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1.5 volt aircraft glow plug supply

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Sopwith321

New Member
Looking for help with the following please.
to provide 1.5 volts to the glow plugs for an rc aircraft engine. 7 cylinder radial each plug needs up to 3 amps.
This would be built into the airframe and powered by say a lipo or life cell or cells, as light as possible. It would only be used for approx max 60 seconds at a time with a gap of 1 to 2 minutes, no need to be very precise with voltage or noise.
The glow plugs on the lower two cylinders of this engine tend to get flooded and go out after a period of low engine rpm, ie on an approach to land, not a good time!
this system will be linked to the throttle and switched on at the appropriate rpm, off again when throttle is increased, average flight times are 15 mins. Be good to have a min of 5 mins worth of battery capacity, plenty of time to recharge on ground.
Ideally the 1.5 volts would be to a common bus, although a separate circuit for each plug would be ok too long they are all powered by same battery pack.
Simple cheap and cheerful
Many thanks for your time.

Link to a video of the engine just for fun.
 

chemelec

Well-Known Member
So you Need 24 Amps at 1.5 Volts, for the 8 Plugs?

You Might consider using a L-ion battery. (3.7V)
Connected between the HOT's of sets of Two Plugs. (So 3.35 Volts per Plug.)
No Connection to Metal casing.
Might Work, But Might require 4 Separate Batteries.
 
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spec

Well-Known Member
Most Helpful Member
Hy Sopwith321,

Welcome to ETO. I see you are from the UK- which county?

Love the radial engine- push rods and all.:cool:

In addition to chemelec's idea here are some additional thoughts:

(1) The best answer I could come up with is to use a buck (step down), switch mode power supply (SMPS). with a high current Lithium Ion (LiIon) 18650 size battery
(2) The power requirements per plug are 1.5V * 3A= 4.5W
(3) Thus total power for 7 plugs = 7 * 4.5 = 31.5W
(4) Total current = 7 * 3A= 21A
(5) Assume a SMPS efficiency of 70%. This means that the input power from the battery would be 1.3 * 31.5W= 40.95W
(6) The nominal voltage of a LiIon single cell is 3.6V. Thus the battery current capacity will need to be 40.95/3.6 =11.375A minimum.
(7) A typical LiIon 18650 battery will have a capacity around 3.4AH, at 340mA current drain, so the nominal battery duration would be 3.4AH/11.375A= 0.289 hours (roughly 17 minutes)
(8) Because of the high current drain compared to the battery capacity this duration would probably be more like 5 minutes. You could put as many batteries in parallel as you like to increase the duration proportionally.

I will have a look for a suitable battery and SMPS, which should be a small printed circuit board or boards if more than one is required to meet the current requirements.

I'm pretty sure that other ETO members will also have some information on suitable SMPS boards.

UPDATE 2016_05_30 It may be better to wire all seven plugs in series and then find a SMPS to give 7 * 1.5V =10.5V at 3A

spec
 
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spec

Well-Known Member
Most Helpful Member
As a first stab this is what I have come up with:

(1) Connect all 7 plugs in series (10.5V @3A)
(2) Use two LiIon batteries (5.69A max) in series to give a nominal voltage of 7.2V
Suitable batteries would be Sanyo (Panasonic) NCR1865GA. Note, beware of forgeries and only buy from a reliable source. Most of the fancy named batteries on eBay for example are rip offs, especially those that claim high Ah ratings.
(3) Use this SMPS board set to 10.5V: http://www.ebay.co.uk/itm/Mini-3A-DC-DC-Adjustable-Step-down-Converter-Power-Supply-Module-LM2596-WT/172213304896?_trksid=p2047675.c100005.m1851&_trkparms=aid=222007&algo=SIC.MBE&ao=1&asc=36841&meid=42382c037a604e3899965368e1a7ad00&pid=100005&rk=1&rkt=5&sd=310951699136

Note, that I have no experience of this SMPS board. It is likely that you would need to use two SMPS boards. They are hellish expensive at 99p each.:D

As a guestimate, the battery duration would be around 20 minutes using 18650 LiIon batteries of 3.4Ah rating.

spec
 
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JimB

Super Moderator
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(1) Connect all 7 plugs in series (10.5V @3A)
I think that is impossible, one connection is to the body of the plug, and the plugs are screwed into the cylinder heads which are all connected together electrically.

So you Need 24 Amps at 1.5 Volts, for the 8 Plugs?
This is a seven cylinder engine, so there will be seven plugs. A series connection of pairs is not possible.

You Might consider using a L-ion battery. (3.7V)
Connected between the HOT's of sets of Two Plugs. (So 3.35 Volts per Plug.)
I think that you mean 1.875 volts per plug.

JimB
 

spec

Well-Known Member
Most Helpful Member
I think that is impossible, one connection is to the body of the plug, and the plugs are screwed into the cylinder heads which are all connected together electrically. JimB
Damn- I was afraid of that. :banghead:

Back to plan A then.

spec
 

alec_t

Well-Known Member
Most Helpful Member
Jim's right. Series connection of the plugs is impossible. Pity.
However, if the problem is only with the two lower cylinders (lets's say four, to be on the safe side), is it necessary to fire all 7 plugs (except prior to take-off)?
 

spec

Well-Known Member
Most Helpful Member
So you Need 24 Amps at 1.5 Volts, for the 8 Plugs?

You Might consider using a L-ion battery. (3.7V)
Connected between the HOT's of sets of Two Plugs. (So 3.35 Volts per Plug.)
No Connection to Metal casing.
Might Work, But Might require 4 Separate Batteries.
I hadn't appreciated the inginuity of your idea chemelec. The eight plugs threw me but I have got it now.

I don't want to take any credit for this, but what could be done is to connect four plug hots together in one bank and connect the remaining three plug hots together in another bank, plus a power resistor of 500mili Ohms in parallel, to effectively make a second bank of four plugs with the same total resistance as the first bank. It is then just a matter of connecting a single LiIon cell across the two banks of plugs.

This would put 3.6/2 = 1.8V nominal across each plug which should be OK as far as I can remember about glow plugs. The voltage across the plugs could be reduced to any valve by including a suitable resistor in series with the battery. This resistor would be 50mili Ohms nominal, but that resistance could easily be the resistance of the interconnecting wires and contacts and the internal resistance of the battery.

A single 3.4AH 18650 LiIon battery would give a duration of 3.4Ah/12A= 0. 283hr (17 minutes) nominal but this would be reduced by a factor proportional to 0.34A/12A which would probably reduce the battery duration to 5 minutes.

I will find a suitable battery. The battery suggested previously can't, in theory, handle the 12A current drain current required by this approach.

Let me say that I am just the application engineer here, chemelec thought up the enabling architecture. :)

spec
 
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spec

Well-Known Member
Most Helpful Member
You could splash out UKP7 and get seven of the buck converters. However, the one above is using an LM2596 that is only rated at 2A without an additional heatsink. Maybe go for 7 of these, http://www.ebay.co.uk/itm/DC-DC-XL4...864362?hash=item4620e9322a:g:xMUAAOxyrrpTiN57

Edit the other advantage of this method is you can use any battery voltage. Even an old 7.2V nimh one.

Mike.
Nice one Mike, :cool:

I hadn't thought of using a separate converter for each plug,

That is a useful converter for other applications too. Shame it doesn't work down to 3V so that it could operate from a single LiIon cell.

Any idea what the main switching chip is; it looks like XL 4015E1 from the eBay image.

Chuck
 
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Pommie

Well-Known Member
Most Helpful Member
It is the XL4015 (mouse over and you can read it) which can handle the 5A claimed.
I've bought a few buck and boost converters of ebay because there's no way you can make them for that price.

Edit, the idea of serially connecting two would reduce the count to 4. Three at 3V and 1 at 1.5.

Mike.
 

JimB

Super Moderator
Most Helpful Member
the idea of serially connecting two would reduce the count to 4. Three at 3V and 1 at 1.5.
But, the XL4015 is not isolated from input to output, so connecting them in parallel to the battery , and have the centre tapped loads commoned together via the engine block could be a recipe for disaster.

JimB
 

spec

Well-Known Member
Most Helpful Member
But, the XL4015 is not isolated from input to output, so connecting them in parallel to the battery , and have the centre tapped loads commoned together via the engine block could be a recipe for disaster.

JimB
Why is that Jim- the battery is floating.

spec
 

chemelec

Well-Known Member
Sorry, Bad Day Yesterday, Can't Count Correctly and Can't Divide the voltage by Two.

But yes I think SPEC's Idea with the Added Resistor on the 7 Plug would work OK.

I don't like the idea of the Buck Converter, Should it FAIL, You Might Burn Out All the Plugs.

One Other Possibility, Use a 6 or 12 Volt Car Battery, and a PWM Circuit.
Connect all Plugs in Parallel and set the Pulse width so to give the Effective Average Total Current.

Take care.....Gary
 

spec

Well-Known Member
Most Helpful Member
Re approach of post #9.

These batteries are recommended:
(1) Samsung INR18650-30Q (3AH 15A) This is one of the most popular batteries for high current applications. Duration 2.4Ah/12A = 0.2Hr= 12 minutes
(2) LG INR18650-HG2 (3AH, 20A) Duration 2.55Ah/12A = 0.2125 hr= 12.75 minutes.

Both batteries weigh 48 grams.

To increase the duration just place more batteries in parallel. Each battery will give you an extra 12 minutes in round figures.

spec
 
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spec

Well-Known Member
Most Helpful Member
Sorry, Bad Day Yesterday, Can't Count Correctly and Can't Divide the voltage by Two.

But yes I think SPEC's Idea with the Added Resistor on the 7 Plug would work OK.

I don't like the idea of the Buck Converter, Should it FAIL, You Might Burn Out All the Plugs.

One Other Possibility, Use a 6 or 12 Volt Car Battery, and a PWM Circuit.
Connect all Plugs in Parallel and set the Pulse width so to give the Effective Average Total Current.

Take care.....Gary
Have I got it wrong?

I thought the plug heater would be on board the aero-model when it was in flight.:arghh:

spec
 

chemelec

Well-Known Member
I Believe the heater is Only Required for Starting the Plane.
Not for Running it.

Back when I was YOUNGER, (50 years ago) guys had single Piston Planes, and the battery was Just for Starting
After that, the Ignition of gas, would keep the plugs Glowing.
 

chemelec

Well-Known Member
Using a Mosfet # STP80PF55 in this PWM Circuit, it will easily Handle the 21 Amps required for the 7 Plugs.

But a 6 Volt Battery is NOT Suitable.
Use 12 Volts, so it gives enough Gate Voltage to Fully Turn on the Mosfet as needed.

http://chemelec.com/Projects/PWM/PWM.htm
 
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JimB

Super Moderator
Most Helpful Member
Why is that Jim- the battery is floating.
True.
But I was thinking about the "seventh" glow plug all on its own. It would be OK if you had a dummy resistor of the correct value to balance the thing up. But then you have this odd resistor floating about in free air, getting very hot just like a glowplug.

I Believe the heater is Only Required for Starting the Plane.
Not for Running it.
Generally speaking, that is correct. But in the initial post there is the statement:
The glow plugs on the lower two cylinders of this engine tend to get flooded and go out after a period of low engine rpm, ie on an approach to land,
So there is a requirement to fly this thing.
Maybe, as has already been suggested, the "fix" should only be applied to the two lower cylinders.

JimB
 
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