EveryCircuit - Single Use Voltage Booster

8 years ago

No blocking diode for the relay?

8 years ago

I think @jason just used the relay to make simulation easier, its not suppose to be a part of a real circuit. If you want to test it in real just use a switch (put two wires together). It does work, nice experiment! Keep it simple.

hurz

8 years ago

Here for maybe better understanding and to rebuild on a breadboard without relay http://everycircuit.com/circuit/6351759550447616

Issacsutt

8 years ago

So how come the switch has to be there?

hurz

8 years ago

To show how everything behave in case of uncharged. So to start from a zero point of energy in all components. Without a switch, the simulator would do DC operating point (DC-op) setup before you can interact with you circuit. That mean after DC-op the cap would be alread charged.

Issacsutt

8 years ago

Ok that makes sense, but one thing that still confuses me a little is; if you simulate it without the switch, it starts with a voltage lower than what it would eventually reach with the switch... so does that mean it will decrease and settle at that lower voltage if you simulate it long enough?

jason9

8 years ago

If the capacitor was imperfect and slightly leaky, like most probably are, then after long enough it would settle at the lower voltage, but this only simulates perfect components.

Issacsutt

8 years ago

Yeah

hurz

8 years ago

@Issacsutt, good observation, right if the switch is closed while startup the transient dynamic is not taken into this scenario! So the resonance effect which makes the voltage doubling is not present!! Its then just a DC analysis and this is limited to the basics of Ohms Law nothing else. Inductivity doesn't matter and the coil is e.g just a zero ohm resistor. And you get something close to the supply voltage at the cap. With switch open its getting interessting, you really start in the moment of close it manual a transient analysis and the effect of resonance takes takes!

Issacsutt

8 years ago

So this is entirely dependent upon the resonance of the inductor and capacitor? ...and nothing to do with an inductor voltage spike?

jason9

8 years ago

@issacsutt, the way I understand this to work is that at first the inductor blocks the current to the capacitor, then it starts to let the current through and by the time the capacitor charges there is still a lot of current going through the inductor which then charges the capacitor further until the voltage from the overcharged capacitor stops the current flowing through the inductor. After that, the diode prevents the capacitor from discharging. If you removed the diode, then it would create a resonant circuit that oscillates between 0V and 24V. The diode just keeps it from going from 24V back to 0V.

Issacsutt

8 years ago

Ok, so what you mean is; the inductor starts out with letting 0 current flow through, and gradually increases the current flow until the cap has fully charged, then the inductor begins to decay the current flowing through it, and because it's still pushing current through, that forces the capacitor to charge a little further, and thus increases the voltage across it?

jason9

8 years ago

Exactly.

Issacsutt

8 years ago

Ok cool

Issacsutt

8 years ago

Hurz, in your description, you said: "with some losses, it could be demonstrated with a simple swith". But, I took those resistors out, and it still functions the same, so do you really need them there? http://everycircuit.com/circuit/4688378846248960

hurz

8 years ago

The little resistors are not all losses, the diode itself has losses in resistance and threshhold and the simulator needs in theorie infinite oversampling frequency or it cause dumpending by rounding errors.

hurz

8 years ago

Set the diode to a almost perfect one and run simulation slower you will come much closer to 20V http://everycircuit.com/circuit/6004241130061824

hurz

8 years ago

The diode is just for your comfort, run simulation much slower and use a switch, when your reach 20V open switch. So you are the diode 😋

Issacsutt

8 years ago

I understand the purpose of the diode, I forgot about considering its resistance though. Thanks, thats pretty neat!

hurz

8 years ago

Right, but to get more then this amount of energy (20V at this cap) you need a different, more advanced strategie. We have to charge the coil before we close the switch to the cap. But this is another topic for a real boost converter.

Issacsutt

8 years ago

I'm in, let's go!

hurz

8 years ago

Continues boost converter are more then needed available on everycircuit, im out.

Issacsutt

8 years ago

Continues boost converter? Thought you said that was a topic for a real boost converter, leading me to believe there is a more advanced way to design them that others may not know? So I guess it would be too time consuming of a topic that I wouldn't get huh...?

hurz

8 years ago

This 'here' is not my thread!

Issacsutt

8 years ago

Right, sorry Jason

jason9

8 years ago

You can go on. I don’t mind. In fact, I’m kind of curious myself.

Issacsutt

8 years ago

Cool, guess well see what he decides to do

hurz

8 years ago

I'm not your drill instructor, you have to come up with ideas, circuits, questions.

Issacsutt

8 years ago

Ok, well then how about this: How would you approach a design for a boost converter that takes a minimum input voltage of 4.5v and boosts it to a common 12v with a maximum load current draw of 5amps (60 watts)? So, what I know is that you cannot draw more power out then what you put in according to laws of concervation of energy.... so does this mean: whatever method you choose to use to boost the voltage, must draw 60 watts of power at the input? So, if that's true, and your using an inductor to achieve this, then the inductor would have to be charged with a current of 13.333... amps (60/4.5)? If that's right, then how do you design such a circuit with components to handle that kind of power, whether your using just an inductor, or just capacitors, or both?

hurz

8 years ago

Nice try, go ahead or start easy.

jason9

8 years ago

That is correct. The main way for an inductor based design using high amperage is with a high power MOSFET switch. I’ll make one myself now with the specifications you gave.

hurz

8 years ago

And try to follow in real!

Issacsutt

8 years ago

Well then what would you recommend hurz?

hurz

8 years ago

Issacsutt, if you do not have any good questions, we are finished

Issacsutt

8 years ago

I thought I already asked a good question, and you shot it down or something. Sorry I bothered

jason9

8 years ago

@hurz, @issacsutt didn’t know how it was supposed to be done so he asked how it should be done and you shot him down. Why don’t you actually help him out?

hurz

8 years ago

Go and help him, I dont see a need to help for crazy projects, before you are ready to understand basics. Keep it simple!

Issacsutt

8 years ago

Ok then, what are the basics?

hurz

8 years ago

I think this is not how EC works. You just stay there and say "feed me feed me" like a little hungry bird. I have made bad experience with this way of knowledge transfer. Cuz, its very limitited in No of students and is somehow dissapointing for the teacher, cuz the students dont care after they got what they want FOR FREE. I have to many other topics/projects to spend lessons. I have to optimize what I can give you, cuz time is limited.

Issacsutt

8 years ago

Ok, I understand. Thank you.