# Can I connect grounds of two 9V batteries?

#### livelongpranav

Joined Oct 18, 2017
19
I want to make a backup battery connector for my project so I can power it externally if the internal battery dies.

For this, I'm planning to use a SPDT switch with the pole connected to the circuit's power input and the throws connecter to both the batteries positive terminals. In this way, the grounds of both the circuits will be connected to the circuit's gnd input all the time.

Is this okay? When I'm powering it from battery one, battery two's ground will also be connected to the circuit's gnd. Won't current flow from one battery to another?

Should I use a DPDT switch to switch both positive and negative wires?

Thanks a ton.

#### WBahn

Joined Mar 31, 2012
26,294
No, current will not flow because the second battery does not have a complete circuit.

When you are standing on the ground you are connected to one side of the electrical grid, both in your house and in the distribution system. Yet no current flows in you because there isn't a complete circuit. Reach out and touch the other side of the power supply (the 'hot' wire) and you complete the circuit.

#### MrChips

Joined Oct 2, 2009
22,085
I want to make a backup battery connector for my project so I can power it externally if the internal battery dies.

For this, I'm planning to use a SPDT switch with the pole connected to the circuit's power input and the throws connecter to both the batteries positive terminals. In this way, the grounds of both the circuits will be connected to the circuit's gnd input all the time.

Is this okay? When I'm powering it from battery one, battery two's ground will also be connected to the circuit's gnd. Won't current flow from one battery to another?

Should I use a DPDT switch to switch both positive and negative wires?

Thanks a ton.
What you have suggested is OK.
Note that there is no GROUND connection on a battery. There is a positive terminal and a negative terminal. It is ok to connect the negative terminals together. There is no need to switch the negative terminals with a DPDT switch.

A circuit schematic is the language of choice for communicating electrical circuit ideas and design.

The first circuit shows what you have described with a SPDT switch which selects either positive battery terminal.

The second circuit shows an automatic change over battery selection using two diodes. There is no need for a switch.The higher voltage battery will power your load.

#### WBahn

Joined Mar 31, 2012
26,294
Note that in the diode switch-over circuit there will be a voltage drop of about 0.7 V (if you use a typical silicon diode). If you can tolerate that, then this is a reasonable way to go.

#### Dodgydave

Joined Jun 22, 2012
9,651
Use the two diode version with BAT 48 Shottky diodes..400mV drop.

#### Alec_t

Joined Sep 17, 2013
11,739
If you opt to use a SPDT switch there will be a very brief period, during the switch change-over, in which the load is unpowered. Is that tolerable?

#### livelongpranav

Joined Oct 18, 2017
19
If you opt to use a SPDT switch there will be a very brief period, during the switch change-over, in which the load is unpowered. Is that tolerable?
Yes, that will be no problem.

Thanks everyone.

#### MrChips

Joined Oct 2, 2009
22,085
Just so that it is understood with regards to your original query, the negative terminals of the battery are interconnected.
This is an arbitrary decision. You could have done exactly the same thing with the positive terminals.

As you inquired, a DPDT switch could have been used to switch both sides of the battery. This is unnecessary.
You only have to switch one side of the battery. Either side will work. There is no strong argument for switching one side versus the other side.

#### KeepItSimpleStupid

Joined Mar 4, 2014
4,287
What typically done without using a switch is to arrange the contacts in typical battery connector (DC barrel) that when you insert the connector the internal supply is removed.

See the "documents" for https://www.sparkfun.com/products/10811

Usually, you have an internal battery and an external wall wart. When the wall wart is used, the battery is disconnected.

You can buy chassis mounted jacks.

Center positive is a common arrangement. There is "some standard" where the OD increases based on voltage to prevent higher voltage adapters to be plugged in. It's rarely followed. You can't easily short out the adapter with a DC barrel connector, but can with a miniature phone plug.

See: https://en.wikipedia.org/wiki/Coaxial_power_connector

#### WBahn

Joined Mar 31, 2012
26,294
If you opt to use a SPDT switch there will be a very brief period, during the switch change-over, in which the load is unpowered. Is that tolerable?
That will happen with the TS's proposed SPST switch solution, as well.

#### WBahn

Joined Mar 31, 2012
26,294
Yes, that will be no problem.

Thanks everyone.
Then your original proposal will not work for the same reason and you need to look at an alternative. The diode-connected approach is the simplest. There are other approaches, including using an appropriate make-before break switch or a similar jack. This will briefly connect the batteries together resulting in a current spike from the fresh battery to the weak battery, but it is likely that it will be sufficiently brief so as not to cause a problem.

Edit: I misread the TS's response as saying that it would be a problem. Since it's not a problem, there's no issue.

Last edited:

#### djsfantasi

Joined Apr 11, 2010
6,955
Then your original proposal will not work for the same reason and you need to look at an alternative. The diode-connected approach is the simplest. There are other approaches, including using an appropriate make-before break switch or a similar jack. This will briefly connect the batteries together resulting in a current spike from the fresh battery to the weak battery, but it is likely that it will be sufficiently brief so as not to cause a problem.
The TS answered no problem if there is a brief moment that the load is unpowered.

How does that make his first proposal not work?

#### WBahn

Joined Mar 31, 2012
26,294
The TS answered no problem if there is a brief moment that the load is unpowered.

How does that make his first proposal not work?
The problem is that I can't read. I saw the "Yes" and then read the sentence as saying, "Yes, that will be a problem" instead of what he actually wrote. Thanks for catching that.

#### livelongpranav

Joined Oct 18, 2017
19
Won't using the automatic changeover circuit be like two batteries in parallel? Due to the diodes, they won't discharge into each other, but won't they supply simultaneously and their current will be added?

#### WBahn

Joined Mar 31, 2012
26,294
Won't using the automatic changeover circuit be like two batteries in parallel? Due to the diodes, they won't discharge into each other, but won't they supply simultaneously and their current will be added?
It's not quite like two batteries in parallel because two batteries in parallel can result in one battery charging the other. But other than that you are correct. Their currents will add, but under most conditions the current of one of them will be negligibly small. The problem is that the available voltage is the battery voltage minus the diode drop. Sometimes this is tolerable and sometimes it isn't.

#### livelongpranav

Joined Oct 18, 2017
19
I read somewhere that if one of the two power sources have a little higher voltage, it will always be the one supplying power. This configuration seems ideal to me. But can you please explain why this happens? When I vary the voltages in the diode switchover circuit, why will it happen that only the higher voltage source will supply power?
It's not quite like two batteries in parallel because two batteries in parallel can result in one battery charging the other. But other than that you are correct. Their currents will add, but under most conditions the current of one of them will be negligibly small. The problem is that the available voltage is the battery voltage minus the diode drop. Sometimes this is tolerable and sometimes it isn't.

#### MrChips

Joined Oct 2, 2009
22,085
An ideal diode is a switch that conducts current in one direction only.
The diode connected to the battery with a lower voltage is reversed biased. Hence it conducts no current.

(Real diodes will leak a very small current when reversed biased. This is so small you don't need to worry about it.)

If you want to reduce the voltage drop across the diode, use 1N5817 diodes or similar.

#### WBahn

Joined Mar 31, 2012
26,294
I read somewhere that if one of the two power sources have a little higher voltage, it will always be the one supplying power. This configuration seems ideal to me. But can you please explain why this happens? When I vary the voltages in the diode switchover circuit, why will it happen that only the higher voltage source will supply power?
It's because of the highly nonlinear voltage-current characteristic of a diode.

For a typical silicon diode, a change in voltage of about 60 mV results in a change in current of a factor of about 10. So if you have a battery that is, say, 8.9 V and a battery that is 8.6 V, there is a 300 mV difference between them. That means that the current from the lower voltage battery will be less than 0.1% of the current from the higher voltage battery.

#### livelongpranav

Joined Oct 18, 2017
19
Sorry but I'm a total noob so need some more clarification. How is the diode connected to the battery reverse biased? According to the diagram, both the diodes are arranged so that they allow current to flow from the battery positive to the circuit.
An ideal diode is a switch that conducts current in one direction only.
The diode connected to the battery with a lower voltage is reversed biased. Hence it conducts no current.

(Real diodes will leak a very small current when reversed biased. This is so small you don't need to worry about it.)

If you want to reduce the voltage drop across the diode, use 1N5817 diodes or similar.

#### MrChips

Joined Oct 2, 2009
22,085
Let us assume that the forward voltage of a 1N5817 diode is 0.3V.
Let us assume that the voltage of battery A is 8.5V while battery B is 9.0V.

Since the voltage of battery B is the greater of the two, battery B is supplying current to the load.
The voltage at the load is 9.0 - 0.3 = 8.7V.

The anode of diode A is at 8.5V (voltage at battery A).
The cathode of diode A is at 8.7V (voltage at the load).
Hence the cathode of diode A is 0.2V higher than its anode. Hence diode A is reversed biased.