317.In a normal state when there is the constant demand of current at the output then 317 will maintain a voltage of 1.25V at its adjust terminal. Therefore the voltage across R is also 1.25V. As the current demand changes at the output then it should also change the voltage drop across R but 317 will adjust the output voltage to compensate a constant 1.25V drop across resistance R.

Hence, the voltage across R is always 1.25V. Therefore a constant current will flow through R. This constant current can be calculated by

Iout = 1.25/R

For changing the output constant current we can change the value of R.

317 as a constant current source

• Output Current

Due to the constant 1.25V reference voltage of the LM317, it provides a constant current at the output. By using this key quality of 317 an adjustable constant current source can be designed. The 317 IC will provides current up to 1.5A if the proper heat sink is applied to the IC.

As the IC will able to provide a minimum current of 10mA and maximum 1.5A.

The minimum value of resistance R in the circuit is – :

Rmin = 1.25/ Imax

Rmin = 1.25/1.5

Rmin = 0.83 ohm

So we can say that below 0.83ohm the current will be greater than the 1.5A and IC will not be able to provide that much current

The maximum value of resistance is – :

Rmax = 1.25/ Imin

Rmax = 1.25/0.01

Rmax = 125 ohm

Internally tolerable power dissipation of LM317

Pout = (Maximum operating temperature of IC) / (Thermal Resistance, Junction-to-Air + Thermal Resistance, Junction-to-case)

Pout = (150) / (65+5) (values as per the datasheet)

Pout = 2W

The 371 internally can sustain up to 2W power dissipation. Above 2W the IC will not tolerate that much of heat and it starts burning. This may cause a serious fire hazard also. So a heat sink is needed to dissipate the excessive heat from the IC.

Practical results

Case 1 :

When a resistance of 50 ohm is connected from the output pin of 317 to adjust pin

Load resistance (ohm) When R= 50 ohm	Output current (mA)
10	24.6
50	24.6
100	24.6

Load resistance (ohm)

When R= 50 ohm

Output current (mA)

24.6

100

24.6

Theoretical observation of output current = 1.25/R

Iout = 1.25/50

Iout = 25mA

% Error = (Theoretical Observation – practical observation) *100 /practical observation

% Error = (0.025 – 0.0246)*100/0.0246

% Error = 1.6%

Power dissipation of regulator when R = 50 ohm

Pout = (Vin – Vout)*Iout

Pout = (12- 1.5)*0.0246

Pout = 0.25W

Case 2 :

When a resistance of 25 Ohms is connected from the output pin of 317 to adjust pin

Load Resistance (ohm) When R= 25 ohm	Output Current (mA)
10	49.8
50	49.8
100	49.8

Load Resistance (ohm)

When R= 25 ohm

Output Current (mA)

49.8

100

49.8

Theoretical observation of output current = 1.25/R

Iout = 1.25/25

Iout = 50mA

% Error = (Theoretical Observation – practical observation) *100 /practical observation

% Error = (0.050 – 0.0498)*100/0.0498

% Error = 0.4 %

Power dissipation of regulator when R = 50 ohm

Pout = (Vin – Vout)*Iout

Pout = (12- 1.7)*0.0498

Pout = 0.51W

In both the cases by varying the R in the circuit, we can change the value of constant currentInternally 317 is capable of handling power dissipation to 2W so here it is less than 2W but it is recommended to use a heat sink to aid cooling to IC and to increase the lifespan of the IC.

Applications

• As a constant current LED driver

• As a constant current battery charger

• As a current limiter circuit

• As a constant current power supply

• Super-Cap Charging

Points to Remember

• The current rating of a step-down transformer, bridge diode and voltage regulator must be greater than or equal to the maximum current required at the output. Otherwise, the circuit will be unable to supply the required current at the output.

• The dropout voltage of LM317 IC is about 1.5V to 2.5V depending upon the output current. Thus input voltage must be 1.5V to 2.5V greater than the maximum voltage drop across the load. The voltage drop across the load can be calculated by, V = I(constant current) * R(Load Resistance)

• For driving high load at the output, heat sink should be mounted at the holes of the regulator. This will prevent the IC from blown off.

• The Proper heat sink is required for power dissipation above 2W

• Current drawn at the output should not be greater than 1.5A otherwise it will damage the IC.

• As our circuit can be adjusted to draw a current of 1.5A at the output. A fuse of 1.5 A is to be connected to the output of the rectifier. This fuse will protect the circuit for current greater than 1.5A. For current above 1.5A, the fuse will blow off and this will cut the input supply from the circuit.