1
\$\begingroup\$

I am looking for some assistance with options in isolating 2 parts of a circuit.

I have a 4G hotspot that I am using in the car, powered via the cigarette lighter socket (rated at 12V/10A) and a 5V USB regulator. The hotspot is intended to be always available even when the car is off. Due to the fact that the car is off for far longer than it is running, the device discharges it's internal battery in a few days as it is not charging long enough.

To this end I have a circuit that will charge several (in my case 3) 12V 2.2Ah Sealed Lead Acid batteries in parallel and operate the hotspot device while the vehicle is operational. The batteries alone will power the device while the vehicle is not operating (ie the Ignition switch is off).

However, I am stumped on how to isolate the rest of the car's accessories within the accessory circuit (as isolated by the ignition switch) from the charge circuit when the ignition is off. Once activated, the charge circuit feeds power back into the vehicle accessory circuit.

One version I have has a relay to enable the circuit when the ignition is on, but the power feeding back from the batteries through the closed relay energises the activation circuit and keeps the relay closed when there is no power from the ignition. The dashed line is roughly the boundary between the part of the circuit that needs to be permanently powered and the part that should only powered when the ignition is on.

wiring diagram

I have another that replaces the relay with parallel diodes in the power path, but the ~1V voltage drop and heat dissipation from regular silicon diodes is undesirable, and I tried several Schottky Diodes in parallel (because they have a smaller/acceptable voltage drop of ~0.13V each (measured)). However the Schottky diodes leave ~1.8V on the 'dead' side circuit that could put extra drain on the SLAs when the ignition is off. The LED on the plug interfacing with the 12V car socket indicates power is there when the ignition is off and I measured the voltage with a multi-meter across the terminals of the plug with it disconnected from the car.

Is there a way that I can have a low or no voltage drop, low impedance circuit, fairly high current (but < 10A) that will sense when the ignition is off and isolate the external batteries and charge circuit, without modifying the car's built-in and under warranty wiring?

I have seen mentioned elsewhere use of a MOSFET(s) and a sensing circuit to control it to replace diodes to make an 'ideal diode' but the references are mostly in the context of rectification of AC current not this circumstance and I haven't worked out how to apply the theory to this situation.

The hardest part in anything I have dreamt up is the fact that once the circuit is first powered from the ignition, you cannot tell when the ignition has been switched off as the batteries in the charge circuit maintain the power on both sides.

Any pointers would be appreciated.

EDIT [2018-05-30 - 22:51 AEST [UTC+10]:

Would something like this work to have the battery circuit connect to the car supply if the car supply voltage is over 13V (ie the car is running and alternator is operating at 14.2V). As the SLAs are only about 12.5V fully charged then no current would flow thru the zener diode and the transistor would be off. If the voltage was above 13V, current would pass thru the zener, limited by the resistor and switch the transistor on to activate the relay. My zener theory is rusty and transistor theory even rustier, but would something like this work? Calculating the resistance values would be the next challenge (for me at least...).

Hotspot Charge Curcuit w/Zenner

\$\endgroup\$
  • \$\begingroup\$ Use ignition enabled Cig lighter circuit to Relay coil only with 25A relay contacts between each V+, with 25A diode in series to gnd on SLA to get 14.2-0.5V with cathode chassis heat sink otherwise hot. \$\endgroup\$ – Sunnyskyguy EE75 May 30 '18 at 2:12
  • \$\begingroup\$ G'Day Tony, I was trying to avoid a diode in series with the power path to avoid both the heat and voltage drop. The difference between the 12.x Volts that the batteries have and the ~13.8V operational voltage see 1/2 of it consumed by the voltage drop over the diode and dissipated as heat. I was looking to maximise the power going to the batteries rather than dissipating that potential charge current as heat. Thanks. \$\endgroup\$ – Braedon King May 30 '18 at 2:38
  • \$\begingroup\$ SLA's need a thermally compensated charge profile that uses a different Vmax than flooded cells in a vehicle. THat's why I dropped it with a diode. Are you sure vehicle is 13.8 or 14.2 or? with no load. \$\endgroup\$ – Sunnyskyguy EE75 May 30 '18 at 2:40
  • \$\begingroup\$ Also SLA don't like fast charges and causes cell inbalance. Like most people here, they fail to understand the datasheets and fail to make a system design spec with tolerances and ask about an implementation without specs. Why use a few puny SLA's when you get an used auto battery with 50Ah. Try again/ \$\endgroup\$ – Sunnyskyguy EE75 May 30 '18 at 2:46
  • \$\begingroup\$ I've not directly measured the car's voltage when in operation, I will however tonight. All the cars I have had in the past were 13.6-13.8V, so I just used that figure. The small SLAs are due to the space under the drivers seat being limited, and the 3 batteries charging normally will then divide their output by 3 (each) when discharging and thus my charge to discharge rate will be greater than for 1 battery. The relay I was using was rated at 30A. \$\endgroup\$ – Braedon King May 30 '18 at 3:02

Your Answer

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

Browse other questions tagged or ask your own question.