# DC to DC Fundamentals

Discussion in 'General Electronics Chat' started by ricardo7890, Jul 2, 2013.

1. ### ricardo7890 Thread Starter New Member

Aug 13, 2012
9
0
I've been trying to learn the fundamentals of DC to DC conversions by reading Wikipedia entries and the book ARRL Handbook for Radio Communications, but neither made sense in several points and then i just tried to come to my own understanding which is the following:
I understand the basic workings, which is that during the on cycle the inductor is energized by the switch current, and immediately afterwards, during the off cycle, this energy is transferred to the capacitor, through a diode to prevent oscillation.The energy in the inductor is a factor of the square of the current times the inductance value times .5 or : (I^2*H)/2=J, and the energy in a capacitor is given by a similar equation (C*V^2)/2, so if the current at the end of one ON cycle is determined then the energy transferred to the capacitor and load is the exact same energy used to energize the inductor, ignoring diode and resistive loses.This happens during in the OFF cycle and ON + OFF =1/F where f is the frequency of the switching square wave form.
The current at the end of the on cycle can be determined using the formula (V/R)(1-e^(-tR/L)) Where t is the time the switch is ON and R is the resistance of the inductor wires and battery internal resistance from these equations i came up with a larger one:

((V/R)(1-e^(-xtR/L))^2 *L/2) *(1/t)=Watts out
This equation given R,V,and L should give me the power put out by the inductor into the load circuitry, the capacitor and resistor, for any frequency (1/t), where x is the duty cycle. Please tell me if this equation is formatted confusingly or if it is just wrong, although it seems to hold up when i run it on a simulator. I pick the frequency which will give me the most power for the given R V and L using the graphing calculator and i know the power out and i determine the voltage out when i determine the resistive load since (power out of inductor)*R load=Vout^2.

Using the duty cycle i can alter the power out, the smaller the duty cycle the smaller the power out.
My main problems now are three things:

1)For one i can't seem to get the maximum amount of power out of a battery.For example using the formula for 5 volts with series resistance of 2 through an inductor of 100uH and a duty cycle of .8 i graphed it and it maxed out at 2 watts at a frequency of 14.3k,meaning at no frequency could i get more power. This same circuit at 5 volts and 2 Ohms could provide 12.5 watts. How can i make a similar circuit that can provide more power than this using a similar topology, or is there some way i can alter my existing circuit to output more power.
2)I wish i could understand all these things i read better for example the Wikipedia page for the buck booster makes several points that i can't understand.
http://en.wikipedia.org/wiki/Buck–boost_converter
The most confusing thing is when it describes the inductor current during the off cycle is shows this equation to describe the rate of change
(Vout*(Off*T))/L the most confusing part is the V out variable.I presume this is meant to mean the voltage of the capacitor, but this isn't a constant values, from what i've found this is completely depended on the R load value. The rest of the article treats Vout like a constant known value that is independent of the Rload, since i don't understand why this is i can't figure any of the rest of the article, the ARRL handbook does the same thing

3)i know i should probably use a specialized ic for practical purposes but i rather design one with discrete components for practice.Of course i need a square wave generator i would use an ic for it but i can't find any that work on really low voltages like 1.5 volts found in AA batteries, do any exist, and if not is there a way to make a square wave using discrete compotents that would provide a really fast response time, comparable to a 555, which needs something like 5 volts minimum to work.

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2. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
508
true of a boost converter. Is that what you are building? A buck works differently.

3. ### ricardo7890 Thread Starter New Member

Aug 13, 2012
9
0
It's an inverter amplifier, i attached an image of the schematic

4. ### crutschow Expert

Mar 14, 2008
16,554
4,466
1) Why did you expect to get more power out of the battery with 2 ohms in series with the inductor (why is that resistor there?) than without the inductor?

2) With a boost circuit the output voltage does depend upon the duty-cycle and the load. That's why you use negative feedback to control the duty-cycle to maintain a constant voltage with changes in load.

3) Few circuits work at 1.5V. Google "Joule Thief" for one example that does.

5. ### ricardo7890 Thread Starter New Member

Aug 13, 2012
9
0
1) The resistor is the to approximate the internal resistance of the battery and of the inductive windings, realistically it should be higher i think. I didn't expect to get more power with the inductor in series, i just want to get close to the power the battery could output with just the resistor.
2) could you explain how feedback can affect duty cycle, in all the stuff i read i never heard mention of that.

thanks for the help

6. ### crutschow Expert

Mar 14, 2008
16,554
4,466
1) The theoretical best you can get with two ohms in series, using the maximum power transfer theorem, is for a matched load of two ohms, giving a maximum output power of (5/4)^2 * 2 = 3.125 W (the same amount of power is dissipated in the two ohm battery and inductor resistance).

2) Most practical switching regulators use a negative feedback loop from the output voltage to control the PWM circuit duty-cycle and maintain a constant output voltage. This typically includes a compensation network in the loop to keep the feedback loop stable. (Read this to start).