[SOLVED] What does the current rating in chargers/adapters mean? (Confused)

Suppose, I have an adapter rated 12V, 800mA. Does it mean that I can draw maximum current of 800mA or does it supply a constant current of 800mA at 12V.
Now, if I want to create a clone of that adapter then shall I design it so that a constant 800mA current is drawn, or maximum of 800mA. I can simply design to draw constant 800mA using emitter biased transistor, but how can I make it such that max 800mA is drawn?
Another thing is, I have got a device that is powered by that 12V, 800mA adapter. I have a SLA 12V, 7.2AHr battery. I do know what that 7.2AHr means. Can I directly connect the battery to the device? Or a really high current will flow from battery that will damage the device? If so then what should be done to get the device powered by the battery.

How can it supply a constant 800mA? If the load is 1k ohms then Ohm's Law says that its voltage must be 800mA x 1k ohms = 800V! What if the load is 1M ohms? Then its voltage will be 800mA x 1M ohms= 800,000V! If there is NO LOAD then what is the output current and will the voltage be so high that it makes a spark all over the world?

So of course the 800mA rating is its maximum allowed continuous output current. Then its output current allows Ohm`s Law to work.

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b.heuju said:

I have got a device that is powered by that 12V, 800mA adapter. I have a SLA 12V, 7.2AHr battery. I do know what that 7.2AHr means. Can I directly connect the battery to the device? Or a really high current will flow from battery that will damage the device? If so then what should be done to get the device powered by the battery.

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Your little battery can supply 360mA for 20 hours or 720mA for 10 hours. It will be listed on its datasheet. It might not be able to supply 7.2A.
It might supply 1.4A for 3 hours or 2.8A for a little less than 1 hour. 5.6A for 15 minutes?
If your device draws 800mA or less then your little battery can power it for 9 hours or more.

The ratings say that the adapter provide 12V output voltage and a current up to 800mA can be drawn. Any lower current is also ok, but a higher current can damage the adapter.

Thanks. I understood now. So it means I should not connect a smaller resistance than 15R. Better not to reach that level too keeping spaces for tolerance.

Audioguru said:

Your little battery can supply 360mA for 20 hours or 720mA for 10 hours. It will be listed on its datasheet. It might not be able to supply 7.2A.
It might supply 1.4A for 3 hours or 2.8A for a little less than 1 hour. 5.6A for 15 minutes?
If your device draws 800mA or less then your little battery can power it for 9 hours or more.

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This is confusing. Doesn't the 7.2AHr means it can provide 7.2A for 1hr? (The sellers here don't provide us with any datasheets so I don't know exactly) Even if it might not be able to supply 7.2A but you are saying 5.6A for only 15 mins. Isn't it too low?

In summary does your ans mean that I can connect it directly to my device without any protection? I am asking this because I have noticed that a small dc motor when connected to this battery runs more strongly then when connected to the adapter. This should mean that high current is flowing when connected to the battery, is it not? So don't I need to do something connecting the battery to device?

b.heuju said:

This is confusing. Doesn't the 7.2AHr means it can provide 7.2A for 1hr? (The sellers here don't provide us with any datasheets so I don't know exactly) Even if it might not be able to supply 7.2A but you are saying 5.6A for only 15 mins. Isn't it too low?

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You should NEVER buy anything without seeing its datasheet because its quality might be garbage.
I have never used an old lead-acid battery so I guessed at its ratings.
Now I looked in Google for 7.2Ah 12V battery and found the name-brand datasheet from Panasonic. Your battery might not be as good.
Notice that it can provide 360mA for 20 hours (7.2Ah) but only 4.9A for one hour.

In summary does your ans mean that I can connect it directly to my device without any protection? I am asking this because I have noticed that a small dc motor when connected to this battery runs more strongly then when connected to the adapter. This should mean that high current is flowing when connected to the battery, is it not? So don't I need to do something connecting the battery to device?

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A motor draws a very high current when it starts and when it is working hard so maybe it draws more than 800mA and is overloading your adapter.
The datasheet for the battery spec's I am posting shows that the 12V battery can supply 12A for a few minutes before its voltage begins to fall. Your adapter's voltage begins to fall when its load is more than 800mA if it is a good one. If the motor or adapter gets too hot then you must reduce its load which will reduce its current.

Please see my answers in line

b.heuju said:

Suppose, I have an adapter rated 12V, 800mA. Does it mean that I can draw maximum current of 800mA? yes.
or does it supply a constant current of 800mA at 12V- No.

Now, if I want to create a clone of that adapter then shall I design it so that a constant 800mA current is drawn, or maximum of 800mA.
I can simply design to draw constant 800mA using emitter biased transistor, but how can I make it such that max 800mA is drawn?
Probably some sort of OCP (over current protection circuit) ie: set up Op-amp with reference voltage with its output to shutdown the biased transistor when output current exceeds 800mA.

Another thing is, I have got a device that is powered by that 12V, 800mA adapter.
I read the above comments, you mentioned the device is a motor.

I have a SLA 12V, 7.2AHr battery. I do know what that 7.2AHr means.
Basically this means that 7.2AHr amount of charge you have at your disposal. Remember, this is not the peak current rating. The peak current could possibly be in 100s of Amps. So, please check the peak current rating. Else, you may destroy your load.

Can I directly connect the battery to the device?
The device is a motor. A motor is basically an inductor. Physics says inductors are short at DC. So, you are basically shorting your battery if you put a motor across it. Now, the load current is limited by the peak current rating of the battery. However, your load (motor) does not limit the current through since there is no current limiting resistor in your circuit. So, to save your motor (load) you must use the resistor. Make sure to calculate the wattage of the resistor or else the resistor will burn red hot.

Or a really high current will flow from battery that will damage the device? If so then what should be done to get the device powered by the battery.
Answered above

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shawnmk123 said:

Please see my answers in line

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Don't be silly. A motor's current is limited by the resistance of its windings. It is normal for a motor to draw the maximum current limited by its resistance when it starts and when it is stalled. Its maximum current is not infinite, it is listed on the motor's datasheet.
A motor usually does not use a series current-limiting resistor because it will reduce the torque then the motor might not start turning.

Thanks.
Sorry for confusing, but the device is not a motor. Its actually an ADSL router. I just mentioned the motor because I noticed the difference in speed when connecting it to battery and adapter.

shawnmk123 said:

The device is a motor. A motor is basically an inductor. Physics says inductors are short at DC. So, you are basically shorting your battery if you put a motor across it.

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But the Physics also says that inductor actually opposes the flow of current.

Audioguru said:

A motor draws a very high current when it starts and when it is working hard so maybe it draws more than 800mA and is overloading your adapter.
The datasheet for the battery spec's I am posting shows that the 12V battery can supply 12A for a few minutes before its voltage begins to fall. Your adapter's voltage begins to fall when its load is more than 800mA if it is a good one. If the motor or adapter gets too hot then you must reduce its load which will reduce its current.

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This still doesn't answer if I can connect battery directly to the device or if I required some protection. Maybe that's because I confused you with the device to be motor. But now you know that its not a motor but a Network Router, what is your suggestion?

Audioguru said:

Don't be silly. A motor's current is limited by the resistance of its windings. It is normal for a motor to draw the maximum current limited by its resistance when it starts and when it is stalled. Its maximum current is not infinite, it is listed on the motor's datasheet.
A motor usually does not use a series current-limiting resistor because it will reduce the torque then the motor might not start turning.

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The torque in a motor is directly proportional to the current. I was not being silly.

You are right that non-ideal motor windings have resistance. However, that winding resistance is very small and hence the current drawn by the load (motor) could easily burn the windings if the power source itself is not current limited. Regarding the torque, use a external variable resistor to control the load (motor) current and hence, vary the torque.

The maximum current mentioned on the datasheet is the maximum current a motor winding can handle, provided an external resistance is there to limit the current.

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b.heuju said:

Thanks.
Sorry for confusing, but the device is not a motor. Its actually an ADSL router. I just mentioned the motor because I noticed the difference in speed when connecting it to battery and adapter.


But the Physics also says that inductor actually opposes the flow of current.
Sorry that is not true, Physics says that inductors oppose only the instantaneous changes to current, not the instantaneous flow of current that you mentioned.
For instance, in power electronics, inductors can be designed to carry large currents like say, 1A. However, say you try to disconnect this inductor, you'll probably get a surprise because of the back emf. This is because the reactive circuit elements like inductor and capacitor don't like any change to their circuit variables.
In the case of inductor, there can be no sudden (instantaneous) changes to the current. However, smooth changes to inductor current are allowed.

This still doesn't answer if I can connect battery directly to the device or if I required some protection. Maybe that's because I confused you with the device to be motor. But now you know that its not a motor but a Network Router, what is your suggestion?

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I hope you would be more straightforward next time you post. To answer your question, you must see the datasheet of the network router. if the datasheet mentions the current required is less than 800mA (adaptor's peak instantaneous current rating), then I believe it should be alright to connect the router to the power supply. A power supply only supplies as much current as the load required. If more current is required, choose a power supply (adaptor) with higher current rating. If it is a off-the-shelf router (I presume), it should be alright to go ahead without any current limiter protection. Else, no harm in using a current limiting resistor with proper wattage.

Please wait for suggestions from the more experienced to respond before you follow the above advice. No one here will be held responsible for any damage.

shawnmk123 said:

I hope you would be more straightforward next time you post.

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Sorry about that. I actually wanted to get an overall answer for any device and just not my router. So I tried not to expose my device.

shawnmk123 said:

A power supply only supplies as much current as the load required. If more current is required, choose a power supply (adaptor) with higher current rating. If it is a off-the-shelf router (I presume), it should be alright to go ahead without any current limiter protection. Else, no harm in using a current limiting resistor with proper wattage.

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I believe you got the wrong concept. I was asking if I can directly connect my battery (SLA 12V, 7.2AHr) to the device [From power supply I presume you are talking about AC mains to DC supply]. From your answer I believe in a similar manner I can use a current limiting resistor just to be safe.

If your router is normally powered from a 12VDC/800mA adapter then it can be directly connected to your 12V/7.2Ah battery (for about 7 hours if it is a good battery) WITHOUT a series resistor.
A series resistor will reduce the voltage and current then the router probably will not work.

Make sure that you connect the polarity correctly.

Note if you load the adapter with a very low resistance then the adapter will no longer provide the same voltage(voltage regulation) but will provide current around max current and if you provide further low resistance then the adapter will damage due to heating of components mainly transformer

Thanks everyone. I think I got the answer I needed.