Hey guys this is my first post on stack exchange so please go easy on me!

I have a hobby project where I was trying to make a few batteries go in series using a micro controller and a floating power supply (in this case an isolated DC-DC converter), the arduino is powered separately.

I've attached a schematic of the rough problem.


simulate this circuit – Schematic created using CircuitLab

The arduino needs to be isolated from the rest of the higher powered battery stuff which is why I have used an isolated DC-DC converter and an opto-isolated mosfet driver

The output of the opto-isolated driver is hooked to the gate of the mosfet and the COM terminal of the dual output converter is hooked to the sources of both mosfets. This lets the mosfets see a +15V and the -15V drive signal. (please dont pay attention to the irf530 i just used the default on circuit labs and have some mosfets that can handle +/-30V for gate to source voltage)

My issue is that instead of buying more DC-DC converters and opto-isolated drivers to drive each next pair of mosfets, is there a way to power the next set using the same DC-DC converter and isolated driver? I need to add six more pairs of mosfets and batteries and thought that someone might know a cheaper way than buying more converters and drivers. Perhaps someone knows how you could use caps, resistors or diodes etc to do it.

Another issue that I have had is that if I were to connect the output of the isolated driver to the gate of the next pair of mosfets that would be okay but if i go to connect the COM terminal to sources of the next pair aswell it would create a short that bypasses the next battery.


  • DC-DC converter is the DPU01M-15
  • Driver is the FOD3150
  • Switching frequency range from 1Hz - 3kHz

I hope that makes sense

Thanks for any help in advance!

EDIT: The switches will turn on and off simultaneously

  • 3
    \$\begingroup\$ What is your project supposed to do? Why are you doing this? What problem do you think you are solving by completing your project? - You have many things you want to do, but I cannot see a good reason for it. \$\endgroup\$ – Harry Svensson May 29 '19 at 6:54
  • 2
    \$\begingroup\$ It's not clear why you have to switch batteries with 1Hz-3kHz? \$\endgroup\$ – Marko Buršič May 29 '19 at 6:56
  • \$\begingroup\$ The system is a bit complicated in what it is actually doing i.e. its driving an inductive load into resonance (which is the reason for the frequency range) but that it is too far off point for what the actual issue is here. The question is simply can you power multiple mosfets using the same floating power supply? Thanks for any help in advance \$\endgroup\$ – NROC May 29 '19 at 7:11
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    \$\begingroup\$ I'm wondering whether you can use a simpler solution that makes use of the switching characteristics of your circuit. Is your minimum switching frequency really 1Hz (higher would be better)? Do the other switches turn on and off simultaneously? If not, is there any synchronicity among them (when one turns one the other turns off, for instance)? \$\endgroup\$ – joribama May 29 '19 at 7:27
  • \$\begingroup\$ Sorry I should have clarified that before - yes the switches will switch simultaneously on and off together. Yes unfortunately 1Hz is the lowest end just due to the resonance of the system and the various inductive loads it could potentially take. \$\endgroup\$ – NROC May 29 '19 at 7:31

Sharing the one converter is "hard" but there is a low cost and easy and simple alternative (choose any 3).

It is common to use a "bootstrap" supply to high side gates using the switching action to provide the reuaisite "AC pump" - BUT that needs the gates to be switched more often than the decay time of the voltage which is produced in one switching cycle. So ...

Have a look at the simple but useful concept used in this SE Q&A .
An AC square wave (can have asymmetric mark/space) is capacitor coupled to the floating node. A +ve half cycle conducting diode provides charge voltage and a negative half cycle diode to local "FLOATING GROUND'/ source" maintains the capacitor voltage at 0V-floating on average.

As seen here it is used to directly drive the gate BUT the same method could be used to provide a floating voltage which is then used by an appropriately coupled floating driver. Do read the answers and comments as they address several relevant issues.

Here are two conceptual circuit diagrams.
Design is required.

The upper circuit provides both drive and drive-power by applying a square wave to "AC signal drive".

The lower circuit uses the diode pump to provide a floating DC voltage and a separate signal line - derived by whatever means suits - perhaps an optocoupler, or ... .


simulate this circuit – Schematic created using CircuitLab

This circuit can be used to provide floating voltage for other applications.

In this application and (probably) yours the gate is driven relative to a steadyish source voltage which is DC referenced to the local ground. If the circuit is fully floating then a capacitor can connect floating ground to local ground or an 180 degree out of phase drive input can be provided by two capacitors. Some low cost panel meter circuits use this arrangement to 'float' the meter supply. Note that when you have multiple driven FETs "standing on top of each other" the switching of one stage may affect others - this needs to be accounted for in the design detail.


Also see SE questions gate driver and [floating gate driver](https://electronics.stackexchange.com/search?q=floating + gate+driver)

This graph and circuit are copied from Selvek's excellent answer to the above reference question.
It can be seen that the gate voltage acquires it's full value within a few cycles of the AC drive waveform. If you wish to use simple high/low drive signals from the microcontroller you could AND these with an appropriate AC waveform - or the uC can provide the AC signal. Do read the cited Q&A as voltage drive levels, gate turn on voltages and more are discussed.

enter image description here

  • \$\begingroup\$ Thanks a lot for that I'll check that out! \$\endgroup\$ – NROC May 29 '19 at 7:20
  • \$\begingroup\$ Ill test that on the bench a bit later on today and post how I get on with it thanks a lot! \$\endgroup\$ – NROC May 29 '19 at 7:46
  • \$\begingroup\$ @NROC I'm about to power a 6 cell balancer and so far am leaning toward using an inductor with an AC driven primary and 6 isolated primaries. Not very hard to implement and provide true floating power supplies. || I also have a number of resonated RC coils on ferrite U cores from a long ago inductive power transfer demonstrator. These use a common single pair of wires current driven. I'd forgotten about them ... :-). Somewhere in the workshop ... :-) \$\endgroup\$ – Russell McMahon May 29 '19 at 7:53

You don't need DC/DC converters at all.

Instead you can use the positive side of the battery to provide the voltage you need to turn on the FETs.

You can use solid state relays (like the CPC1002N optical relay for instance) between the positive of the upper battery and the combined gates of the FETs. You will also need a shunt resistor between the gates and the drains of the FETs to prevent the leakage current of the relay from turning the FETs on.

This assumes the batteries will never fully discharge and their voltages will be always higher than the threshold voltage of the FETs you use.

If you use optical relays you can wire their LEDs to your micro controller the same way you were originally planning on connecting the FOD3150 driver. In fact, the only reason why you can't use the FOD3150 itself as the relay is that it requires the battery voltage to be 13.5V or greater, what is a little too high in your case.


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