tips on starting an aircraft with a dead battery

So, the discussion comes up, from time to time, about how to start an aircraft when the battery is dead, and what systems can be used once the engine is running. If you like/dislike hand propping, this is NOT the thread for that discussion. This thread is for sharing tips on getting your electrical system up and running without the normal circuits available.

First off, if you have steam gauges, you probably have them because they operate without electrical power so you only need to get the engine running. That leaves the guys stuck hand propping and the guys that have a plug in power pack specifically for starts. Hand propping assumes no battery power. Once the engine is running, there will be no way to bring the battery on line with the master switch as the master relay relies on battery power to operate. Even if you have an old school generator, the power from the generator will not energize the master relay. And you guys with power packs that plug in to an external power plug, be aware that these will energize your start relay and power your bus right up until you unplug. Then you are still stuck without the ability to energize the master relay.

What you need in these cases is a circuit that will temporarily provide power to the master relay to allow the battery to be connected to the main bus circuit. Look at the picture I've attached to this post. It's from a Cessna battery power diagram and shows how they wired their circuit to fix this issue. I've added color to help me explain the operation. The red lines are the heavy battery cables. You can see one marked 'to battery' and one marked 'to start solenoid'. Self explanatory. Now look at the orange wires. You'll see that one end of the circuit is connected to the 'battery' side of the master relay. It connects to a fuse (5 amp as per Cessna) through a resistor, through a diode, and finally connects to the 'start solenoid' side of the master relay. The reason for this particular order is that if the battery is dead and power is applied to the start solenoid side of the circuit (as would be the case with external power) the diode/resistor arrangement will allow a very small amount of power to the 'hot' side of the coil on the master relay. Note that the green line represents the master switch side of the master relay control circuit. When the master switch is closed, it connects this wire to ground. This works pretty slick when you close the master switch and plug in external power. That orange circuit applies just enough power to close the master relay which, in turn, removes that orange circuit from use. In the blink of an eye and all automatic.

NOW. Lets say you have that orange circuit installed but no external power available. The procedures are similar but a bit more time consuming. Close the master switch and get the engine running. Now you'll need to get power to the master relay in order to close it. To do this, you can spoof the alternator into coming on line as long as you have some type of battery. And I mean pretty much any battery. I'll use a 9V transistor battery as an example. With the engine running and master switch on, pop the alternator field breaker. With jumper wires connect the transistor battery ground to airframe ground and the battery positive to the REGULATOR side of the field breaker. The alternator will come on line in just a second or two. When it does, push in the field breaker and remove the transistor battery. The moment the alternator comes on line, it will energize the master relay, so that the aircraft should be running, battery on line, and the alternator charging. Just remember that it's important to make sure that the master switch is closed any time you are trying to get the aircraft up and running with this circuit installed on the master relay. If you don't it usually blows the fuse when power is applied.

Of course, you could just get yourself a Microstart kit and jump the battery directly, lol.

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Watt is the rating of the resistor and the diode in your bypass circuit?

Jerry

The diode needs to be rated for at least five amps (five amp fuse used here).

Cessna calls out a .75 ohm resistor. With a five amp limit, the voltage drop would be .75Ω X 5A = 3.75volts.
3.75V X 5A = 18.75 watts, so go with a 20 watt resistor.

In operation the resistor/diode circuit will be in series with the coil of the master relay. Using a 'perfect circuit' we will assume 12VDC - .6VDC for the diode voltage drop. This leaves 11.4VDC to push a current through .75Ω + 14.5Ω (resistance of relay coil). 11.4VDC ÷ 15.25Ω = approx .75A.

And I know that technically the battery itself is in parallel with the relay coil while this circuit is in operation. But remember that a depleted battery has a very high resistance. High enough to be ignored anyways. Also remember that In the first post, I cautioned to make sure the master switch was closed when external power or 'spoofed' alternator power was applied. This means that even if there was a momentary high current draw, the power applied to the master relay coil will cause it to close immediately, taking the resistor/diode circuit out of operation.

Web

I assumed both would be rated fairly beefy in an ideal circuit but was curious if Cessna had figured any electric fudgery due the momentary demand. Even a 1N400x will hold the coil for a while before it goes pop. I also wondered about the standardization of the the coils used as mine has a 8.5 ohm coil compared to your 14.5. Maybe mine is a truck part installed in an airplane.

Informative as always - Jerry

The 14.5Ω is from the spec sheet on a Lamar, 12 volt, master relay. As always, real world rules mean nothing is exact, lol.

Web

Web>

How about a way to excite the alt once the engine is started?

That first post has some spacing issues but read through the second half. Once you get the engine running, you can use a small battery to excite the alternator field and get the charging system on line.

Sometimes my explanations are clear as mud so don't hesitate to ask questions.

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