Hi guys

what is the effect of having a diode in parallel with a voltage source (a battery cell)

I want to use it to prevent reverse charging of a cell, and also to divert current away from a damaged cell.

how does the voltage drop work out for example with a string of 4 dc cells of 3.6 volts with diodes connected in parallel with each one?

I have a 4S3P configuration of cells and want to roughly calculate the drop lost by the diodes and also the blocking diode (a schottky, drop is about .43 volts)).

so there is 1 schottky diode and then the 4 parallel ones to contend with.

what would i get from the output? 3.6v x 4 - blocking = 14.1 volts plus the parallel ones?

cheers!

 

Hi guys

what is the effect of having a diode in parallel with a voltage source (a battery cell)

If a diode is wired in parallel with a battery, that means the cathode is connected to one battery terminal, and the anode to the other battery terminal. If the anode is connected to the battery positive terminal, the battery will effectively be shorted; high power dissipation in the cell and diode will result. If the battery does not rupture and the diode not burn up, then the battery will discharge down to ~0.6v or so.

If the battery is connected positive terminal to cathode, only a small leakage current will flow.

I want to use it to prevent reverse charging of a cell

That won't work; the battery would have to be charged to ~-0.6v before the diode would start conducting.

... and also to divert current away from a damaged cell.

That's not really going to work, either.

how does the voltage drop work out for example with a string of 4 dc cells of 3.6 volts with diodes connected in parallel with each one?

The Vf of a diode depends on the chemistry/construction of the diode, current and temperature. You'll see 0.7v referenced very commonly around here, but that's for the sake of simplicity.

I have a 4S3P configuration of cells and want to roughly calculate the drop lost by the diodes and also the blocking diode (a schottky, drop is about .43 volts)).

Look at the datasheet of the diode you're thinking of using. If it has plots, it should show the Vf over current. It may have thermal data as well.

You really need to draw a schematic and post it, otherwise lots of dithering will go on and not much will be accomplished.

A schematic is considered a basic tool that's really as important as words in the beginning of a thread; it helps immensely to get the idea of what's going on.

 

I should point out these are thionyl chloride batterys... ie non rechargable cells.

The parallel didoes are needed to stop any current from flowing from + to -

they are also needed to divert current away from a cell if any are damaged.

so for a 4S3P configuration there would be 3 parallel strings each with 4 diodes in parallel with a cell.

My question is what would the voltage drop be on the output, assuming a cell of 3.6 volts.?

also as i mentioned each string needs a blocking diode.. usually a schottky, this stops the series strings from being charged up.. no current is allowed to charge up any cell.

all of these are safety prevention, thionyl cells are very dangerous, like almost all batterys.

I know I have a loss of about .4 volts from the blocking diode but I am unsure if the parallel diodes have any effect?

the will be wired up as follows:

cathode connected to cell positive.


thanks again.

 

Well, here goes nothing.

You gave me very little to go on, so I just threw together four "batteries" (actually, voltage sources with a fixed resistance; not very accurate) in series, each with parallel diodes, and a series diode in the string to prevent reverse current/charging.

I elected to use MBR520L Schottky diodes.
Having no idea what the actual parameters of the batteries are, I simply used fixed voltage sources with an internal resistance of 1.

I then ran a simulation over temperature (10°C to 40°C) and with a load varying from zero to 250mA in decade.

The resulting schematic and simulation are attached.

Note that V(BAT1) remains the same for a given load over temp, where V(OUT) varies significantly. This is due to the diode temperature.

The difference in output voltage for BAT1 can be attributed strictly to the resistance of the voltage sources (4 Ohms cumulative per string) vs the load current.

I don't know what the characteristics of lithium thionyl chloride batteries are, nor what size you are planning on using, nor what diodes, nor what temperature range, nor load range. There are so many unknowns, that basically all you can tell from this simulation, is that according to the model, the Vf of the MBR0520L diode varies over temperature and forward current.

But, I told you that already.