Battery ?

SgtWookie

Joined Jul 17, 2007
22,230
Or look at it another way.

The adapter is rated for 9v output with a 400mA load.

You measured it with no load at 11.5v.

So, what's the internal resistance of the adapter?

And, what's the output voltage of the adapter when a 150mA load is on it?

Hint: apply Ohm's Law

[eta]
You are trying to make this much more complicated than it really is.
 

Thread Starter

Mathematics!

Joined Jul 21, 2008
1,036
Like Bmorse said, You could hook up a 9v battery rated at 2,000,000,000,000,000,000,000,000,000,000 mAh and it won't damage your circuitry, think of mAh as the capacity and ability to discharge, not the actual output. the output current is determined by the circuitry of the device, it won't take more than it needs unless of course something were shorted. Also, if the camera uses a rechargeable li-ion, then most of the current it is referencing is charging current. I have never heard of a 9v li-ion battery either, are you sure its not a 3.6 or 7.2v?
Bmorse never in his last post said anything about mAh he was talking about mA current not amp-hours which is a different story. And I agree more mAh means longer battery life the battery would beable to supply the current that the device draws for a longer period of time....

I agree but this does not tell you that picking any random 9volt battery the current the battery delievers to the device could be to high for the device to dissipate. For instance you can use any 1ohm resistor with a 9volt battery to get a current of 9amps to be drawn but I doubt if the resistor wasn't rated to dissipate more then 9watts would work . (it would blow the resistor if not the correct power ratings)

This is where I am confused you would need to know the resistance of the camera which can be measured but also the power that this resistor equivalent camera can dissipate...........???
How to do that is beyond me without seeing the circuit design.
Does anybody know how to figure out a resistors wattage rating with just a DDM ?


Or look at it another way.

The adapter is rated for 9v output with a 400mA load.

You measured it with no load at 11.5v.

So, what's the internal resistance of the adapter?

And, what's the output voltage of the adapter when a 150mA load is on it?

Hint: apply Ohm's Law

[eta]
You are trying to make this much more complicated than it really is.
11.5V/9V = x / 400mA => x = 511mA approx

then internal resistance is 11.5v/511mA = 22.5 ohms of internal resistance for the adapter.

output voltage of the adapter = 22.5ohms * 150mA = 3.375 volt approx.

Ya , but then why is the output voltage 9volts , 400mA on the sticker?
Shouldn't it be either the max ratings or the what is going to be supplied to the device?
The 9volts , 400mA is meaningless...
 
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SgtWookie

Joined Jul 17, 2007
22,230
You don't need to know the resistance of the camera.

You know that it requires 150mA current.

You know that the adapter is rated for 9v @ 400mA current.

You know that the adapter has an 11.5v output with no load.

Now you can figure it all out by starting with the adapter knowns.

It's just Ohm's Law. Addition, subtraction, multiplication, division.

Nothing complicated.
 

Thread Starter

Mathematics!

Joined Jul 21, 2008
1,036
You don't need to know the resistance of the camera.

You know that it requires 150mA current.

You know that the adapter is rated for 9v @ 400mA current.

You know that the adapter has an 11.5v output with no load.

Now you can figure it all out by starting with the adapter knowns.

It's just Ohm's Law. Addition, subtraction, multiplication, division.

Nothing complicated.
look at my previous post I updated it.

You know that it requires 150mA current.
no this is the max consumed ratings not what it is actually going to draw?

You know that the adapter is rated for 9v @ 400mA current.

You know that the adapter has an 11.5v output with no load.

Now you can figure it all out by starting with the adapter knowns.

It's just Ohm's Law. Addition, subtraction, multiplication, division.

Nothing complicated.
I just would think it would be more convient if they put for the output voltage/current of the adapter with the camera load. Just makes more sense to me. Since what the hell is 9vdc , 400mA mean somebody could interpert this as a max rating , ....etc etc I shouldn't have to use a DDM to measure the voltage/current of the output of the adapter it should be clear what the voltage/current would be for the device it is supplying....
 

SgtWookie

Joined Jul 17, 2007
22,230
Nope.

11.5v with no load.
9v with 400mA load.
So, what resistance drops 11.5v-9v=2.5v when a 400mA current flows through it?
R=E/I = 2.5/400mA = 6.25 Ohms. That is the internal resistance of your adapter.

You know that your camera presents a 100mA to 150mA load to the adapter.
So, how much voltage will be dropped across the internal resistance of the adapter?

E=IR, so E = 150mA * 6.25 = 0.9375V. 11.5v-0.9375v = 10.5625v across the camera when it's drawing 150mA.
E = 100mA * 6.25 = 0.625v; 11.5v-0.625v = 10.875v across the camera when it's drawing 100mA.
So, what's the camera power dissipation?
P=EI, so 10.5625V * 150mA = 1.584375 Watts maximum; 10.875v * 100mA = 1.0875 Watts minimum.

See how easy that was?
 
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SgtWookie

Joined Jul 17, 2007
22,230
I just would think it would be more convient if they put for the output voltage/current of the adapter with the camera load. Just makes more sense to me. Since what the hell is 9vdc , 400mA mean somebody could interpert this as a max rating , ....etc etc I shouldn't have to use a DDM to measure the voltage/current of the output of the adapter it should be clear what the voltage/current would be for the device it is supplying....
They're selling these things in huge volume. To keep the price competitive, they will select a standard adapter that is made in HUGE quantities to get a lower price, rather than build a custom adapter with your 10.56v @ 150mA rating on it.

Normally, a consumer wouldn't worry about the voltage @ current rating of an adapter, just whether it worked or not. If it didn't work, they would send it in for repair or buy a new one.

If you were the manufacturer, wouldn't you want someone to buy the replacement from you instead of a different manufacturer?
 

Thread Starter

Mathematics!

Joined Jul 21, 2008
1,036
Nope.

11.5v with no load.
9v with 400mA load.
So, what resistance drops 11.5v-9v=2.5v when a 400mA current flows through it?
R=E/I = 2.5/400mA = 6.25 Ohms. That is the internal resistance of your adapter.

You know that your camera presents a 150mA load to the adapter.
So, how much voltage will be dropped across the internal resistance of the adapter?

E=IR, so E = 150mA * 6.25 = 0.9375V. 11.5v-0.9375v = 10.5625v supplied to the camera.
So, what's the camera power dissipation?
P=EI, so 10.5625V * 150mA = 1.584375 Watts.

See how easy that was?
Ya, thanks

So the output ratings on a power adapter is just a frame of reference.
You then have to at least measure the voltage with no load and from their calculate the rest.

Seems to me if I make 10.5 or 11volts with the abilty to supply 150mA or more with the only restriction being in keeping it lower power then 1.5watts. It should work to run the camera without damageing it.

I.e the camera should run fine provided that the bottom sentence is taken into consideration as well.

The only other then is I have to look into how much mAh this battery I created will allow the camera to run for.
(i.e how many hours can the battery suppy 10.5volts with 150mA ...)

correct me if I am wrong in anything I am saying.


Since they use this power adapter for many different devices probably the output voltage/current ratings where in reference to some other device or something.... either way with a frame of reference voltage and a DMM it is good to know that I can figure everything I can to produce the same battery equivalent for any device.... Cool

Thanks again
 
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SgtWookie

Joined Jul 17, 2007
22,230
So the output ratings on a power adapter is just a frame of reference.
You then have to at least measure the voltage with no load and from their calculate the rest.
Usually, people just want to know what the output of a "wall wart" type supply is going to be at a particular current.

It would help a good deal if either the internal resistance were given or the no-load voltage as well.

You know, you can double-check the math I used if you go down to Radio Shack, and pick up a 100 Ohm resistor that's rated for at least 5 Watts, and use that as a load for your adapter. When you measure across it, you should get about 10.82v. How much current do you think would be flowing through that 100 Ohm resistor?

Seems to me if I make 10.5 or 11volts with the abilty to supply 150mA or more with the only restriction being in keeping it lower power then 1.5watts. It should work to run the camera without damaging it.
You need a voltage regulated source. You could use current limiting along with that if you wanted to, but the camera will probably require extra current when it's first powered on.

I.e the camera should run fine provided that the bottom sentence is taken into consideration as well.
The voltage regulation is the key item.

The only other then is I have to look into how much mAh this battery I created will allow the camera to run for.
(i.e how many hours can the battery suppy 10.5volts with 150mA ...)
Battery mAH ratings are based on a 20 hour discharge rate.
So, if you want to be close in your estimates, use 20x150mAh = 3000mAH rated batteries.

A 3000mAH 11V battery should power your camera for a good portion of 20 hours. However, at some point it will stop functioning properly due to low voltage.

Since they use this power adapter for many different devices probably the output voltage/current ratings where in reference to some other device or something....
9V is a convenient voltage rating, as many items are powered by 9v.

either way with a frame of reference voltage and a DMM it is good to know that I can figure everything I can to produce the same battery equivalent for any device.... Cool
Note that the 9V @ 400mA rating is approximate. It may vary up to 10%, which is generally considered acceptable for unregulated supplies.

If you really want to get a more accurate idea of what the supply is capable of, performing that load test I mentioned with the 100 Ohm resistor will give you accurate data points where you can re-calculate everything.
 

Audioguru

Joined Dec 20, 2007
11,248
Wall warts with a higher current rating have a bigger transformer with thicker wire that has a lower resistance. so their voltage regulation is better than one thet is small with a lower current rating.

I have a small 9V/100mA wall-wart. Its voltage without a load is almost 18V!.
I have many larger 9V/500mA wall-warts. Their voltage without a load is about 11V.

The current drawn by the camera is not always 150mA. Sometimes it is less so the voltage from the wall-wart goes up and down as the current changes.

So the camera manufacturer selected a 9V wall-wart that had a voltage of 10.5V when the camera draws 150mA and its voltage does not rise too high when the camera draws less. The voltage from a 9V/200mA wall-wart might be 9.5V with a 150mA load but might be too high at 15V when the camera's current is less.
 

Paulo540

Joined Nov 23, 2009
191
Bmorse never in his last post said anything about mAh he was talking about mA current not amp-hours which is a different story. And I agree more mAh means longer battery life the battery would beable to supply the current that the device draws for a longer period of time....

I agree but this does not tell you that picking any random 9volt battery the current the battery delievers to the device could be to high for the device to dissipate. For instance you can use any 1ohm resistor with a 9volt battery to get a current of 9amps to be drawn but I doubt if the resistor wasn't rated to dissipate more then 9watts would work .


This is where you seem to be getting sidetracked. He was talking about the output of a plugin power supply and Im referring to batteries. power supplies are rated in mA or A at a certain voltage and batteries, being a finite energy source, are rated in mAh, The same basic laws apply though.

The battery only 'gives' as much current as it is allowed to.

They have maximum discharge rates as well, which is part of the mAh rating. another topic tho.

If you have a car battery or 8 AAAA batteries in series (12v), a resistor connected to either one will (ideally) produce the same current through it. and thus the same wattage dissipation. Right? A 1k resistor will allow 12 mA of current through it at 12v, regardless of the amount of 'available amperage'. I = E/R = 12/1000 = .012A. Wattage (.012 x 12) = .144W That really is all you should think about.


The camera has all sorts of power management circuitry built into it, including charging the li-ion battery if it has that feature. A variety of different voltages and currents are flying around inside, some for the back screen, some for the flash, some for the zoom, some for the sensor, etc etc. The battery or power supply is merely a well that the little camera workers draw from when their respective operation needs energy.

You always want the wall wart to be rated higher than the highest your device will ever use. The quickest way to melt one is to use an underated one. (I know this from theory as well as practice, lol)
 

BillB3857

Joined Feb 28, 2009
2,570
How many amps do you think the generating plant that supplies your house can produce. Probably many thousands of times more than your house could possibly use. Your hous will only use as many as it needs at any given time. Your camera will only draw as much current as it needs to operate as long as the voltage is within its operating range. Any external supply must be rated at or above that CURRENT rating with the proper voltage output. As others have said, the typical walwart is unregulated and will have various voltage outputs at various loads. That is the reason most unit manufacturers that use walwarts supply one they know will properly support their product.
 

Thread Starter

Mathematics!

Joined Jul 21, 2008
1,036
This is where you seem to be getting sidetracked. He was talking about the output of a plugin power supply and Im referring to batteries. power supplies are rated in mA or A at a certain voltage and batteries, being a finite energy source, are rated in mAh, The same basic laws apply though.

The battery only 'gives' as much current as it is allowed to.

They have maximum discharge rates as well, which is part of the mAh rating. another topic tho.

If you have a car battery or 8 AAAA batteries in series (12v), a resistor connected to either one will (ideally) produce the same current through it. and thus the same wattage dissipation. Right? A 1k resistor will allow 12 mA of current through it at 12v, regardless of the amount of 'available amperage'. I = E/R = 12/1000 = .012A. Wattage (.012 x 12) = .144W That really is all you should think about.


The camera has all sorts of power management circuitry built into it, including charging the li-ion battery if it has that feature. A variety of different voltages and currents are flying around inside, some for the back screen, some for the flash, some for the zoom, some for the sensor, etc etc. The battery or power supply is merely a well that the little camera workers draw from when their respective operation needs energy.

You always want the wall wart to be rated higher than the highest your device will ever use. The quickest way to melt one is to use an underated one. (I know this from theory as well as practice, lol)

Ok, I get what you are saying but if you use a 1kohm resistor in the main 120vac house outlet it would draw 0.12 Amps of current but will any 1kohm resistor do? I would think those 1kohm resistor would have to be rated to dissipate 14.4 watts of power or more .... so you couldn't use 1/4 , 1/2 , 1/8 watt rated resistors....

So my point was if we considered the camera as a resistor then you would need to know the wattage this resistor is rated at so you don't damage it.
Right?

Like a 12volt car battery and a 12 volt battery you made by putting batteries in series is their no difference in using one or the other provided if both where rated to deliver more current the the device could consume?

And if this is the case the only difference is the car batteries are made with less internal resistance then the series batteries....which gives them a much higher max current rating. (like 700A or so ....)

Because I doubt a 12volt battery that can produce max 150mA can start a car. But curious to know if I made a battery which was 12volts and could supply 700amps max would that beable to start the car?

I.e what would you need for the current ratings/internal resistance of a 12volt battery to be equivalent to a car battery to start the car?
 
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SgtWookie

Joined Jul 17, 2007
22,230
Car batteries are rated for their AH capacity, and also something called CCA, or "cold cranking amperes". Some automotive batteries can exceed 900A for short periods.

It usually takes well over 100A current to get an auto engine cranking fast enough to start.
 

Thread Starter

Mathematics!

Joined Jul 21, 2008
1,036
modified my previous post a little can somebody full answer the previous post
Thanks
more concerned about the wattage rateing question.....
 

steveb

Joined Jul 3, 2008
2,436
... if you use a 1kohm resistor in the main 120vac house outlet it would draw 0.12 Amps of current but will any 1kohm resistor do? I would think those 1kohm resistor would have to be rated to dissipate 14.4 watts of power or more .... so you couldn't use 1/4 , 1/2 , 1/8 watt rated resistors....
Yes, you need the proper power rating, or else the resistor will overheat. Greater power rating is OK, but less is a problem usually.


...
So my point was if we considered the camera as a resistor then you would need to know the wattage this resistor is rated at so you don't damage it.
Right?
No, you don't need to know that. The camera is designed to run with a certain specified voltage range, and to draw a certain specified current. You can estimate effective resistance from this, but a camera is likely a nonlinear resistive load. First of all, it probably has regulators in it. Linear regulators look like constant current loads, and switching regulators look like constant power loads. Neither of these is an "Ohm's-Law" type of resistance. Both types of regulators have a working input voltage range that must be maintained.

...
Like a 12volt car battery and a 12 volt battery you made by putting batteries in series is their no difference in using one or the other provided if both where rated to deliver more current the the device could consume?
Of course there are differences. But if the battery can supply the current and maintain the rated voltage range, then it should work.

...
And if this is the case the only difference is the car batteries are made with less internal resistance then the series batteries....which gives them a much higher max current rating. (like 700A or so ....)
That's not the only difference, but it is a big difference if you are comparing 8 or 10 D-cells to a car battery.

...
Because I doubt a 12volt battery that can produce max 150mA can start a car. But curious to know if I made a battery which was 12volts and could supply 700amps max would that beable to start the car?
Yes, that should be able to start your car. If the battery maintains the voltage and provides the current, it will work.

...
I.e what would you need for the current ratings/internal resistance of a 12volt battery to be equivalent to a car battery to start the car?
As an estimate, take 100 A and assume 3 V drop. This is about 30 mOhm. Just a rough guess. It would really depend on the type of car and the temperature. I expect a good car battery would be much less than 30 mOhm, but you could look in up easily.
 

SgtWookie

Joined Jul 17, 2007
22,230
If you use a 1kohm resistor in the main 120vac house outlet it would draw 0.12 Amps of current but will any 1kohm resistor do? I would think those 1kohm resistor would have to be rated to dissipate 14.4 watts of power or more .... so you couldn't use 1/4 , 1/2 , 1/8 watt rated resistors....
Safety note: you don't use resistors to reduce line voltage/current; you use transformers. That is for both safety (isolation from mains power) and efficiency.

A general "rule of thumb" is to use a resistor rated for at least twice the expected power dissipation. This is for the sake of reliability.

And if this is the case the only difference is the car batteries are made with less internal resistance then the series batteries....which gives them a much higher max current rating. (like 700A or so ....)

Because I doubt a 12volt battery that can produce max 150mA can start a car. But curious to know if I made a battery which was 12volts and could supply 700amps max would that beable to start the car?

I.e what would you need for the current ratings/internal resistance of a 12volt battery to be equivalent to a car battery to start the car?
Cold cranking amps (CCA) is a measurement of the number of amps a battery can deliver at 0°F for 30 seconds and not drop below 7.2 volts.

So, if a battery were rated for 500 CCA, it's initial voltage was 12.7v, end voltage was 7.2, the internal resistance would be (12.7v-7.2v)/500 = 0.011 = 11mOhms.

CA is cranking amps measured at 32°F. This rating is also called marine cranking amps (MCA). Hot cranking amps (HCA) is seldom used any longer but is measured at 80°F.
 

Paulo540

Joined Nov 23, 2009
191
What I was trying to convey in my last post, is that the wattage resistor chosen has nothing to do with the size of the current 'pool'. at any specific voltage, a resistor will dissipate a SET amount of wattage based solely on its resistance.

So, your ac example, 120v rms across a 1k resistor. 120v divided by 1000 = .12A like you said. so, you then multiply the current that the resistor is allowing, by the voltage (120 * .12 = 14.4 watts) So, yes, you will need a 15w minimum resistor to safely handle that amount of power. But notice that you didn't need to know the aH rating of the outlet or the MW/h rating of the power plant behind it.

This is what I really really want to make clear for you. When figuring out the wattage of a resistor, leave the current rating of the supply out of it! :)

Here's an analogy I'll throw out....

Remember in science class or Mr Wizard or Bill Nye, etc, where they took a tin can and filled it with water, then poked holes in it at different levels?

This demonstrated the beauty of water pressure, right?

Well, it also makes a lovely electricity analogy. The holes are all same size 'resistors' and the potential energy of the stack of water is the voltage. The lowest hole streams farther, there is more power there. You have the same resistance 'hole size' but more water pressure ( V ).

I know you know about this but it is just to set up my next zinger.

So, think of this, would the same pressure be available at the bottom hole if the can depth and hole location were identical but the diameter was 10x as large? Would the hole be allowing the same amount of water through as the smaller can?

If you answer yes to both questions then go get yourself a cookie.

The resistor (or hole) has no clue of the diameter of the can. It's not that smart, it failed geometry in high school.
 

Thread Starter

Mathematics!

Joined Jul 21, 2008
1,036
I see your point given the battery voltage and a fix resistor resistance their will be a fix amount of current determined by the voltage of the battery and the resistor value (i.e V/R = I )

This is what you are basically getting at so no matter what 12volt battery I use the same current will be drawn thru any 1kohm load.

But my problem is you also have to be sure the 1kohm load must beable to dissipate the power = 12volts * 0.012A = 0.144watts

Put it another way P = V^2/R if you have a 1kohm resistor and the power rateing was 0.5watts then 0.5watts = V^2/1kohm
=> V = 22.360679774997896964091736687313 volts so you could not use a battery no higher then approx 22volts with this 1kohm.

So my point is you need to know what the loads max power dissipation
is to know what range of batteries can be used with it and what the max battery that can be used with this load before you will overheat the load device.

I do understand the voltage and the load (resistance) determine the current it will draw. But the max power ratings determine the max voltage /current used with this load....

That is what I was getting at.

But I really didn't have to worry about the max power ratings with this camera since I already knew from the adapters output 9vdc , 400mA and calculations with no load on the adapter I was able to find the load resistance of the camera approx. And from that particular resistance the camera load will draw a specific amount of current from V/(cameras load).
And obviously since the adapter was meant to be used with this camera the power ratings must be correct. And so any 9volt battery that can supply 800mAh or more will power the camera fine for at least 4 hours.
Right?
Since 800mAh/4h = 200mA and if the camera only needs around 120 to 150mA then this should be more then enough to power the camera for 4hours
 

Audioguru

Joined Dec 20, 2007
11,248
The camera is designed to work with the "9V" adapter that actually has an output voltage of 10.5V when its current is only 150mA. A 9V battery voltage might be too low.

Many rechargeable batteries have their capacity rated at the 10 hour or 20 hour rate.
Then a 800mAh battery can supply 80mA for 10 hours or 40mA for 20 hours. It might provide 800mA for half an hour or less.
 

Thread Starter

Mathematics!

Joined Jul 21, 2008
1,036
Ok , I final bought the option battery for the camera instead of using the 6 ft adapter cord with the power adapter...
Remember the adapter output on the sticker of the adapter was 9volt , 400mA.

When I got the battery it said on it 9volts , 500mA And I already told you or you could have read it in the pdfs/links that the battery was 800mAH and would last approximately 3 to 5 hours.

The camera consumes a max of 150mA

What I am curious about is why can a 9volt , 400mA adapter fully charge a 9volt 500mA rechargeable battery... Wouldn't you always want to recharge a rechargable battery with a voltage source slitly higher then with the voltage being recharged? Could you every use a smaller voltage to fully charge something like an 8volt rechargeing a 9 or 10volt battery...

Or will this just charge the battery up to a certain % of it's max rechargeable capacity.... (<- this is what I am thinking)

I am assuming once the battery recharges enough to be able to be as strong voltage/current wise as the rechargeing source then nothing will happen and the batteries will be equal in strength which means no net work will be done and the batteries will sit their doing nothing....
Curious if this is when the voltage , or current or power reachs the same for both the supplier and reciever sources ????

Thanks
 
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