SMPS Output Voltage Control Using PWM Technique Using IC TL494 Based SMPS. 12-0-12 V Ac to 220 V Ac @ 20 Watt. Frequency 63 KHZ

Hello everyone, in this article Im going to demonstrate SMPS output voltage control using Pulse width modulation (PWM) technique. My SMPS input is 12V DC and output is 220V AC @ 63 Khz. With PWM adjustment output varies between 220V to 158 V AC.

Supplies

Bill of materials

C1 22PF

Q1 25N120

Q2 25N120

R1 100K

R2 100K

R3 100K

R4 100K

R5 10K

R6 10K

R7 330E 1W

R8 330E 1W

R9 100E

R10 100E

R11 20W LAMP

R12 1K

R13 1K

R14 10k

T1

U1 TL494

Step 1: Construction

The work involves 220V AC mains line voltage available on power transformer T1 output. This can cause risk of electric shock. Hence this activity must not be performed with electricity handling awareness.

The IC TL494 input pin 3 is the feedback pin. The voltage at this pin is 0.85 V to 2.27 V DC. This voltage at pin 3 (feedback pin) controls the IGBT Q1 and Q2 gates. The PWM output finally controls the voltage fed to the transformer. Since the input to the transformer varies between 9.5 V AC to 6.5 V AC, the output voltage also varies.

Proper heatsinks are needed to dissipate the heat on IGBTs

Please follow the circuit diagram and transformer design sheet for building this project.

Step 2: Data Obtained

Supply voltage = 12 V DC

Voltage on IGBT 1 and 2 drain = 9.5 V AC to 6.5 V AC (Transformer input at each center tap)

Transformer secondary voltage = 220V to 158 V AC (PWM)

Frequency = 63 KHZ

TL494 IC pin 3 = 0.851 to 2.27 V DC

pwm data.pdf
Download

Step 3: Transformer Design

Transformer Primary = 12 V @ 60 W

Transformer Secondary = 230 V

Primary no. of turns = (Primary volts X 10^10)/(4 x frequency (Hz) X Core area (sq. mm) X flux density (Gauss) )

Primary no. of turns = (12 X 10^10) / (4 X 65000 X 69 X 1500 ) = 4.45 or 5 turns

Secondary voltage = 230 X 1.414 = 325.22 V

No. of turns in secondary = (Secondary Voltage / Primary voltage) / Primary no. of turns

= (325.22 / 12) X 4.45 = 120.6 turns

Wire used = 38 SWG for both primary and secondary windings.

38 SWG wire current capacity = 0.350 A

30 W @ 12 V ==> 30 /12 = 2.5 A

No. of conductors in primary = 2.5 / 0.350 = 7.14 , say 8 conductors.

Secondary can have 1 conductor to serve this power.

Transformer design summary

Core area = 69 sq. mm

Operating frequency = 65 khz

No. of turns in primary = 5 ( 5 + 5 as it is bifilar winding, center tap required for push pull configuration. )

Secondary turns = 120 or 121

38 SWG copper wire

Primary has 8 conductors with 5 turns two windings in bifilar winding.

Secondary has 1 conductor 120 turns.

Step 4: Conclusion