Battery switching without tears

Agreed.

On 4/16/2020 1:16 PM, 2G wrote:
On Thursday, April 16, 2020 at 12:09:32 PM UTC-7, Dan Marotta wrote:
Well, Tom,Â* I don't have a scope and you do, so I ask again:Â* Why don't
you measure it?

But I will say that it's absolutely impossible to have a 5 volt
difference between the batteries.Â* I've never seen a 12 volt SLA higher
than 13.6 volts and that's fresh off the charger.Â* By the time you plug
it into your system, it's closer to 12.2 or 12.4 volts.Â* An 11.4 volt
battery will run a variometer, but likely won't transmit over your
radio.Â* A DPST switch will switch over in mili seconds, not seconds.

We're not talking about bridges that carried heavy trucks for 40 years
with a design defect.Â* If your wiring is 40 years old, I'd suggest
changing it.

They also said that a jet fuel fire couldn't weaken a steel beam
sufficiently to cause a building to collapse.Â* Come toÂ* Moriarty and
have a look at the steel post and beam hangar that slumped to the ground
last week after a fire.

On 4/16/2020 11:23 AM, 2G wrote:
On Thursday, April 16, 2020 at 8:56:54 AM UTC-7, Dan Marotta wrote:
Really?

What is the shorted time when flipping the switch?Â* What's the voltage
difference between the two batteries?Â* What's the total circuit
resistance, including the internal resistance of the batteries?
Theoretical math and practical application do not always agree.Â* It
might be fun to set up such a demonstration and use your o'scope to
measure that current and it's time duration.Â* Compare that to the "blow
time" of any fuses.

Seriously, I've done it for years without any problems, but I recognize
that past performance is no guarantee of future results. I'd be curious
about the results and you have the equipment to do it.

On 4/16/2020 12:21 AM, 2G wrote:
On Wednesday, April 15, 2020 at 6:18:08 AM UTC-7, wrote:
While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.

If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.

After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
A make-before-break is also called a "shorting" switch. If you use such a switch you WILL short the two batteries together, which could result in a large current flow from the battery with the higher voltage to the battery with the lower voltage. This large current could blow your protection fuse(s). This is especially the case if you have two separate switches.
--
Dan, 5J
Dan,

The I-35W bridge in Minneapolis worked fine for 40 years before it collapsed. The problem was a design error that was there since Day 1.

How much current flows between your two batteries? A lot! The internal resistance of the batteries is probably 0.01 ohm apiece, 18 AWG wire is 0.06 ohm/ft and your switch contact is typically 0.01 ohm. The big unknown is the location of your batteries and the length of the wire. However, the longer the run the more likely they used a smaller gauge wire, so let's start with 10 ft. This totals 0.1 ohm. Looking at the worst-case scenario, you may have a 5 V difference which results in 50 A of current. Since you are manually flipping switches, this current could last a few seconds. The largest factor here is the length and gauge of the wire. If I were you I would measure the actual current the way I did: with a scope and current probe or shunt.

Tom
--
Dan, 5J
I will measure it. The 5V, which I said was worst case, comes from a 9V battery being connected to a 14V battery. These voltages will equalize very quickly and impossible to see with an ammeter or a DVM because they don't have the bandwidth to observe millisecond events.

Tom

--
Dan, 5J

CH.........I have switched on the fresh battery just before switching off the low battery for over 40 years and counting. No resisters, no capacitors, no nothing!
I documented this with my little test! Now, I’m using the same batteries, same capacity and same type batteries!
Hope this helps,
JJ

JJ, unfortunately, the previous owner installed a 3 position switch in bat1/off/bat2 so the switiching takes just a fraction of a second longer than if It was just a 2 position. And nothing else in my panel is affected other than the powerflarm and then only once in a while ( last time was in a nationals

On Thursday, April 16, 2020 at 12:07:53 PM UTC-7, 2G wrote:
On Thursday, April 16, 2020 at 10:35:35 AM UTC-7, wrote:
I got so mad the last time we fought this battery switching battle, I went out and hooked a fresh 12v battery to a low 10v battery with the parallel circuit completed with an amp-meter! I saw no spike and only 300m/a current flow as the fresh battery tried to charge the low battery.........the switches didn’t melt, the amp-meter didn’t explode and the fuses didn’t blow!
I have recently solved the issue by only using one battery............15a/h lithium iron battery............have t seen voltage below 12.2 v, even after 4 hours at 2a average current flow!
Please don’t stop arguing this issue, it’s so entertaining to watch the EE’s chase imaginary amps around!
JJ

I will add that Jon's concern about arcing is misplaced; arcing is associated with inductive loads when the current is interrupted. What happens in an inductor is that there is a magnetic field that is built up that has to have a place to go when the current is suddenly interrupted. The voltage in the circuit goes to very high levels as a result and will cause an arc in a mechanical switch. This is usually dealt with by a fly-back diode that allows the current to continue to flow and die off gradually.

If arcing had been taking place I would have seen it on the scope waveforms as a series of spikes in the current waveform. This did not occur in any of the many times I tested it, and it should not occur as capacitors are effective in minimizing or eliminating spikes.

Tom

Arcing will occur on break in an inductive circuit and on make in a capacitive circuit. It is quite easy to strike an arc with a 12V battery. It will happen without the capacitor - the increased capacitance will increase the energy and/or duration. You are very unlikely to see that on an oscilloscope - it will not create the voltage spike that an inductive load will. Open the switch body up and switch it on in a dark room - you'll see it. I'll be convinced when you show me the manufacturers spec that says 90A is ok on a 2A switch "as long as it doesn't last too long" or recommending it to switch large capacitive loads.

By far the simplest solution is to get a 12AH LFP and fly all day with the voltage above 13 the whole time. (By the way, you do get a warning near the end of the LFP's charge, on my 12AH it takes about 1.5 hrs to go from 12.5 to 11.5V at about the 9 hour mark).

On boats we use make-before-break switches to switch between battery banks (to preserve alternator diodes). The banks are hundreds of AH, with many thousands of Amps available. There are no problems with current when switching, even if a 10V bank is continuously connected to a 14V bank by selecting "Both" on the switch. The reason is the charge acceptance rate on LA batteries is quite low, even from a constant voltage, unlimited current source. Typically less than 1/2C, so in a glider with 12AH batteries, you might get around 6 amps for a few minutes between a fresh and completely discharged battery worst case. At about 1/2C charge rate, the battery terminals will be at max charge voltage even starting from completely flat. The IV characteristics of an LA battery are very different than capacitors. Switching LFP batteries this way may create problems - it depends on the particular batteries. Switching a charged LFP battery onto a discharged LA will again hit the same 1/2C current limit. Switching a charged LA battery onto a discharged LFP could create high current, depending on the batteries.

So it isn't a mystery why this has been done for 40 years with no problems. But those gliders don't have huge capacitors on the bus.

On Thursday, April 16, 2020 at 10:27:48 AM UTC-7, 2G wrote:
On Thursday, April 16, 2020 at 9:29:50 AM UTC-7, jfitch wrote:
On Wednesday, April 15, 2020 at 11:36:35 PM UTC-7, 2G wrote:
On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially..

Andy

KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.

Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.

- Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.

I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).

Tom

Tom,good plan to increase the value of the resistor to limit the current inrush. One could even make the resistor 100 ohms and have a reversed diode in parallel with it so that when the capacitor is needed to sustain the instrument during switching, the current would flow in the reverse direction through the diode, bypassing the resistor. You now have the best of both worlds - slow charge of the capacitor when the power is turned on, and a fast discharge to support the instrument without IR drop across the resistor, instead the drop would just be the bias voltage of the diode.

I don't really think that the resistor is necessary, but offer it to those that are overly concerned about the inrush current. The bottom line is the energy that is transferred from the battery to the capacitor heating the switch contacts. Switches have current ratings to limit the temperature rise to tolerable levels when the current is flowing continuously; this short current pulse will not raise the switch contact temperatures to any significant level. Remember, the SAME amount of energy will be transferred between the battery and the capacitor REGARDLESS of the resistor value (joules = C * V). This translates to the SAME amount of switch contact heating. If you make the resistor insanely large so the time constant is on the order of minutes, the heat will dissipate and lower the maximum temperature. A better approach is to use a smaller capacitor that still maintains voltage during switching.

Tom

The concern isn't heating in the switch contacts due to their specified resistance. It is high current arcing during switching. This can cause erosion of the contacts, in more extreme cases welding them together. A 12V battery is quite capable of generating a vey high energy arc, one can be used to arc weld steel. Arcs are peculiar phenomena with negative resistance, not easy to measure their presence or characteristics. All DC switched arc on make, the current of the arc is limited by the impedance of the circuit - in this case very low.

There is little mortal danger, the switch isn't going to catch fire or explode. It probably will have a markedly shorter life. In the worst case it may weld itself on one day. The capacitor may be the easiest solution, but not the most elegant, and it may not be without tears. The best solution is to parallel the batteries always so that routine switching is unnecessary. This has higher reliability, will result in longer battery life, and requires no operator action. If circumstances make that impossible then as suggested above a select switch shunted with diodes, followed by an on-off switch is safe, zero energy loss, and keeps all components in spec.

As I have mentioned several times now, a small series resistor will reduce the current down to acceptable levels. How many times must I repeat this?

Tom

A small resistor will reduce the inrush current to the capacitor, but it will also use power all day. A 1 ohm will reduce your effective battery capacity by 8% - making the need for switching all the more likely!

On Thu, 16 Apr 2020 23:00:28 -0700, jfitch wrote:

On Thursday, April 16, 2020 at 10:27:48 AM UTC-7, 2G wrote:
On Thursday, April 16, 2020 at 9:29:50 AM UTC-7, jfitch wrote:
On Wednesday, April 15, 2020 at 11:36:35 PM UTC-7, 2G wrote:
On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7,
wrote:
On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4,
wrote:
On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy
Blackburn wrote:
I used Shottky diodes plus power resistors plus
capacitors. I'm no EE but I took enough circuits courses
to handle this problem. The Shottky diodes keep the
batteries from cross-discharging each other, the
capacitors keep the instruments powered when the switch
is disconnected from battery 1 and before it is connected
to battery 2 and the resistors keep the capacitors from
drawing too much current when you power them up since
they make the circuit (even with the diodes) look like a
direct short initially.

Andy

KISS. Just the two diodes (and no switch) should be
enough. Whichever battery is stronger (higher voltage)
would take the load. Automatically. No switching needed.
With the higher voltage of LiFePO4 batteries (relative to
lead-acid) the voltage drop in the diode is acceptable,
especially if it's the Schottky type.

Or, if you really want to remove the voltage drop in the
diodes, add an SPDT switch (perhaps one with also a
center-off position) IN PARALLEL to the diodes. No matter
which position that switch is in, both batteries will still
be connected. But the battery the switch leads to will
feed the avionics with no voltage drop since the switch
bypasses the diode on that side. The other diode will
meanwhile prevent current from going INTO the other
battery. The middle-off position (or no switch at all) is
the safest though, since if either battery develops a
shorted cell (or shorted or loose wiring, blown battery
fuse, etc) without your knowledge, it won't affect the
other battery and the avionics, thanks to the two diodes.

- Clarification: I meant a diode between each battery and the
avionics bus as a whole. Not separately for a specific
instrument.

I measured the inrush current once again and found that the
vertical of the scope was set for a 1X probe instead of the 10X
actually being used. This meant that the peak current was 90A
instead of 9A, which is a bit high. I added a 1.1 ohm resistor
and the peak current dropped to 6A. A simulation shows that a 2
ohm resistor drops it to 3A. This is a good value to use if you
have a 1A current drain as the voltage drop will be 2V. The
wattage of resistor is unimportant because so little energy is
being dissipated by the resistor. The energy transferred
remains constant regardless of the resistor value as it is the
energy required to charge the capacitor (the current pulse
lengthens for larger resistor values).

Tom

Tom,good plan to increase the value of the resistor to limit the
current inrush. One could even make the resistor 100 ohms and
have a reversed diode in parallel with it so that when the
capacitor is needed to sustain the instrument during switching,
the current would flow in the reverse direction through the
diode, bypassing the resistor. You now have the best of both
worlds - slow charge of the capacitor when the power is turned
on, and a fast discharge to support the instrument without IR
drop across the resistor, instead the drop would just be the bias
voltage of the diode.

I don't really think that the resistor is necessary, but offer it
to those that are overly concerned about the inrush current. The
bottom line is the energy that is transferred from the battery to
the capacitor heating the switch contacts. Switches have current
ratings to limit the temperature rise to tolerable levels when the
current is flowing continuously; this short current pulse will not
raise the switch contact temperatures to any significant level.
Remember, the SAME amount of energy will be transferred between the
battery and the capacitor REGARDLESS of the resistor value (joules
= C * V). This translates to the SAME amount of switch contact
heating. If you make the resistor insanely large so the time
constant is on the order of minutes, the heat will dissipate and
lower the maximum temperature. A better approach is to use a
smaller capacitor that still maintains voltage during switching.

Tom

The concern isn't heating in the switch contacts due to their
specified resistance. It is high current arcing during switching.
This can cause erosion of the contacts, in more extreme cases welding
them together. A 12V battery is quite capable of generating a vey
high energy arc, one can be used to arc weld steel. Arcs are peculiar
phenomena with negative resistance, not easy to measure their
presence or characteristics. All DC switched arc on make, the current
of the arc is limited by the impedance of the circuit - in this case
very low.

There is little mortal danger, the switch isn't going to catch fire
or explode. It probably will have a markedly shorter life. In the
worst case it may weld itself on one day. The capacitor may be the
easiest solution, but not the most elegant, and it may not be without
tears. The best solution is to parallel the batteries always so that
routine switching is unnecessary. This has higher reliability, will
result in longer battery life, and requires no operator action. If
circumstances make that impossible then as suggested above a select
switch shunted with diodes, followed by an on-off switch is safe,
zero energy loss, and keeps all components in spec.

As I have mentioned several times now, a small series resistor will
reduce the current down to acceptable levels. How many times must I
repeat this?

Tom

A small resistor will reduce the inrush current to the capacitor, but it
will also use power all day. A 1 ohm will reduce your effective battery
capacity by 8% - making the need for switching all the more likely!

How do you work that out?

There is no inflow to the capacitor once it is charged to the same
voltage as the prime battery. This happens when you first connect the
battery and set the switch to connect the prime battery to the panel.
Don't forget that the capacitor is on the PANEL side of the switch.

When you switch power over from prime to backup battery the capacitor
will discharge for a millisec or two when no battery is connected and the
capacitor is running the panel, followed by an equally quick inflow as
the backup battery tops up the capacitor to match its voltage.

If you want to be really picky, there will also be a very small outflow
from the capacitor: as the battery voltage slowly drops under load, the
capacitor will discharge slowly to match the battery voltage.


--
Martin | martin at
Gregorie | gregorie dot org

On Thursday, April 16, 2020 at 10:46:57 PM UTC-7, jfitch wrote:
On Thursday, April 16, 2020 at 12:07:53 PM UTC-7, 2G wrote:
On Thursday, April 16, 2020 at 10:35:35 AM UTC-7, wrote:
I got so mad the last time we fought this battery switching battle, I went out and hooked a fresh 12v battery to a low 10v battery with the parallel circuit completed with an amp-meter! I saw no spike and only 300m/a current flow as the fresh battery tried to charge the low battery.........the switches didn’t melt, the amp-meter didn’t explode and the fuses didn’t blow!
I have recently solved the issue by only using one battery.............15a/h lithium iron battery............have t seen voltage below 12.2 v, even after 4 hours at 2a average current flow!
Please don’t stop arguing this issue, it’s so entertaining to watch the EE’s chase imaginary amps around!
JJ

I will add that Jon's concern about arcing is misplaced; arcing is associated with inductive loads when the current is interrupted. What happens in an inductor is that there is a magnetic field that is built up that has to have a place to go when the current is suddenly interrupted. The voltage in the circuit goes to very high levels as a result and will cause an arc in a mechanical switch. This is usually dealt with by a fly-back diode that allows the current to continue to flow and die off gradually.

If arcing had been taking place I would have seen it on the scope waveforms as a series of spikes in the current waveform. This did not occur in any of the many times I tested it, and it should not occur as capacitors are effective in minimizing or eliminating spikes.

Tom

Arcing will occur on break in an inductive circuit and on make in a capacitive circuit. It is quite easy to strike an arc with a 12V battery. It will happen without the capacitor - the increased capacitance will increase the energy and/or duration. You are very unlikely to see that on an oscilloscope - it will not create the voltage spike that an inductive load will. Open the switch body up and switch it on in a dark room - you'll see it. I'll be convinced when you show me the manufacturers spec that says 90A is ok on a 2A switch "as long as it doesn't last too long" or recommending it to switch large capacitive loads.

By far the simplest solution is to get a 12AH LFP and fly all day with the voltage above 13 the whole time. (By the way, you do get a warning near the end of the LFP's charge, on my 12AH it takes about 1.5 hrs to go from 12..5 to 11.5V at about the 9 hour mark).

On boats we use make-before-break switches to switch between battery banks (to preserve alternator diodes). The banks are hundreds of AH, with many thousands of Amps available. There are no problems with current when switching, even if a 10V bank is continuously connected to a 14V bank by selecting "Both" on the switch. The reason is the charge acceptance rate on LA batteries is quite low, even from a constant voltage, unlimited current source. Typically less than 1/2C, so in a glider with 12AH batteries, you might get around 6 amps for a few minutes between a fresh and completely discharged battery worst case. At about 1/2C charge rate, the battery terminals will be at max charge voltage even starting from completely flat. The IV characteristics of an LA battery are very different than capacitors. Switching LFP batteries this way may create problems - it depends on the particular batteries. Switching a charged LFP battery onto a discharged LA will again hit the same 1/2C current limit. Switching a charged LA battery onto a discharged LFP could create high current, depending on the batteries.

So it isn't a mystery why this has been done for 40 years with no problems. But those gliders don't have huge capacitors on the bus.

Arcing is a chaotic process that produces erratic current spikes - go do your research and you will see. I saw none of that on the waveforms I recorded. Again, for what must be the FIFTH TIME add a series resistor to reduce the current level if you wish. I REPEAT, add a series resistor to reduce the current level if you wish. Do you got it this time, Jon?????

Continuing the diversion...

Not a problem here and maybe optimal given the times, but
what is is about the field of EE?

We seem ready, willing and able to have a discussion unencumbered by reality.
Other sorts of E's seem more grounded.
I wonder if it is because with SW and IC's we can make things with so many frilly details that we forget the real?

Or perhaps making really new stuff is about imagining the unreal and then making it happen. That might explain being wired differently?

I think we’re arguing over two different events. In a typical SH bird, that has one switch for Bat-1 or Bat-2, all power will be shut off for a brief moment as batteries are switched.........Yes, this will cause a voltage spike as electronic fields collapse across coils, etc in the radio and other electronic equipment, that was drawing about 2 amps before the power was removed. Several ways to prevent the momentary power loss have been recommended and I believe they will all work!

The procedure that Hank, Chip and I use does not remove all power, it only adds about 2 volts to the power supply! We turn on the fresh battery with its own SPST switch, just before we switch off the low voltage battery with another SPST switch. I have tested this and observed no voltage spike, and read 300 m/a (1/3 amp) as the fresh battery tried to recharge the low battery. This event is little more that what the old “voltage regulator� did when cars had generators. The voltage regulator would shut off generator voltage when the battery was fully charged, then turn on generator voltage as it sensed the battery voltage was getting low. The same event occurs when your buddy offers to “give you a jump� when your car won’t start. Using jumper cables, he adds about 2 volts to your car battery.
Does this help to put to rest this never ending argument?
JJ

Does this help to put to rest this never ending argument?
JJ

Of course not! This is RAS! I expect this argument to go on until the Sun burns out. (Because of a lack of diodes and resistors.)