Help with boost converter circuit

[billybob]

I was messing around with a boost converter circuit and I'm trying to understand how I can milk more power from it without changing the input voltage.

I have a 3.7V lithium ion battery input and I would like the output to be a steady 9-12V able to draw at least 0.3 Amps.
Here is the schematic I used:
I ended up bridging the potentiometer completely, replaced R2 with 100ohm, Q1 is C9014, and I'm not sure how to accurately measure L1.

The output I'm getting is around 40-70V, but drops to 2.6V with a 100ohm load. The current drawn from the power supply says 0.27A.
What can I do?

 

[AnalogKid]

There are a couple of problems with the circuit.

For an output voltage of 11.1 V, the output power is 3.33 W. At the impossible efficiency of 100%, the average (not peak) current through Q1 is 900 mA, while the BC547 is rated for an absolute max. collector current of only 100 mA.

The out-in voltage ratio is 3:1, so the current ratio must be 1:3. That is, the current through the switch must be 3x the output current, or 2.7 A. Again, this is at 100% efficiency. The actual current will be greater. The inductor also must be able to handle this current.

The output diode (reference designator?) is too slow a part for the application. A diode has a performance parameter called it switching speed. Your circuit will need a fast or ultra-fast diode. Note that these diodes have a higher Vf (forward voltage) than a typical power rectifier. This must be taken into account when calculating the voltage and current transfer ratios.

There are many boost converter design websites on the innergoogle. Attached is an app note from Linear Tech. It is long and complicated, but it covers *every* aspect of boost converter design. The basic equations are universal, and apply to your circuit.

ak

 

[billybob]

Thank you for the quick response.
My question is, say I replaced Q1 in the circuit about with a more suitable transistor or mosfet, and have ultra high switching diodes, why isn't the circuit drawing more amps from the input power supply? I would think that if the voltage is dropping on the load side the current would have to increase on the input side, right?

 

[Nigel Goodwin]

Thank you for the quick response.
My question is, say I replaced Q1 in the circuit about with a more suitable transistor or mosfet, and have ultra high switching diodes, why isn't the circuit drawing more amps from the input power supply? I would think that if the voltage is dropping on the load side the current would have to increase on the input side, right?

You have no feedback, so it won't work in any usable fashion - it should work by turning itself up as the load increases - and needless to say, a 555 is a really poor chip to try and use anyway.

 

[billybob]

So a circuit that works off it's resonant frequency might be better?

 

[Nigel Goodwin]

So a circuit that works off it's resonant frequency might be better?

No, you need feedback for it to work usably.

 

[Papabravo]

Your initial post reveals that you have never learned or do not remember the immutable rule of DC-DC conversion schemes. It is this: The output power will ALWAYS be less than the input power. In some cases, it will be a great deal less. Depending on the efficiency you have achieved it may be worth your while to see if you can make incremental improvements by other means.

 

[crutschow]

Boost converters are difficult to build without a dedicated controller IC.
Why not just buy a cheap boost module such as this?

 

[billybob]

Boost converters are difficult to build without a dedicated controller IC.
Why not just buy a cheap boost module such as this?

Yeah the purpose isnt really for perfection, I'm just enjoying building it.

 

[billybob]

I'm trying to understand the principle at the moment, and I'm curious why as soon as I install a load the voltage drops from 50-80V to 2-5V with little to no current.
Why won't the current move up instead of the voltage dropping?

 

[Papabravo]

I'm trying to understand the principle at the moment, and I'm curious why as soon as I install a load the voltage drops from 50-80V to 2-5V with little to no current.
Why won't the current move up instead of the voltage dropping?

It is the necessary tradeoff. If you are going to boost the voltage, you must of necessity lower the output current. When the load requires more current than the device can provide, the output voltage will drop like a turd in the rain.

It might help to assume, for the sake of argument, that your converter is 75% efficient. Now estimate your input power. Let us say it is 12 watts. Multiply that power by 75% and you have 9 watts. Now let's say you boosted the voltage by a factor of 5. Now divide the 9 watts by 5 times the input voltage and that will be your maximum output current.

If it were otherwise, you would have unlimited free power available, and we know that just aint happenin' - not is this universe.

 

[crutschow]

Yeah the purpose isnt really for perfection, I'm just enjoying building it.

Okay.
But I know of no simple regulated boost converter that can be built with common parts.
And if a boost converter is not regulated, the output will vary widely with load, as compared to a buck regulator.

 

[audioguru]

The datasheet for the LM555 says its minimum supply voltage is 4.5V. With a 5V supply, its output high is only 3.4V with a 100mA load (0.34W).

 

[AnalogKid]

I'm trying to understand the principle at the moment, and I'm curious why as soon as I install a load the voltage drops from 50-80V to 2-5V with little to no current.
Why won't the current move up instead of the voltage dropping?

Because without any kind of feedback and regulation, what you have is not a constant-voltage output; it is a constant energy output. And since energy (in watts) equals voltage times current, if one goes up, such as increasing current caused by a lower load impedance, the other goes down, the output voltage sag you are seeing.

The max energy available at the output is set by the energy available at the input and the switch duty cycle. Other factors that reduce this are the inductor size and the switch parameters.

Your components indicate an oscillator freq of over 2.5 MHz. A bipolar 555 is not characterized for operation at that speed. It might oscillate, but the output will not be very "square". Also, at that frequency the inductor impedance is so high that I'm surprised there is any output at all.

The original circuit has a nominal freq of 150 kHz. At that freq, and assuming the inductor is supposed to be 80 uH (as shown in the parts list) rather than 80 mH, the inductor impedance still is around 75 ohms. That still is too high for the output voltage and current you want.

NOTE: There is no simple fix for your circuit; it will not work for a variety of independent reasons. And even if you get it to work as the original designer intended, it will not do what you want. It is the wrong circuit for your application. Until you do some studying of how and why boost converters work, no amount of try-this / try-hat will work.

ak

 

[Diver300]

I'm trying to understand the principle at the moment, and I'm curious why as soon as I install a load the voltage drops from 50-80V to 2-5V with little to no current.
Why won't the current move up instead of the voltage dropping?

In real life there will be losses, but it's always good to work out the ideal result first. If the ideal result doesn't do what you want, the real life result will always be worse, so you need a better idea.

If you ignore losses for now, this is what you get.

Unless the output voltage is very low and the off time is very short, the current in the coil will drop to zero while the transistor is off.

When the transistor turns on, the current starts increasing in the coil.

During the time that the transistor is on, the increase in current is V * Ton / L
Where V in the input voltage, Ton is the time that the transistor is on, and L is the inductance.

As the current increases linearly, the average current is 0.5 * V * Ton / L and the energy input is 0.5 * V² * Ton / L

If the switching frequency is F, the input power is 0.5 * F * V² * Ton / L

If there are no losses, that power is transferred to the output, and what you've got is a power supply that runs at constant power (if the output voltage is large enough that the current drops to zero each cycle). So at no load you get infinite output voltage. If you put a resistive load, there will be an output voltage, but halving that resistance will reduce the voltage to around 71% of what you had before. At quarter the original resistance you get half the original voltage.

So that is the best that the power supply could possibly perform without losses and it seem pretty terrible, so the basic idea needs changing. As others have said, you need feedback.

Also, I don't know what the resistance and current rating of your 80 mH coil is. If you are wanting to take 0.3 A at 12 V, that is 3.6 W, so the average input current needs to be 1 A. To get an average current of 1 A, you will need a peak current of at least 2 A and probably more.

Digikey sell one inductor that is around that. They sell one made by Hammond, rated at 100 mH and 5 A, so it's rated a little larger than you want, but not much. It would be a good starts if you want to use somewhere around 80 mH.

https://www.hammfg.com/electronics/transformers/choke/195-196

It costs over $100 and weighs 14 lbs.

Even that monster has a resistance of 0.64 Ohms so the resistive loss will be 0.64 W at 1 A, giving the best possible efficiency of 85%, and there will be other losses as well.

My take from that is that 80 mH is far too much. A lower value inductor will be much smaller and cheaper and have less resistance to get the current rating you want, but will need a high switching speed. To get anywhere near a constant voltage output you need voltage feedback.

 

[Nigel Goodwin]

I'm trying to understand the principle at the moment, and I'm curious why as soon as I install a load the voltage drops from 50-80V to 2-5V with little to no current.
Why won't the current move up instead of the voltage dropping?

I've already explained twice - here's a third:

feedback, feedback, feedback!!!!!!

 

[billybob]

Because without any kind of feedback and regulation, what you have is not a constant-voltage output; it is a constant energy output. And since energy (in watts) equals voltage times current, if one goes up, such as increasing current caused by a lower load impedance, the other goes down, the output voltage sag you are seeing.

The max energy available at the output is set by the energy available at the input and the switch duty cycle. Other factors that reduce this are the inductor size and the switch parameters.

Your components indicate an oscillator freq of over 2.5 MHz. A bipolar 555 is not characterized for operation at that speed. It might oscillate, but the output will not be very "square". Also, at that frequency the inductor impedance is so high that I'm surprised there is any output at all.

The original circuit has a nominal freq of 150 kHz. At that freq, and assuming the inductor is supposed to be 80 uH (as shown in the parts list) rather than 80 mH, the inductor impedance still is around 75 ohms. That still is too high for the output voltage and current you want.

NOTE: There is no simple fix for your circuit; it will not work for a variety of independent reasons. And even if you get it to work as the original designer intended, it will not do what you want. It is the wrong circuit for your application. Until you do some studying of how and why boost converters work, no amount of try-this / try-hat will work.

ak

That is very helpful information thank you. The sound the circuit made with the original resistors was too annoying so I tried to increase the frequency. How did you calculate the frequency output? Just curious.

 

[billybob]

I've already explained twice - here's a third:

feedback, feedback, feedback!!!!!!

Hmm I think I'm hearing feedback but I'm not sure...

I will probably look into an IC that provides feedback and has more efficiency than I do now, thank you.

 

[AnalogKid]

How did you calculate the frequency output? Just curious.

The equation is in the 555 datasheet.

ak

 

[eTech]

Hmm I think I'm hearing feedback but I'm not sure...

I will probably look into an IC that provides feedback and has more efficiency than I do now, thank you.

Hi
"Feedback" is a term used in switching regulators that describes the output signal being "fed back" to the input circuit so the circuit can adjust its output level(s). This forms a feedback loop. Your circuit has no feedback circuit. There are different types of switching regulators. Voltage mode regulators control the output by changing the duty cycle of the switching frequency using a "feedback" signal sent back to an error amplifier.