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Power Questions

battery power

Best Answer Mario Vernari, 06 May 2015 - 04:06 AM

A diode is a semiconductor component which allows the current flowing only to one direction:

  • when drawn as a symbol, the arrow indicates the direction
  • when a real component, there's a striped which indicates the outgoing lead.

However, when the current is flowing there's a small voltage drop across the leads, roughly 0.6V (it's an exponential formula, though).

 

About what you want to do, consider the internal point where the power is needed: this point is Vin.

 

Suppose to feed the power through the barrel. Let's say exactly 10V, but there's a diode on the board (D_board), so the Vreg is expected to 10-0.6 = 9.4V. Everything's working fine.

There's a 9V battery connected with a diode (D_bat) to the same Vin point. Now, the voltage on the positive lead is 9V MINUS the voltage on the Vin is 9.4V, yields -0.4V which is lesser than the minimum required for the D_bat to allow the current flowing. That is, under these conditions the battery can't feed any energy, and all the power is given by the barrel.

 

Now detach the power barrel.

The voltage on the Vin pin WOULD drop to zero, but AS SOON the point falls below 9-0.6=8.4V, the D_bat begins to allow the current flowing. The battery now is the only one feeding energy.

Why 9-0.6=8.4? Because 9V is the battery voltage (on the positive lead) and 0.6V is the drop of the D_bat diode.

 

NOTE: I used 10V as the barrel voltage to emphasize the circuit's behavior. However, 10V is not a standard voltage for an adapter. If you have a common 9V adapter, you'll experience kinda "challenge" between the adapter and the battery: the one having the highest voltage will actually feed the board. Bear in mind that a new battery has typically a voltage slightly higher than the nominal, but decreases also easily.

I suggest to make some experiment, but if you want to be super-guaranteed there's no "challenges" anytime, just use TWO diodes in series for the battery. At this point the drop is 0.6+0.6=1.2 and the barrel should "win" all the times!

FW3RYJJGA0O8T9C.LARGE.jpg

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4 replies to this topic

#1 tridy

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Posted 05 May 2015 - 01:07 PM

1. I was wondering if there are 2 sources of power of which one is a battery and another is an adapter. Lets say the power goes down, will the board autoswitch to the battery to get the power?

 

2. I have seen things like this:

 

Attached File  9v.png   37.02KB   0 downloads

 

If I connect a 9v battery to be a power source for the board, how long will it last. Assuming no LCD screen (battery drainer?) is connected to the board.

 

Thanks!



#2 Mario Vernari

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Posted 05 May 2015 - 04:21 PM

The short answer to the first question is: yes, but...

The long answer is: there is the "Vin" pin on the header which is basically the same as the power barrel pin. However, you can't wire a battery directly to the Vin pin, but you must add a diode in series to the positive lead. The ground of the Netduino should be connected to the negative lead.

 

The second question leads to a simpler answer.

The current required by a Netduino may span from 50mA to 200mA: it depends on the model, on the state, on the hardware connected.

Then, check the nominal capacity for a 9V battery here: http://en.wikipedia....ne-volt_battery

Finally, divide the capacity by the current flowing.

Let's say you have a Zinc-Carbon battery (400mAh) and the average current for the Netduino is 100mA. You'll have 400mAh / 100mA = 4 hours.

Bear in mind that a more practical result would lead to the time halved or a bit more.

 

Good luck.


Biggest fault of Netduino? It runs by electricity.

#3 tridy

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Posted 05 May 2015 - 05:13 PM

Thanks for the answers!

 

 

add a diode in series to the positive lead

 

Hopefully one day I will get there to understand how this could work :)

 

 

Let's say you have a Zinc-Carbon battery (400mAh) and the average current for the Netduino is 100mA. You'll have 400mAh / 100mA = 4 hours.

Bear in mind that a more practical result would lead to the time halved or a bit more.

 

Ok, so it can do around solid 2 hours.

 

 

thanks!



#4 Mario Vernari

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Posted 06 May 2015 - 04:06 AM   Best Answer

A diode is a semiconductor component which allows the current flowing only to one direction:

  • when drawn as a symbol, the arrow indicates the direction
  • when a real component, there's a striped which indicates the outgoing lead.

However, when the current is flowing there's a small voltage drop across the leads, roughly 0.6V (it's an exponential formula, though).

 

About what you want to do, consider the internal point where the power is needed: this point is Vin.

 

Suppose to feed the power through the barrel. Let's say exactly 10V, but there's a diode on the board (D_board), so the Vreg is expected to 10-0.6 = 9.4V. Everything's working fine.

There's a 9V battery connected with a diode (D_bat) to the same Vin point. Now, the voltage on the positive lead is 9V MINUS the voltage on the Vin is 9.4V, yields -0.4V which is lesser than the minimum required for the D_bat to allow the current flowing. That is, under these conditions the battery can't feed any energy, and all the power is given by the barrel.

 

Now detach the power barrel.

The voltage on the Vin pin WOULD drop to zero, but AS SOON the point falls below 9-0.6=8.4V, the D_bat begins to allow the current flowing. The battery now is the only one feeding energy.

Why 9-0.6=8.4? Because 9V is the battery voltage (on the positive lead) and 0.6V is the drop of the D_bat diode.

 

NOTE: I used 10V as the barrel voltage to emphasize the circuit's behavior. However, 10V is not a standard voltage for an adapter. If you have a common 9V adapter, you'll experience kinda "challenge" between the adapter and the battery: the one having the highest voltage will actually feed the board. Bear in mind that a new battery has typically a voltage slightly higher than the nominal, but decreases also easily.

I suggest to make some experiment, but if you want to be super-guaranteed there's no "challenges" anytime, just use TWO diodes in series for the battery. At this point the drop is 0.6+0.6=1.2 and the barrel should "win" all the times!

FW3RYJJGA0O8T9C.LARGE.jpg


Biggest fault of Netduino? It runs by electricity.

#5 tridy

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Posted 06 May 2015 - 12:25 PM

Thanks for the explanations, Mario.

I am starting to get a picture.







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