DC to DC Fundamentals

[ricardo7890]

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.
https://en.wikipedia.org/wiki/Buck%E2...oost_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.

 

[ted]

Hi,

At least you have tried hard to understand and calculate things. That is good, I think you have a good attitude for learning the stuff!

I am sorry, but just now I didn't have time to fully analyze your equations and the long text. However, I try to help a bit anyway.

First, the topology in your drawing is a sort-of fly-back configuration, or variation of inverting buck-boost converter. So there you indeed "charge" the inductor with certain amount of energy, and then "dump pout" it through a diode to the load. (And, BTW, the diode is not to prevent oscillations, it is an essential switching component, transferring the energy).

The voltage conversion in this configuration is controlled by on-off ratio of the switch. Also, note, that the peak currents are significant, as all energy is transferred one way at t-on, other way at t-off time. On-off ratio of about 50% gives thus lowest current pulses, as there current amplitudes are descent (then for example the transistor and coil carry, in best case, "only" a little more than 2x the average current from the battery).

When you move more far away from 50% on-off ratio, things become more demanding.

I think your problem of getting more output is largely due to the series resistance of 2 ohm. You see, the peak current demand is very high. If that resistor of 2 ohms represents battery internal resistance, it is real enough, and has to be handled. However, it WILL always seriously limit the output power even when everything is well-engineer! The normal way to help the situation is to have a relatively large (here maybe 100uF or more) capacitor between top end of the inductor and transistor emitter. That capacitor would supply the peak currents, when the transistor is "ON", essentially averaging the battery current. At higher current levels that 2 ohm resistor will be still a serious problem.

Another challenge you mentioned is to make this (or any other DC-DC converter) to work well with only 1.5V input:
It is not easy to a) get significantly large power and b) descent efficiency with so low input voltage. I would not try it as the first exercise, except for maybe for a very small output power. The real-life components are not at all close to ideal on so low voltage, and hence for example transistor saturation voltage and diode drop become quite a nuisance. I have developed several small converters for 1.5V battery as input and providing for example a few ten mA output on 3.3V. However, getting several watts out is VEEERY challenging.

I believe a paper in https://w5jgv.com/hv-ps1/pdf/topologies_for_switched_mode_power_supplies.pdf is one of the better application notes I have seen.
I can see from your text, that you know the basics pretty well already. Anyway, maybe having a look in this might be useful, as in my opinion this AN is quite comprehensible, but still pretty readable.

Good luck,
ted