# Design of two interchangable power sources

Discussion in 'The Projects Forum' started by Chris T, Dec 15, 2012.

1. ### Chris T Thread Starter New Member

Dec 15, 2012
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0
Hello, everybody. I'm new here and trying to learn new things as this is my new hobby.

Basically, I want to create a charging system for battery, but it involves two power sources.
Only one power source is used at a given time.

The power sources both have the same volt output from adaptors, so volt is not the problem.
What I want to know is: what is the best design for this, so the Ampere is not decreased when it reaches the device?

I will try to give a drawing:

Power A ------>------x--->- device
...........................|
Power B ------>------|

> flow
x junction
. just spacing, please ignore it

Because only one power is going to be used at one time, I think it is okay.
However, if Power A is used, the current will be flowing into the wire of Power B (I believe this will reduce the Ampere to the device).
Same thing will happen if I use Power B, the junction will reduce the current.

Or should I do this:

Power A -x---->---x--->-- device
............^..........|
............|...........|
Power B -x---->---|

The question is what design is the best for this situation? Should I use diode? Or the normal parallel is already efficient for this?
Thank you.

Chris

2. ### wayneh Expert

Sep 9, 2010
14,780
5,271
Two, actually, with one on each power source. This prevents either power supply from powering the other.

You need to be aware that a diode drops about 0.6V across itself, more or less depending on the current, and so you need a bit more on the supply side to get enough voltage to your load (the battery). The diode will waste a bit of power, about 0.6 x A watts, where A is the current in amps.

The only way around that is a more complicated power "OR"ing scheme that eliminates the diode drop. That is widely done, but overkill for your application I think.

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3. ### Chris T Thread Starter New Member

Dec 15, 2012
6
0
Thank you for the info. However, I cannot afford for it to be losing the volts because of the diodes. So I guess, I will not go with diodes.

I might go for it.

Last edited: Dec 15, 2012
4. ### Dodgydave AAC Fanatic!

Jun 22, 2012
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5. ### wayneh Expert

Sep 9, 2010
14,780
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My understanding is that you use the body diode of a MOSFET as your blocking diode. Then, if/when a small voltage appears across it - as it would for any normal diode - you turn on the MOSFET. This provides a very low resistance path for current and eliminates the voltage drop of the diode. Well, it bypasses it actually.

Sounds simple enough, but the devil is in the smarts of the circuitry to detect the right times to turn on and off, and to do so without any chance of getting it wrong. No intermediate state, no ringing or oscillation, and so on. For low frequency ORing like your application, this may not be too bad. It gets harder as you go faster. The same idea is used in active rectifiers, again to eliminate the voltage drop normally caused by a passive rectifier.

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6. ### Chris T Thread Starter New Member

Dec 15, 2012
6
0
Thank you guys for trying to help.

@Dodgydave: I read from the pdf item description that ultra fast recovery diode still reduce the voltage (* Low forward voltage drop). So I guess, I cannot use any diode for this as this project uses only low voltage and I try to make it as small and as simple as I can.

@wayneh: I just searched the net about MOSFET and Reverse Battery Protection. But it's really too heavy for a beginner like me. Especially the thing about smart circuitry you were telling me...

I think, the best thing for me at the moment as an amateur is to accept the loss of Ampere by using basic paralleling.

However, to minimize the Ampere loss, I was thinking to use something like this:

Scheme 1:
Power A -x---->---x--->-- device
............^..........|
............|...........^
Power B -x---->---|
So, if the Ampere from Power B is 2mA, the device will get 1+0.5 = 1.5 mA

Compared to....

Scheme 2:
Power A ---------<---x--->- device
...........................|
Power B ------>------|
The device will only get 1mA

Is Scheme 1 is the best simplest-scheme for this purpose? Is there any danger by using the first scheme?

7. ### wayneh Expert

Sep 9, 2010
14,780
5,271
What are A and B? I don't think your question can be answered without knowing that detail. If they are truly at identical voltages, current flow from one to the other will be small or zero, in which case you can indeed parallel them to combine the current available. Just like using two batteries in parallel.

But a small voltage difference might cause an unacceptable imbalance. If they are power supplies that include some ripple in their output, undesirable things can happen when you connect them.

8. ### Chris T Thread Starter New Member

Dec 15, 2012
6
0
Power A and Power B are basically adaptors that have same Volts but a lil bit different in Amps.

I'm rarely going to use both adaptors at the same time, mostly I'm going to use only one adaptor at a given time.

What I was trying to do is to minimize loss in Ampere because mostly I'm going to use one Power at a time...and because of that the junction of the parallel is gonna reduce the Ampere...right? That's why I think Scheme 1 is better than Scheme 2.

9. ### wayneh Expert

Sep 9, 2010
14,780
5,271
Change the highlighted words to never and always, and you're done.
In general, no. Adding power in parallel increases available current as long as there is not a lot of current flowing from one to the other.

10. ### Chris T Thread Starter New Member

Dec 15, 2012
6
0
Okay, I will do that. But which scheme is more efficient?

But I'm not going to add power, as I'm going to use only one power each time.

I think I'm not clear enough. The parallel cables will always be connected to the device. The adaptors can only be switch off the wall, but not unpluged from the device.

Because only one power is on at a given time, there will only be 2 situations...

Situation 1:
Power A (on) ------>---x--->- device
....................................|
Power B (off) ------<---|
Power A is on; Power B is off the wall (no current flow from Power B, so any current from Power A will flow to Power B) but the adaptor and cables are still connected to the device.
Because no current flow from Power B, the current from Power A is divided to device and to Power B.

Situation B:
Power A (off)------<---x--->- device
...................................|
Power B (on)------>---|
Power A is off the wall, so no current flow from Power A (the Power A adaptor and cables are still connected to the device tho); Power B is on.
The other way around from Situation A, current from Power B is gonna flow to device and to Power A.

If both Power A and B have output 2mA, in Situation A and B the device will get 1mA only, right?

In Scheme 2:
Power A (off)-x---->---x--->-- device
.....................^...........|
.....................|............^
Power B (on)-x---->---|
Now, with this scheme (additional cable that connects Power A and Power B)...if Power B output is 2mA, the device will get 1.5mA...considering the other scheme only gives 1mA to the device.

So it means the last scheme is better to minimize Ampere loss, right?
Thank you so much and sorry for not able to explain things clearly.

11. ### wayneh Expert

Sep 9, 2010
14,780
5,271

If they are conventional DC wall-warts (transformer + full bridge rectifier + capacitor), your scheme will work fine since no reverse current will flow to the secondary coil of the transformer. Current is blocked by the rectifier diodes.

Modern power adapters tend to be switched mode power supplies (SMPS). These are lighter, smaller and more efficient, and tend to have a better regulated voltage output.

I'm not skilled enough to speculate how a SMPS would handle a voltage applied to its output while it is "off". It might also be fine, since this happens in normal usage, but you'll need confirmation.

One thing you could do is test it for yourself if you have a multimeter. Put your meter in ammeter mode and put the meter leads in series with a AA battery and the poles of your (unplugged) adapter, battery + to adapter +. If there is reverse current flowing when you make the connection, that's your answer.

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12. ### Chris T Thread Starter New Member

Dec 15, 2012
6
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Thank you, wayneh. I will get on to it.