# 6v & 12 v batteries

1. Jan 12, 2010

### Alkemist

I was wondering if I could connect 6v DC [B1] and 12 v DC [B2] batteries in parallel using two Diodes [D1 & D2], connected to +ve end of both batteries in forward biased configuration and then connecting them to the load, by forming a T junction, to get more current through the load [L1] and voltage across it. Do I need to add more resistors on the return paths from Load to each batteries, if so how to figure out their values?

PS : Please ignore the dots, I used them for text alignment.

....................B1..........D1..........D2...............B2
...Ground------|+--------|->-------<-|----------+|---------Ground
..........................................|
..........................................|
........................................{|} L1 10 Ohm
..........................................|
..........................................|
..........................................|
......................................Ground

2. Jan 12, 2010

### gnurf

The voltages add up to at the junction (minus the diode drop on each side), and then you can apply Ohm's Law on the resistor and figure out the current running through it, cross check it against the rated capacity of your batteries, and the ability of your resistor to dissipate power.

3. Jan 12, 2010

### vk6kro

D1 would have +11.4 volts on its cathode (12V - D2 drop) and +6 volts on its anode so it would be reverse biased.

This means the 6 volt battery would not be doing anything.

4. Jan 12, 2010

### gnurf

Right ... not pretend that the voltage sources were equal is probably a better way to do it if you insist on getting the right answer. Sorry about that.

5. Jan 14, 2010

### Alkemist

I haven't done the experiment and measured the Voltage values yet. But what I was thinking is the voltages will add up. Since Diodes won't allow reverse current flow though them and similarly about voltage, they will add up and will be around 18 (minus Voltage drops across each diode) volt. Similarly currents will add up. As far as D1 & D2 ratings, I'll be using power diodes, which can withstand against 100V & 10 Amp.

6. Jan 14, 2010

### sophiecentaur

Alkemist
Why 18V?
The polarities are 'against each other' so you will get 6 - 12 = -6V across the reverse biased diode.

7. Jan 14, 2010

### Alkemist

Hi sophiecentaur, thanks for comments. Do you mean to say if I join + end of B1 and -ve end of B2 to D1 & D2 respectively will make it approximately 18 v? At present +ve +ve are joined to the Diodes.

8. Jan 14, 2010

### sophiecentaur

I mean that Volts add up as Vectors - i.e. they are directional. Of course you can get an 18V supply with a 12V and a 6V battery - every torch uses a number of cells to produce 3V, 6V etc.
If you happen to have diodes in your circuit, they may or may not conduct - according to the way round they are connected. If they happen to be 'facing' in opposite directions, there will be no current flow because one of them will always have an 'infinite' resistance.
I can't understand your description of the second arrangement but I suggest that you look at Kirchoff's Second Law if you want an answer.

9. Jan 14, 2010

### vk6kro

Do you mean to say if I join + end of B1 and -ve end of B2 to D1 & D2 respectively will make it approximately 18 v? At present +ve +ve are joined to the Diodes.

If you reverse one of the batteries, without the diodes that is the same as connecting the batteries in series and then shortcircuiting them. So, you won't get 18 volts like that.
If you then add the diodes, depending on the diode polarity, you would either get huge currents and blown/damaged diodes or no current at all.

You can't connect batteries of different voltage in parallel safely, because the higher voltage one will discharge into the lower voltage one.

If you want 18 volts, just connect the batteries in series and put a suitable load on the combination. There is no need for diodes if you do that.

10. Jan 14, 2010

### Alkemist

As per reply #8, series arrangement of batteries will add up volts, but current through 6 V battery goes up, which might lead to "I don't know".

Per reply #9, if we have very high power Diodes, which can withstand against very high currents & voltages, i.e. above the VAC capacity of both batteries combined, so the diodes won't be damaged.

I'm planning the arrangement as shown in above figure only and add extra resistors L2 & L3, of suitable values after L1, which will be connected to -ve ends of B1 & B2, so that different current flows through them [think of current divider arrangement]...

11. Jan 15, 2010

### vk6kro

In the above diagram, can you see that the top drawing represents a short circuited 18 volt battery? That is what happens if you reverse your 12 volt battery without the diodes. If the batteries are lead acid types, they can certainly destroy any diodes.

In the lower diagram, D1 is reverse biased and stops any current flowing into or out of the 6 volt battery. If the diode wasn't there, the flow of current would be INTO the 6 volt battery from the 12 volt battery.

As it is, there will be about 11.4 volts across the 10 ohm resistor so a current of about 1.14 amps will flow in this resistor.

Putting resistors in series with the 6 volt battery won't change anything but doing it to the 12 volt battery will reduce the current in accordance with Ohm's Law.

The best way to convince yourself is to try it.

12. Jan 15, 2010

### sophiecentaur

But no need to use a hot 10 Ohm resistor. Use something higher and save burning your fingers.

13. Jan 15, 2010

### Alkemist

So is there any other circuit arrangement, which might work and add up currents as well as voltages across L1, when we use different voltage batteries?

14. Jan 15, 2010

### sophiecentaur

Of course there are. Feed the 12V output through a potential divider in which the mid point is at lower than 6v (a slightly higher resistor at the top and a smaller one at the bottom) then connect the 6V battery to this point, via another resistor. Both batteries will then be passing current into the junction of the three resistors (the node). The sum of the currents in the 'upper' two resistors will equal the current in the lower (common) resistor. This is Kirchoff's First Law.
Then you can verify Kirchoff's Second Law around each loop in the circuit. You will find that the sum of the I.R s will equal the battery volts, in each case.
Use resistors of greater than 100 Ohms, to ensure nothing gets hot and the batteries last some time, but you can still measure Currents and Volts easily with a DVM.

I recall blowing up a regulated power supply in my distant youth when I tried something like you are doing. Things are usually safe from harm as long as you pass low currents - I didn't and got a bollocking for ruining Lab equipment.

15. Jan 15, 2010

### Alkemist

In that case, I won't need Diodes at all, right?

Do you have any circuit diagram available for this?

16. Jan 16, 2010

### sophiecentaur

This was the sort of thing I had in mind. There are many variations. You can play for ever with combinations but it is safer to put a diode in series with the 6V battery so you never risk charging the 6V with the 12V (if the volts on the node get higher than 6V).

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17. Jan 21, 2010

### Alkemist

So, Is there any standard way [circuit] to connect different voltage batteries in parallel, which will add up voltage and currents in the load?

18. Jan 21, 2010

### Averagesupernova

No. This is not possible. I'm not sure you have a grasp yet of series and parallel batteries/cells. If you isolated a 6 volt battery and a 12 volt battery with diodes while hooked in parallel as discussed earlier in this thread what would happen is that the 12 volt battery would have to run down to the point that its voltage is at 6 volts before any current could be drawn from the 6 volt battery.

19. Jan 27, 2010

### sophiecentaur

It is actually possible - Kirchoff's Laws would tell you that. But it is difficult to arrange it to work unless you know the value of the load you are supplying. You just need an appropriate resistor in series with the 12V battery which will give you 6V, when in circuit. Current could then be supplied from both batteries. However, this would be very impracticable and the added resistor would be getting very hot as it would be dissipating half of the power that the 12V battery was actually supplying. The precise share of current from each battery, supplied to the load, would be very dependent upon actual resistor values - an unstable situation, which could result in either the 6V battery getting charged or the 12V battery supplying no current.
If you really needed to combine currents from two separate batteries of different voltages (say, where you had two existing supplies but neither could supply enough power on its own) you would probably use a sophisticated switch-mode system which took dollops of charge from each battery in turn. This would also eliminate the wasted power from the 12V battery - but that's another story.

20. Feb 1, 2010

### Alkemist

My intention is to get approximately 6v + 12v = approximately 17 v at the junction and then currents will be added in the load. On the return path to each batteries, we can add suitable resistors to divide the current which is coming from the Load. Any thoughts in line with this?