GH2 GH2 Power - specs, tests, DIY

So I will not need the PTN78000WAH to make the reduction from 12 v to 8.6 v? I need only the part that you mentioned?

Or this part will work in conjunction with the PTN78000WAH?
 
the LM317T is a voltage regulator. It will output 8.7Vdc when fed 12vdc after being configured properly. You don't need the PTN78000 part when using it.
 
I already bought PTN78000WAH and have to give him useful.

Who will arm the system is a professional, I just want to facilitate his work, purchasing all equipment necessary to attach to PTN78000WAH

The next project I will use the LM317T, I plan to buy my second GH2.

Thanks Svart, We thank you more if you can list in detail the parts for PTN78000WAH
 
wow, you guys are making things much harder than they need to be.

Eugenio is, not me. I did warn him about trying to make electronic devices at home: http://www.dvxuser.com/V6/showthrea...cs-tests-DIY&p=2313358&viewfull=1#post2313358


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.

Yes there is: a switcher loses less power and does not require a heatsink. To power the GH2 externally you need a supply of >8.4 volts rated at 1.2 Amps that typically puts out 740 mA (this measured current figure comes from the first post in this thread and I assume it is correct). Let's design a regulator to provide 8.6 volts at a current of 800 mA, driven by a certain number of Panasonic, Li-ion, 3.7 volt, 1950 mAh batteries. (http://www.panasonic.com/industrial/includes/pdf/Panasonic_LiIon_CGA103450A.pdf).

LM317 design
Minimum input/output differential: 2.5 volts. Let's say 3.0 volts.
Thermal resistance (junction to case): 65ºC/Watt.

For this design to regulate properly, it requires a minimum input voltage of 8.6 + 3.0 = 11.6 volts (3.9 volts/cell). Would a three-cell Li-ion pack work? Not really. The "Discharge Graph" of the PDF for this battery (see above), at an interpolated 800 mA, indicates it would provide only about 400 mAh down to 3.9 volts, below which the output voltage of the regulator would start to fall. So we'd only get 400 mAh from these cells. Not good.

Let's try 4 cells. That gives us a starting voltage of 16.4 volts, and a cut-off point of 2.9 volts (11.6/4) before the regulator output begins to fall. That's good. It means the batteries provides all their capacity. Are there any problems with this design? Yes – power dissipation!

Power dissipated in LM317 for fully charged batteries = (16.4 - 8.6) x 0.8 = 6.2 Watts. It can be shown (too complicated to explain here) that the LM317 has a maximum power dissipation, without a heatsink, of about 1.5 Watts. We need a heatsink on the LM317 when run from fully charged batteries. My calculations show that the heatsink would have to be rated at less than 15ºC/W, i.e. something like this: http://www.jaycar.com.au/productView.asp?ID=HH8504

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. But worst of all, is the power wasted in the LM317 over the complete discharge cycle of the Li-ion batteries – on average 4.5 Watts (800 mA at an average volta drop across the LM317 of 5.6 volts). In providing about 6 Watts to the GH2, you need 10 Watts from the battery. 40% of the battery capacity is wasted.

There are switcher equivalents of the LM317 (http://products.cui.com/adtemplate_...tky=328060&subcatky1=521999&subcatky2=568080#), and they would not need a heatsink, and would have minimal power loss. To me, the ideal external power for a GH2 would be to have a switcher LM317 set to 7.2 volts, housed inside a DCC8.


The next project I will use the LM317T, I plan to buy my second GH2.

The LM317 may not be a good idea.. There are better, off-the-shelf solutions.

SUMMARY
It is not easy to get the best out of an external power supply when trying to assemble the components yourself. And by best I mean it has to be reliable, look aesthetic, and uses all of the battery capacity available. Do you choose Li-ion, NiMH or Lead acid? What capacity? What voltage? How are you going to charge it? How are you going to connect it? You could end up with a hotch potch of components and leads that look a mess and which only uses 25% of the battery capacity. Check out my Batteries in the Wild post for more discussion about external power for the GH2 and other devices.

And for those interested in batteries: the best single document I have come across is: http://www.national.com/appinfo/power/files/f19.pdf.
 
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 :)).
 
Heh. I design switching power supplies for a living. I can tell you guys about some stuff that's not trivial.

This is trivial. :)

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.

I personally have to make choices every day about what I'm going to use to power devices with these things. Sometimes I use switchers, sometimes I use linears(of various types). It all depends on quality of the output, cost, ease of use, and waste.

Something like this doesn't need to be extremely efficient, but it DOES need to be quiet (no hash, no ground bounce, no transients, etc), a linear excels here. Maybe not even a LM317T, maybe a linear pass element would work better as you can scale your darlington to meet dissipation requirements.

In any case, I don't understand where you are coming from with the battery/LM317T discussion. I thought this was simply a wallwart to camera adapter discussion? In either case, you are doing math for increasing the voltage to 16.4V, which I do not understand. I would PARALLEL the cells up and get more current and leave the voltage the same. Anyway, you just can't do that with LI-ION as the battery watchdogs are usually only designed to monitor 2-3 series and 2-3 parallel cells. In this case, any battery I want to upgrade, I would choose higher mAh for each cell.

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 $10 dollars of components, and 5-6 hours to build.
nah it's more like 2$ 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.

Besides, for the switching design, you also need (minimum) 2 caps (probably expensive low ESR tantalums for output stability of the switcher), 1 resistor (for output V setting) and the same case/mounting you would need for the linear supply for a net difference between the designs of 1 resistor but possibly add a couple more $$ for the caps. So you really aren't gaining anything much, just lower dissipation(at the cost of needing more filtering or gaining more noise).
 
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I need to know the names of parts that will be part of the voltage regulator PTN78000WAH.

I'm going to use 12v battery with 6800mah and I want the regulator to convert 8.6 v 1.5 A

Please someone help me.
 
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 :)).

Don't reckon there'd be any money in it, except if the Chinese took it on. I don't know why they haven't. The made a battery for the GH2, and they should be churning out DCC8s and 8.6 volt regulators. What I will be doing in the next few weeks is buying one of those $5 converters from HK, running it for a week or two at maximum output, and testing it thoroughly.
 
"I need to know the names of parts that will be part of the voltage regulator PTN78000WAH."

The text below is a list of parts that I need?

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


If anyone knows any site where to find these parts, and the quantity of each part, tell me !!!
 
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Heh. I design switching power supplies for a living. I can tell you guys about some stuff that's not trivial. This is trivial. :)

Sure, for someone with your experience. Not for the hobbiests and non-pro hackers that tend to hang out here.

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.
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.

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.
 
if you can get that module for 10$ then, it's considerably more viable than the 20$ I saw it for on mouser's site. Good find!

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.

If it were me, I'd just get a 12V 2A smps and adjust the feedback divider around the tl431(what most use for voltage monitoring on the outputs) until you get ~8.6v. I've done this before for one-off projects that need odd voltages.

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.
 
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.

I would use generic 12V 1A switchers like these (ultra cheap and easy to replace if it ever dies).

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.

Agreed - the circuit boards can be seen here, see if you can spot the part.
 
I'll buy the components for PTN78000WAH in the site of Texas Instruments. Please someone post it here for me the link for each component that must surely have in stock of Texas Instruments

http://www.ti.com/
 
So on another note, I got my chinese batteries the other day. I've opened one up and found two flat/rectangular cells in series and a small PCB with mosfet, controller IC and supporting parts. I have not tried these batteries yet as my camera is on loan to a director I'm working with right now.

Good news:
The PCB and parts are seeming good quality. The PCB is actually silkscreened and in english. It seems to be of higher quality than a lot of production chinese batteries I've come across.

Bad news:
The cells are unbranded and unidentifiable. I have not found anything similar at all from any of my sources.
The controller IC is also completely unmarked.


Neutral news:
The data(D) pin leads to the IC. Not sure if any communication happens or not. When I get the camera back I'll hook it up to the scope and see.
The thermistor(T) pin leads to a voltage divider and some sot23 mosfets. Not what I expected. Usually there is either a thermistor or a dummy load connected here as most battery controllers need to see some load here to turn on.
 
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?
 
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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?

I'll get some pics today, however the soldermask is white and you cannot see the traces at all. Tracing through with a meter shows that the PCB is mostly just an over/under voltage protection circuit. There is a single mosfet that makes/breaks the - side of the battery.

I got it on ebay. I'll link to it later. I wanted to hold off linking to it in case it was a poorly designed battery so that folks won't buy junk. So far it seems decent enough though.

Yes, after removing the batteries(99% of the whole volume of the pack) there is plenty of room for other things.
 
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I am using it too. Works fine and the connector is a perfect fit.

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).
 
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).
The adapter outputs 9.61V when open.
I also measured the voltage coming ffrom 8 NiMH (Eneloop) cells. It comes to 10.46V open.
When connected to GH2 body (no lens) the voltage drops to 10.08V. With the 14-140 lens attached the voltage drops to 9.8V. The camera seems to be fine with it for over 10 minutes.

So it seems even 10V is safe. I tried to connect a DC adapter rated at 9V that displayed 12.5V when open and the camera did not even turn on.

Unfortunately my current meter is maxing out at 200mA so I cannot measure the current going through.
 
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