Eh?lectricity

BullwinkleII

SuperVeg wrote:

Charge pumps are never used for high power applications. The biggest I have seen is a 30W one (just in an application note for a controller IC). To be honest Im not sure why exactly high power ones are not used, but I would guess it is because the components would be very expensive, far more expensive than just building an ordinary boost switch mode power supply. A rudimentary one would look like this (this is a design im using in something I am designing for a client)

The PWM_Vol comes from the microprocessor. The micro drives a PWM signal to vary the voltage on the power supply.
The orange wire PSU_V goes to the ADC of the micro and provide a voltage feedback to the micro so it can adjust the PWM signal to get the correct voltage. My application requires a variable voltage supply and this is the simplest way to do it. The voltage can vary from maybe 150v down to just a few volts. There are limitations to such a simple boost supply but they are acceptable to my design.

Attachment:

basic boost supply (Custom).png

EDIT: forgot attachment :whistle:

I thought anything with a coil would generate heat and be extra lossy.

And on a more general note...There is something really fundamental that I cant wrap my head around with this el;ectronics stuff. It just doesn't seem to make sense. Just when I think I've understood something, it slaps me in the face.

BullwinkleII

DuiNui wrote:

BullwinkleII wrote:

If you are using circuits like a voltage doubler to get to the desired voltage, how much actual useful power is lost by the diode? is it proportional to the amount of voltage lost?

Standard diodes, typically 0.7V
LP Schottky ~0.3V

Power is always proportional to voltage P=VI and V=IR

I take it the smile is because you can guess I have the V / i | R triangle thing stuck to my screen, but dont think about it before asking a question

DuiNui

Coils behave in a very similar way to a Cap.
As I said before, Caps are very lossy, really only any use to store volts for a very short time, unless you have very very expensive ones.
They can however deliver massive currents in a very short time.

werdna

An inductor works to stop change.
It is generally used to stop noise or to resist change in this kind of circuit.
It stores energy in the form of a magnetic field and releases it when the voltage drops, working similar to a capacitor.

At the end of the day it is a long wire, they are not lossy at all and shouldnt get hot.
They can vibrate if there is a frequency running through them, causing bad solder joints over time.

Charlie

Are you sure about that andrew?

SuperVeg

Everything generates heat and wastes power, we just do our best to compromise between loss, size and cost.
With that boost circuit I drew, if I wanted it to supply much more current than it does I would use a much larger inductor with thicker wire. Also the switching FET would have to be changed for one with a lower on-resistance to further minimise losses.

Inductors are certainly harder to understand than capactors. Capacitors can be easily visualised to simply "store voltage" but inductors while they "store current" do things slightly non-intuitively.
If a current is running through an inductor a magnetic field exists through it. When you disconnect the circuit, the magnetic field cannot just dissapear and so "forces" the current to keep flowing. Since the circuit is disconnected it is very hard to make the current flow and so the voltage increases dramatically and so a spark occurs across the part of the circuit that has just been disconnected. It is this creation of the high voltage that the boost power supply uses. Everytime the FET is turned OFF the voltage goes up, and is stored in the CAP. When the FET turns ON the diode stops the cap discharging back through the FET and current flows through the inductor again. Repeat.
The faster this cycle occurs the smaller the inductor and cap is required, the smaller the peak currents and generally the higher the efficiency (to a point...complex)

Im not sure if anyone actually wanted that explained... it just happend

Charlie

I was trying to stir andrew... hehe

werdna

I noticed you havent commented on my videos Charlie...

Charlie

yea you win... Im still confused though I must admit.

Im glad I learned something, I always thought of power is that you never get something for nothing... Ive never looked into the design of a pond pump. All my schooling is in relation to hydraulics so not quite the same

but yes, you win all the same.. just dont tell anyone ok

werdna

I didnt win, our discussion has meant there is visible proof.
It is a very common theory that people get wrong

BullwinkleII

SuperVeg wrote:

Everything generates heat and wastes power, we just do our best to compromise between loss, size and cost.
With that boost circuit I drew, if I wanted it to supply much more current than it does I would use a much larger inductor with thicker wire. Also the switching FET would have to be changed for one with a lower on-resistance to further minimise losses.

Inductors are certainly harder to understand than capactors. Capacitors can be easily visualised to simply "store voltage" but inductors while they "store current" do things slightly non-intuitively.
If a current is running through an inductor a magnetic field exists through it. When you disconnect the circuit, the magnetic field cannot just dissapear and so "forces" the current to keep flowing. Since the circuit is disconnected it is very hard to make the current flow and so the voltage increases dramatically and so a spark occurs across the part of the circuit that has just been disconnected. It is this creation of the high voltage that the boost power supply uses. Everytime the FET is turned OFF the voltage goes up, and is stored in the CAP. When the FET turns ON the diode stops the cap discharging back through the FET and current flows through the inductor again. Repeat.
The faster this cycle occurs the smaller the inductor and cap is required, the smaller the peak currents and generally the higher the efficiency (to a point...complex)

Im not sure if anyone actually wanted that explained... it just happend

You see.

Now thats just the kind of stuff that makes plumbing seem so "real world" compared to the E-Lec_tronics

DuiNui

Plumbing is hammers and nails.
Electronics is all mathematical, highly predictable, providing you know the right equations.

Sleepe

I'm not sure whether this is the right thread but I just got a small solar panel for nix.

It wasn't producing voltage but there are two diodes on the back which were going both ways. I cut the leg on one diode and its now producing voltage in proportion to light exposure. On the back it has:-
Maximum power (pm) 35Wp
Open circuit voltage (voc) 21.96v
Short circuit current (isc) 2.19a
Maximum power voltage (vmp) 17.82v
Maximum power current (imp) 1.97a
I am assuming the diodes were for protection and failed. It also has three terminals the two outers are now positive and negative respectively.
I know nothing about solar panels but if I replaced the diodes what could I use it for and what extra would I need (running an air pump or small water pump would be nice)

SuperVeg

got a close up pic of the back with the terminals and diodes ?

Even better, I have this link that explains diodes and solar panels

http://www.solar-facts.com/panels/panel-diodes.php

Not sure why what the configuration of your diodes are but if you only have one panel you only need 1 diode on the positive terminal (stripe on diode away from positive terminal) to stop the battery sending current back into the panel at night.

Maybe your panel is actually 2 panels ?

To replace it I would use this
http://www.jaycar.com.au/productView.asp?ID=ZR1048

Sleepe

This is as good as I get with a camera

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