Questions regarding DIY boost converter
I have always wanted to make my own boost converter(Just to know how to to), so i bought a few 220 uH inductors from mouser and plugged them into a breadboard setup... I have a 555 timer wired in astable mode. I have a 1.5k pot hooked up to adjust the output waveform(which forms a voltage divider with another 1.5k resistor and charges a 220 uF cap). The boost converter consists of a TIP3055, a diode and a couple of capacitors. When the 555 is outputing a lower frequency, i can measure up to 112 volts on the output. On the highest frequency, i can get about 13 volts or so. The transistor gets really hot, but the circuit worked well, until i discovered something. The output voltage was decreasing and when i applied a load it dropped right off and stopped working, until i took the load off. I read something online about the circuit needing some kind of feedback system to adjust the output frequency to keep the voltage constant? Or maybe i need a true PWM and not just a square wave that has a fixed duty cycle? I am also going to conclude that i need a bigger value capacitor on the output to supply voltage while the inductor is charging?...
Also, my understanding of a Henry(Unit of inductance) is a voltage of one volt being produced, if the current is changing at 1 amp per second? How does that relate to boost converter design? I just can't wrap my head around it.... Thanks for any info!!! I would submit a schematic, but i can't draw at all:) I hope i am not shooting in the dark with this one?
Also, my understanding of a Henry(Unit of inductance) is a voltage of one volt being produced, if the current is changing at 1 amp per second? How does that relate to boost converter design? I just can't wrap my head around it.... Thanks for any info!!! I would submit a schematic, but i can't draw at all:) I hope i am not shooting in the dark with this one?
Comments
One major consideration in PSU design is efficiency. Especially in switchers where they can do rather well. As you said your TIP3055 was getting hot.
Why is that? Well clearly if it is fully OFF no current flows and no power is consumed. Also if it were fully ON max current can flow but the voltage across it is low, again little power is consumed. Problem comes when it is transitioning from ON to OFF and OFF to on. If that switching is slow it will spend a significant amount of time with both a large voltage across it and a large current flowing through it. That is to say consuming power and getting hot. Power W = IV.
Another problem is how well ON is it when it is on. In that state you want the voltage across it to be as small as possible such that the IV is low and power is not wasted.
So, that transistor has two requirements here:
1) A high switching speed.
2) A minimum ON voltage.
Many modern MOSFETs can do this. Look for the Ron parameter in the data sheets. As the flyback article linked to above says:
Whilst we are at it, it's good to not have much resistance in the inductor so as to minimize power loss. Again from the article:
Now, if you ever get this working nicely with the Prop replacing the 555 as the control element I would be very interested in the results.
I wonder if you could make a simple voltage detector with just a resistive divider feeding a prop pin? A 5 to 1 divider might be able to detect about 8 volts at the input based on detecting the prop input pin switching at half the 3.3 volt CMOS supply. Then the 8 volts could feed a conventional 5 volt regulator to power low current 5 volt parts.
BTW bipolar transistors are inherently slow to turn off from saturation due to charges stored in the junction. MOSFETs are universally used for fast switching these days as they don't have this 'minority carrier' issue.
Interesting. I have an old 2 by 16 LCD that needs a minus 5v or so supply to get any contrast out of it. Looks like a cap and diode or two on a Prop pin could do it.
I think the output capacitor has a great deal to do with the circuit being able to store the charge until the next swing is approaching.
Well, imagine that the switching transistor is like a resistor connecting your
inductor to ground. That resistance is controlled by the current into the base
of a reguralr transistor or the voltage applied to the gate of a MOSFET.
Ideally when the thing is ON, the resistance from collector to emitter (or
drain to source for a MOSFET) is zero. Thinking in terms of Ohms law V = I * R.
So if R is zero so is V. Then the power dissipated is W = I * V. Which is also
zero as V is zero. All is good.
Alternatively when the thing is OFF ideally the resistance it presents is
infinite and we have from Ohms law I = V / R. So I is zero because dividing by
infinity gives zero. Then the power dissipated is W = I * V again which is
again zero. All is good.
So we see that if the transistor can switch instantly the power dissipated in
it is zero.
But what happens if it switches slowly? Then it spends a lot of time with some
intermediate voltage across it's terminals and significant current flowing.
Resulting in significant power dissipation and heat.
Of course no transistor is perfect. So for example in the ON state it's
resistance, collector to emitter or drain to source is not zero. So in the ON
state it will be dissipating power and generating heat. Best to find transistors
with low ON resistance specs, Ron.
Actually when I said "minimum ON voltage" I meant the voltage, drain to source,
when it is switched ON. Which as we see above we would like to have as low as
possible.