4
\$\begingroup\$

I want to power a motor with a battery, however the complication arises as I want to have two power sources and be able to switch between them. The context of this problem is for an electric vehicle. Sensing circuitry on the vehicle will measure speed and driver input and decide which of the two batteries to use under the conditions. i.e. when acceleration is needed the higher power battery is selected and power is drawn from that. When the vehicle is driving at slow speeds or at constant velocity the so called "cruising battery" will be switched to and used to power the motor. Both batteries will be at 48V and the motor power will be at a maximum of 5kW giving a maximum current of around 100A. 2 possible switching circuits

The above two circuits were suggested to me as possible ways of switching between the two batteries. I believe that a Schottky diode or capacitor of the correct rating can be found for fairly cheap however the main problem I seem to be facing is what type of switch to use and how to control this switch. On the circuits I have drawn a relay however am I right in thinking a contactor would be needed for a high power application such as this? In this case is it reasonable to control a contactor with a MOSFET as shown in the diagrams?

If anyone has any suggestions or advice for components that would be really useful, as well as suggestions for controlling the MOSFET/contactor. Using my current circuits the system would require 4 power sources, 2 for the motor, one for the contactor and one for the MOSFET!

Apologies for my inexperience or any unclear parts of my question.

Thanks!

\$\endgroup\$
  • \$\begingroup\$ How are the two batteries different? \$\endgroup\$ – Dan D. Feb 5 '15 at 13:23
  • \$\begingroup\$ Your first circuit is not going to work, that series capacitor is an open circuit at DC, and a battery is DC indeed. And tyo answer your question, just search for a 50V 100A relais. You could use a couple of power mosfets too, that would be much better. \$\endgroup\$ – Vladimir Cravero Feb 5 '15 at 13:28
  • \$\begingroup\$ @Dan : see his other question electronics.stackexchange.com/questions/151341/… \$\endgroup\$ – Brian Drummond Feb 5 '15 at 13:28
  • \$\begingroup\$ @VladimirCravero apologies it's been a while since I've studied electronics! Am I right in thinking that if the capacitor was in parallel with a resistor, when the switch is turned off the capacitor would discharge and hence power the controller for a short time? \$\endgroup\$ – Joe Feb 5 '15 at 14:19
  • \$\begingroup\$ @Joe You have to put the capacitor between the +48V and the 0V. Capacitors storage energy, after charging, it will have 48V between its terminals. When the battery is disconnected, it will gradually discharge pumping energy to the motor. You should also put a schottky diode between motor terminals, otherwise it will fry everything up due to its inductance when switching. \$\endgroup\$ – gstorto Feb 5 '15 at 14:46
3
\$\begingroup\$

A number of issues.

In the first circuit, of course, the capacitor should be

schematic

simulate this circuit – Schematic created using CircuitLab

Just as importantly, don't get hung up on the value just yet. Its' value is determined by switching speed of the switch, and the allowable droop in voltage when there is no source connected.

Using a mechanical contactor for switching a DC current is generally a bit tricky. The problem is that, unlike AC, when you try to break a current flow an arc will develop between the contacts, and without the voltage reversal inherent in AC, the arc can persist and damage the contacts. DC contactors do work around this, but they tend to be expensive. (Actually, contactors in general tend to be expensive, but you probably already know this.) Solid-state DC relays are probably your best bet if you want to go the contactor route. Digikey has some 160 amp units. For $150 +.

Going MOSFET is probably your best choice, and for this application you need a p-type high-side unit configured like so

schematic

simulate this circuit

And you're in luck. The MOSFET I've shown is available from Digikey for $25.

M2 is almost any n-type with a voltage rating greater than 50 volts.

I show the input drive as 0/5 volts. If you must use a lower logic level (like 3.3) that's certainly possible, but you must remember to get "logic-level" MOSFETs, since otherwise they may require 4 volts to turn on fully.

This circuit ought to be easily capable of 1 usec switching. That means that you will have to learn the fine art of protecting against inductive surges, but that's a story for another time.

You'll also note that this circuit does not require a separate "contactor" supply. If, for some reason (like you can get one for free) you decide to go the contactor route, note that you don't need a separate supply for it. You just use the battery you're switching to drive it.

You can use two of these circuits to create a double-throw effect, but you have to make sure that there is a delay between releasing one before activating the other. The delay should be on the order of a microsecond or so, but check your actual circuit operation first.

\$\endgroup\$
  • \$\begingroup\$ Thanks for the response and sorry about the late reply, I've been building some of the circuits you suggested with low power components and a 12V power supply to test them. I've just started looking into the MOSFET's and have a few questions. Why does the circuit you suggested require two seperate FET's? In other words why can't you just control the power FET with the driving input 0/5V? Also what is the purpose of the resistors and diode? \$\endgroup\$ – Joe Feb 26 '15 at 13:31
  • \$\begingroup\$ M2 doesn't NEED to be a MOSFET - you can use an NPN bipolar instead. But M1 needs a gate drive which varies from 50 V (off) to ~40 V (on). 0 to 5 volts wont' do it. Research p-type MOSFET. R2 keeps the circuit well-behaved if the logic input gets disconnected. R1 and D1 (which is a zener diode) set the gate level at ~12 volts below Vcc. More than ~20 volts will kill the MOSFET. \$\endgroup\$ – WhatRoughBeast Feb 26 '15 at 13:49
  • \$\begingroup\$ Added R3 to 2nd schematic - it is needed. \$\endgroup\$ – Dwayne Reid Mar 7 '15 at 15:24
  • \$\begingroup\$ Thanks @dwayne , could you elaborate as to why this extra resistor is needed? \$\endgroup\$ – Joe Mar 9 '15 at 20:22
  • \$\begingroup\$ Actually, it may not be necessary, depending on what your 0/5 volt drive looks like. It's pretty straightforward that the 5 volt level will be driven by a voltage source, but what about the 0 volts? If 0 volts is produced by actually connecting the input to ground, you don't need R2. If it's just sort of floating, such as being driven with a PNP to +5, or if it's possible that the input may be disconnected, then you need R2 to force the gate to ground. Otherwise, it will float to some intermediate voltage and M1 and M2 may not be turned fully on or off. \$\endgroup\$ – WhatRoughBeast Mar 9 '15 at 22:13

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.