## Wednesday, July 10, 2013

### DESIGN DC-DC CONVERTER USING DOUBLE TRANSISTOR

In the past we have discussed about how to design DC to DC converter using empiric method, where we calculate the current output and voltage desired. In this chapter we will make an experiment how to design Boost converter using double transistor. Transistor used to drive the FET to generate switching processes.
In this design, we want make a boost converter from 5 V to 12 Volt, with current output 100mA. First remember the basic concept of DC-DC diagram, look at the picture 1.
 Picture 1, Basic Diagram of Boost Converter

In this case, FET (Q1) was drive by double transistor.
 Picture 2, Double transistor as FET driver

I used the push pull transistor which part of VFD driver. The frequency that generate about 108kHz, base on this form, can calculate the value of Inductance (L) and Capacitance (C), then matching impedance (R) as well.
Empiric solution:
Inductor Calculation
T=1/f, then T=1/108 kHz= 9.25 us
Vout= Vin/ (1-k), then 12=5/(1-k), so we can obtain duty cycle(k)=0.583
ton +toff=T and k=ton/T
in our last meeting, we have figure out the value of ton and toff by using mesh analysis method. So we can do a same thing as before to get the final equation.
ton +toff=T, so
L∆i/Vin +L∆I/(Vout-Vin) =T
L∆I (Vout/ Vin(Vout-Vin)= T
And then, we can enter data that we have,
L(100mA) (12/5(12-5)=9.25us, and we obtain value of Inductance(L) =270 uH.

Capacitor calculation:
Vc=Vout
Ic Xc=12
Ic. 1/(2*phi*f*C)=12 V
(100 mA)* (1/2*(3.14)*108kHz*C=12V
So we obtain value of Capacitance = 12.3 nF

Diode, for diode you can check the datasheet of the diode with reliable value of current and voltage, because we will use 12 V output, we recommended you using diode with voltage rating 12 V too.

In the end we need to ensure whether our DC converter is work or not. In this case, we need a dynamic load, where this device can pull the current that we set. In general we can illustrated the dynamic load as bellow:
 Picture 3, Dynamic Load was applied.
First time when you operate the Our boost converter without load, the Dynamic load will measure the Voltage only, and when we loaded the DC to DC with specific current, in this case 100mA, the DC to DC will work. If your Design is good enough, the voltage which measured by Dynamic load would not drop or drop but just a little (not more than 5%).

I have check our design above, without load the voltage is about 58 Volt, and when it loaded with 100 mA, the voltage is drop to 8 Volt.

Case 1, Voltage 58 Volt
I don’t know what exactly happened with the voltage, why the voltage measured is 58 Volt?, is our calculation is fail?, even the voltage output is 58 Volt without load, we still can manipulate this by put the Zener diode on voltage output, with Vz=12 Volt. Look at picture bellow,

Case 2, Voltage Drop to 8 volt
When voltage increased to 58 volt, we can put zener diode on output voltage, but when the voltage dropped, it means your design in problem. The voltage drop is caused by our system, right now we using the open loop concept, se there is no feedback, so the duty cycle absolutely depend on the inductor performance. In the open loop system, we need adjust the value of inductor the get the better value.

 Picture 5, Adjust Inductance value to increase performace

Conclusion,
1.       It’s ok, even our calculation did not match with reality.
2.       Our design is good, but not perfect, because there is no feedback to control duty cycle.
3.       We can put the zener diode to clamp the voltage output, in fact our design has voltage output 58 volt without load, and 8 Volt with load 100 mA.
4.       Adjust the inductor value to get better performance, ensure the value of current is over of 100 mA.
5.       Need more improvement to make our design perfect.

FTo be Continued....