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wow, you guys are making things much harder than they need to be.
What I'm trying to say is that there is just no good reason to use a 20$ switcher where a .50$ linear will do.
The next project I will use the LM317T, I plan to buy my second GH2.
nah it's more like 2$ in parts(that you buy), (not sure where you get your parts) and you could do away with the pot if you used the resistor values I mentioned earlier and save more money. I'd just thermal glue to the LM317 to an old computer CPU heatsink and use the 12v input to power the fan too. But I'd also do this because I have all this laying around my lab and can build it for free.An LM317 design sounds simple, but really, it is not in the running. You need an LM317, a resistor, a pot, 2 capacitors, a heatsink, a circuit board, a mounting case, and input and output sockets. About $10 dollars of components, and 5-6 hours to build.
As guyburns is showing, it's not trivial to build electronics devices. The devil is in the details (hell, I just spent two days reading MOSFET datasheets).
guyburns, maybe you should build and sell voltage good regulators here, might be some money in it (nobody ask me, I'm too busy as it is).
C1
Um microfarad 2.2, 25 volts, capacitor cerâmico multicamadas, conforme descrito na página 13 e Tabela 5 da página 15. É um dispositivo de montagem em superfície (uma dor de utilização) e custou cerca de 20 centavos. Este é um componente complicado pegar. Eu nunca tive de usar e amadores balcões provavelmente não vendê-lo. Aqui está o C1 será parecido com: http://australia.rs-online.com/web/c...294635735&No=0
C2
Um microfarad 100, 16 volts capacitor eletrolítico. Comumente disponíveis.
R (conjunto)
A 1%, 1 / 2 ou 1 / 4 watt resistor cujo valor pode ser calculado a partir da fórmula (1) na página 10. Eu trabalhei para fora de 4,3 kohms. Felizmente, este valor padrão na faixa de 1%. Você pode calcular a tensão de saída a partir desta fórmula, mas rearrrange (1) se a certeza de que essa é correta:
V = (82,1 + 2,5 R) / (6,49 + R) = 8,61
Heh. I design switching power supplies for a living. I can tell you guys about some stuff that's not trivial. This is trivial.![]()
The TI part can be gotten as a free sample from the TI website, or the 3A version is on Ebay for a decent price. It also has all kinds of protections that are useful (including overheating), especially for people not super-schooled in making their own circuits, thermal management etc. and therefore liable to make mistakes which could (at worst) kill their camera - and it's efficient, saving money on wasted battery juice. And it also just needs two caps and one resistor (better quality ones, sure), but not even a heatsink. Worth the extra money considering (as you point out) you need parts and an enclosure either way.I know what you are saying Guyburns, you seem like a very knowledgeable person and are certainly on the right track with your ideas. If this were some production design, I would agree 100% about the use of a SMPS for this. However, this is just a quick and cheap project for most and cash is king.
And then you can use cheap AC and your own batteries (people care because the official ones are small and bloody expensive). So as you get both options from the one part, to me it's worth the investment, even if you buy the part.
You'd still need some way of stepping down line level and then rectifying so you can't get around using a wall wart or switcher of some kind.
What I'd really like to know is what IC panny is using to control the battery. That way we'd really know what we can safely feed the camera with. Chances are that 9vdc is perfectly fine and that all this talk about getting exactly 8.6V is really just academic but not realistic.
Q1: Can you take a close up of battery internals and both sides of the PCB, and gmail them to gdburns? Jpeg >2 MB if possible.
Q2: Where did you buy this battery?
Q3: Does it look like it would be possible to remove the batteries, keep the existing PCB, and leave enough room for another PCB? What would be the maximum size of an additional PCB?
For reference: this thread http://www.dvxuser.com/V6/showthread.php?246528-alternative-to-the-dmw-ac8 states that a particular AC-DC adaptor, 9.5 volts, 2A, works acceptably into the DCC8 to power the GH2.
I am using it too. Works fine and the connector is a perfect fit.
The adapter outputs 9.61V when open.Could you (or someone else using a similar external supply) measure the open-circuit voltage of the power supply and the current drawn from the supply by the GH2? I want to confirm that 9.5 volts is not too-high a voltage, and that the GH2 draws about 800 mA when powered externally (according to post 1).