1
\$\begingroup\$

Background: For a university final project, my student group is investigating the possiblility of powering a motorcycle with two seperate batteries, each of a different chemistry (Lead-acid and Lithium-ion). The idea behind the design is to try and achieve a design which costs less to produce but has the same range as one that uses exclusively Li-ion. We want to power the bike with li-ion at low power demands (e.g. cruising or gentle acceleration) and then switch from the li-ion to lead-acid powered for times when a higher power is needed (harsh accelerationa and high top speeds for a short time). Initially we were hoping to simply disconnect the first power source then switch over to the second. However after speaking to a professor here he advised that the controller would register a 0A current and switch off meaning the whole system would need to be reset before it could start working again. Obviously this is not suitable for an electric vehicle.

So my question is: Is it possible to power a single motor with one power source then quickly switch to another power source (of equal voltage) without damaging any components and ideally not losing power for an extended amount of time while switching? Also we would ideally want to make sure that the unused power source is completely disconnected and not using any power when not in use.

Sorry for my inexperience and if any of the above didn't make too much sense, electronics isn't my strong suit (mechanical engineer).

I've had a read of existing questions related to mine but can't seem to find any relevant ones, if anyone has any knowledge of some that would be very helpful. Also if anyone knows of any existing work that has done something similar that would also be really useful.

Thanks!

\$\endgroup\$
  • \$\begingroup\$ Solid state relays have fairly fast switching times these days, I'd look at that route. When switching you'd probably want a few decoupling capacitors to help with short-term power drops at the controller end. \$\endgroup\$ – Jarrod Christman Jan 28 '15 at 14:09
  • \$\begingroup\$ What current is being taken under load conditions? How quick is "quickly"? \$\endgroup\$ – Andy aka Jan 28 '15 at 14:23
  • \$\begingroup\$ I wonder if you don't have the batteries backwards. Use Lithium for the high current requirements, and lead acid for cruising. A fist sized lithium will comfortably start a car. Lithium is IMHO electrically equivalent or better in every way than lead acid. See here for a choice quote about lithium supporting up to 20C continuous discharge. \$\endgroup\$ – tomnexus Jan 28 '15 at 14:40
  • \$\begingroup\$ Can't you make it 3 state supply instead of 2? I mean creating a 3rd phase where both batteries are supplying the system at the same time for a short period if time while switching over? \$\endgroup\$ – dscharge Jan 28 '15 at 15:34
2
\$\begingroup\$

I suggest that you talk to another professor.

I'm assuming you use an H-bridge motor controller to actually drive the motor. Without actually trying to implement it (that is, it may not work), I'd suggest that the controller has an enable input which will disable its internal drive. If you disable the controller just before you make the switchover, then reenable it just afterwards, the controller ought to handle the job.

Alternatively, if you provide a decent-sized capacitor at the controller input, it will provide current during the switchover. And while this is obviously impractical for prolonged outages, using MOSFET switches the switchover period should be measured in microseconds, and it's not hard to get a capacitor do that, even for fairly high currents.

Yet another possibility you might consider is to provide the two batteries at different voltages, with the high-demand battery (oddly enough) providing a slightly lower voltage than the cruise battery. Then you provide a switching circuit which looks like

schematic

simulate this circuit – Schematic created using CircuitLab

In cruise, the switch is closed, and since the cruise battery has a higher voltage than the acceleration battery, D1 is reverse-biased, and no current is drawn from the accel battery. If the switch is opened, the accel battery provides all the power.

The switch is used only to maximize efficiency. Some power is wasted during acceleration due to the drop across the diode (which is why a Schottky is specified), but this is only true during the relatively brief periods when high acceleration is called for. It also has the virtue of allowing the accel battery to act as reserve battery, just in case the cruise battery is exhausted during use.

It's perfectly possible to replace the switch with another diode. In this case, high current draw will hog down the cruise battery until the accel battery takes up the slack. This has a couple of drawbacks. The first, and most obvious, is that there is a continuous power loss due to diode heating. If you use a high voltage (and therefor lower current) battery, this will minimize the problem, but it may still be objectionable. A second, and less obvious difficulty is that as the cruise battery is discharged its voltage will lower, and the accel battery will be accessed more and more, until both batteries run out.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks so much for the response it was really useful. I've spoken some professors and they agree with the methods you've described. We've decided to further investigate the 2nd and 3rd ideas (a capacitor based circuit and the one with the Schottky diode). Some further information: we are planning to use two 48V (approximately) batteries, powering a 5-10kW rated motor. As this is a fairly high power application I imagine that the components needed will need to be rated specifically for high power applications. Can you advise if the circuits you suggested are possible with these kinds of values? \$\endgroup\$ – Joe Feb 3 '15 at 15:39
  • \$\begingroup\$ Oh, sure. It's just a matter of money. Which may be a problem in your case, I agree. For instance, you've already computed the current levels you'll be working at, right? No? Well, it's fairly simple. P = V x I, and with V equal to 50 volts and P at 10,000, current is (obviously) about 200 amps. Go to Digikey, and search on Discrete Semiconductors, Diodes Rectifiers(single), then use the filters for 400 amp diodes (safety factor of 2). Include the "in stock" criterion, and you'll see you're looking at $70 a pop. 400 amp DC switches (why specify DC? - ask around) ain't cheap, either. \$\endgroup\$ – WhatRoughBeast Feb 3 '15 at 16:23
  • \$\begingroup\$ Oh yeah, and the capacitor. dV/dt = i/C, right? Let's say (just as a starting point), 200 amps, 10 usec and a 10 volt drop. Do the math and you get 200 uF. Piece of cake, right? Well, this is sort of the poster child for low ESR caps (ignoring things like fusion research lasers and such). You can't use bog-standard electrolytics - they won't take the current surge. And you also need to start thinking about wire sizes. And connectors. \$\endgroup\$ – WhatRoughBeast Feb 3 '15 at 16:31
  • \$\begingroup\$ And another oh yeah - start thinking about inductive spikes when you switch 200 amps in less than a microsecond. L di/dt is not your friend. \$\endgroup\$ – WhatRoughBeast Feb 3 '15 at 16:33
  • \$\begingroup\$ Thanks. We have decided to reduce the max power to 5kW meaning voltage will be 48V still and current will be around 100A. I have found a few diodes can you advise if they are suitable? uk.rs-online.com/web/p/rectifier-schottky-diodes/6247766 uk.rs-online.com/web/p/rectifier-schottky-diodes/6518614 The Schottky is over double the price so could the other one be used even with the increased voltage drop for the sake of saving money? Also is there a specific type of DC switch you could recommend for this application? \$\endgroup\$ – Joe Feb 4 '15 at 16:45
1
\$\begingroup\$

Here's a short answer which could be improved upon if we knew more about the currents and voltages you're working with. Data sheets, links, etc., to the motor and batteries would help.

:enter image description here

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ The voltages will be around 48V and the maximum current will be around 100A however this will only ever be for a few seconds and not very often. The motor we will be using is goldenmotor.com/goldshop/product/171.html however we are not running it to it's full power capacity. The main problem we are struggling with is what type of switch to use in an application with fairly high power, i.e. MOSFET, relay...etc. \$\endgroup\$ – Joe Feb 4 '15 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.