Fixed Output Boost Converter Using UC3843¶

Design boost converter to get output voltage 14.2 (to charge lead-acid batteries) or other nearby voltages by adjusting resistor divider, from voltage source around 12V.

It is based on document Basic Calculation of a Boost Converter's Power Stage. Also read UC3843 datasheet from Texas Instruments, which has lots of useful information for designs around UC3843 IC.

Here is circuit diagram. Below I will try to explain functions of most of the parts used in it.

Parameters¶

Minimum input voltage, V_IN(min) = 11.5 V
Maximum input voltage, V_IN(max) = 12.5 V
Average input voltage, V_IN = 12 V
Output voltage, V_OUT = 14.2 V
Maximum output current, I_OUT = 4 A
Expected efficiency, η = 80%
Frequency, f = 100 kHz
Forward voltage of the rectifier diode, V_F = 0.75 (of 1N5825)

Inductor Selection¶

Inductance, $L = \frac{V_{IN} \times (V_{OUT} - V_{IN}) }{ ΔI_L \times f_S \times V_{OUT} }$

Thus inductance, L = 9.8 µH

Calculate Maximum Switching Current¶

Duty cycle, $D = 1 - \frac{ V_{IN(min)} \times η }{ V_{OUT} }$

Thus, D = 35.21%

Inductor ripple current, $ΔI_L = \frac{ V_{IN(min)} \times D }{f_S \times L }$

Inductor ripple current, ΔI_L = 1.89 A

Maximum switching current, $I_{SW(max)} = \frac{ ΔI_L }{ 2 } + \frac{ I_{OUT(max) } }{ 1- D }$

Thus, I_SW(max) = 7.1 A

Rectifier Diode Selection¶

To reduce losses,Schottky diodes should be used.The forward current rating needed is equal to the maximum output current:

I_F = I_OUT(max)

Maximum average rectified forward current of 1N5825 is 5 A. So it is within the limits

Power dissipation of the diode, $P_D = I_F \times V_F$

Thus P_D = 3 W

Resistive Divider for Output Voltage Setting¶

Feedback voltage for UC3843 is 2.5 V.

Current through the resistive divider should be at least 100 times as big as the feedback bias current. $I_{R1/2} \geq 100 \times I_{FB}$

So, $R_2 = \frac{V_{FB}}{I_{R1/2}}$

Info

Feedback input bias current of UC3843 taken as 2 µA. That value is taken from datasheet of UC3843 from On Semiconductor

So, R₂ = 12.5 kΩ

To find R₁, equation is $R_1 = R_2 \times (\frac{ V_{OUT} }{ V_{FB} } - 1)$

Thus R₁ = 58.5 kΩ

Output Capacitor Selection¶

Desired output ripple voltage ΔV_OUT = 0.1 V

Minimum output capacitance $C_{OUT(min)} = \frac{ I_{OUT(max)} \times D }{ f_S \times }$

Thus C_OUT(min) = 140.8 µF

However using larger capacitance will be good. Larger capacitance will reduce ripple voltage and ESR because of larger plates.