Question

Consider an ideal down converter supplying load resistance of 80 ohms from a voltage source rated 9V and 250mA. The converter operates at 70-kHz and is designed for 5% voltage ripple and 10% current ripple with a duty time of 60%.

a) Calculate the output voltage, output voltage ripple, output current and output current ripple.

b) Calculate the cycle period, T on and T off.

c) Calculate the required inductance and capacitance values

d) Calculate the output voltage as functions of time

e) Graph the ideal output voltage and inductor current waveforms.

Answer 1

From the provided data , the down converter is a ideal converter , which implies that the down converter has no losses, that is to say that it draws no power or no voltage drop across pass transistor and other components.

a) Output voltage calculation :

Formula to calculate output voltage is Output Voltage = Duty Cycle * Input Voltage

From the data, Input voltage is 9 Volts and the duty cycle which is duty time is 60% = 60/100=0.6

Substituting the values, the output voltage calculates to

Output voltage = 0.6 * 9 = 5.4 Volts

The calculated output voltage is 5.4 volts.

Output voltage ripple calculation :

From the data the output has 5% voltage ripple i.e 5/100 = 0.05

The calculated output voltage is 5.4 Volts. So 5% of 5.4 Volts equals to 0.27 Volts ( 5.4 * 0.05)

The calculated ripple voltage is 0.27 Volts.

Output current calculation :

The data provided is that the converter is ideal, which implies there is no loss.

So, the power input equals to output power.

Formula to calculate power is voltage * current i.e P=V*I

In Ideal down converter Input power = output power

Input V *I = Output V*I

From data 9 volts * 250 mA = 5,4 * I mA

Output current I mA = (9 * 250)/5.4 = 416.666

The calculated output current is 416.666mA.

Output current ripple calculation :

From the data , the output has 10% ripple current.

So , 10% of 416.666mA equals to 416.666* (10/100) = 416.666 * 0.1= 41.666mA.

The calculated output ripple is 41.666mA.

b) Given the converter operates at 70Khz.

That is the frequency is 70Khz, So, in time (seconds ) = 1/70Khz =0.01428ms

So the converter is operating at 0.01428ms =14.2 micro seconds.

The calculate cycle period is 14.2 micro seconds.

The converter is operating at 60% duty time , that is the ON time

60% =60/100 = 0.6.

So T_ON = 14.2 * 0.6 = 8.52 micro seconds.

The off time is 100%-60% which equals to 40%

40% = 40/100 = 0.4

So T _OFF = 14.2 * 0.4 = 5.68 micro seconds.

The calculate cycle time is 14.2 micro seconds, T_ON = 8.52 micro seconds. and  T _OFF = 5.68 micro seconds.

c) The formula to calculate Inductance is     

L=((Vin-Vout)*DC)/(ΔIL*Fs) where L=Inductance

Vin = Input Voltage

Vout = output voltage

  ΔIL = Inductor ripple current

Fs = operating frequency.

DC = Duty cycle

From the data and calculated values

L = ((9-5.4)* 0.6)/(41.66*10^-3 * 70 *10³) = 2.16/2916.2 = 7.40*10^-4Henry = 0.074 micro Henry.

The calculated inductance is 0.074 micro Henry.

The formula to calculate the capacitance is

C = ΔIL/(8* Fs * Vout) when the values are substituted

C = (41.666* 10^-3)/(8*70 *10³*5.4) = (41.666/3024)*10^-6 =0.01377 micro Farads = 13.7 nano farads.

The calculated capacitance is 13.7 nano farads.

d)

Formula to calculate output voltage is Output Voltage = ON time * Input Voltage

From the data, Input voltage is 9 Volts and the ON time which is duty time is 60% = 60/100=0.6

Substituting the values, the output voltage calculates to

Output voltage = 0.6 * 9 = 5.4 Volts

The calculated output voltage is 5.4 volts.

e) Given below the output voltage and inductor current waveforms.