How do I limit current flow from a Car battery to an amp?

[t-head]

Reply to email question before my post from respected member/moderator to whom I was able to send manual with schematic of amp in question (Elekit TU8100):

"I see no theoretical reason why IC1 would see a difference between a battery source or an SMPS since its power rail is governed from a 5V chip reg anyway (IC2). The total power demands will be determined by the amp's power components, IC1 is just a microcontroller."

Suggestion from new member who deals with 12v car audio:

Cincon Isolated DC/DC Converters | Mouser

So all is good and I maybe learning something...now to solder gazilions of resistors, caps, connections and both ends of a 26-conductor ribbon...the fun begins...

 

[AndrewT]

Indeed what the original poster is trying to do does not make sense. What are you trying to do in the big picture?

There are commercial battery protection devices, for example current limiters for the 2nd battery in a dual battery system, or battery isolators that isolate the battery if voltage drops below a certain threshold. Are you conflating what these devices do and what your needs are?

you can use a Constant Current Source CC as a current limiter.

Fitted after the SMPS, it will limit the current drawn from the SMPS. But the output voltage after the CCS will be variable.

Fitted before the SMPS, the SMPS will try to maintain output when it's input voltage drops and will pull more current until you reach the CCS limit.

In both cases you can size the main dropper/passing device to survive the max current draw with max voltage differential.

The CCS does not blow when it enters limit it just reduces the available output voltage until the load circuit effectively stops working.
When the overload is removed the circuit goes back to normal. No reset is required.

I use similar as a battery charger. CCS followed by a voltage regulator. This limits the maximum charge current and as battery voltage rises the regulator takes over to prevent overcharging.

 

[t-head]

Thanks AndrewT,

I am actually trying to get rid of the mains connected external SMPS (like a computer power supply) and run an Elekit TU8100 amp directly from my sailboats storage battery. I have been educated on basic electric theory to correct my lack of understanding and have also found some DC-DC regulators that should work if I am hesitant to run directly from battery...

 

[infinia]

correct my lack of understanding and have also found some DC-DC regulators that should work if I am hesitant to run directly from battery...

I hear your concerns about eliminating the SMPS brick , its probably fine directly connected. post or PM the Elekit schematic and I can advise better.

one good thing you can do is monitor the 12V supply current ( I ) with 10A setting on a cheap DMM . > ideally the load should draw constant power P= E * I! calculate a few settings as the battery voltage changes > like when the battery charger comes on and drops out again. make sure the current reduces as battery voltage rises. post your results.

 

[t-head]

@ Infinia...thanks for the thoughts. I will post measurements when amp is in my hands and built (it is a kit.) Since the amp is running Class A at 2 amps, that is the maximum draw and is constant.

Victor Kung (the supplier of the Elekits in N.A.) decided (with my urgings and his curiousity) to connect his TU8100 directly to a 12v battery. He reports that after 10 hours operation it is still running fine and sounds better than the supplied smps (no surprise). So, thanks for all your replies and thoughts. My intuition was better than my electrical knowledge. You can run an Elekit TU8100 2watt PCL86 amp from a 12v battery directly...

T (richard)

 

[infinia]

"directly" implies no fuse, please > add a inline DC fuse located as close as possible to the battery consistent with your load connections (wire ampacity). ( you can NOT shut off a battery)

Class A at 2 amps, that is the maximum draw and is constant.

under what input voltage range ? I suspect it may act as a resistive load, not the desired constant power.

 

[t-head]

When I have my amp set up, a fuse will definitely be added (all my circuits on the boat are fused). As far as a resistive load, I cannot reply with knowledge. Victor has not told me how he connected the amp to the battery, only that he did and the amp did not 'fry'. I will send you the schematic if you are interested in such ...

 

[infinia]

well we haven't progressed from post 24 so...
my main concern is when you equalize the battery monthly... you could / should disconnect all loads. FWIW inverters usually send alarms past 15V which is my problem.

 

[Chris Daly]

Hi
After the battery yes a fuse,

but then a current regulator
On your shopping list will be

1. a T03 case heatsink,
2 T03 mounting hardware
3 LM338 T03 case type
4.2 x 1watt 1 ohm resistors
5.A 3A fuse and fuseholder
6 Wire

Recipe see attached schematic will be to mount the LM338
www.ti.com/lit/ds/symlink/lm138.pdf
on a suitable heatsink making sure each leg and particularly the main body of the
LM338 is thermally transfer insulated, and each leg cannot in any circumstance
touch ground. ( this is why we use T03 case hardware. Best to test
insulation with a multimeter to ensure isolation.

In electronics steady state current conditions have to be designed into circuits
and this is what the LM338 will do for you, here it is being designed to be
a current regulator of exactly 2.5 amps.

How the current regulator works is the LM338 adjustment leg carries an
output reference of 1.25 volts, to calculate output current then we just divide
this reference by the parallel combination of 2x 1 ohm resistors which is 0.5 ohm
so 1.25/0.5 is 2.5 amps. Your amplifier whilst it may state 2 amps current draw
will invariably have a small margin over this, so designing for 2.5 amps is
sensible.

The LM338 will regardless of the voltage whether it be 13.8v or 12 volt
continue to deliver exactly 2.5 amps.

Cheers / Chris

 

[infinia]

but the constant current source will have a voltage burden associated with that design reducing the voltage for the amp. 12V -1.24 - 2.5V (at 2A ) leaving ~ 8V at the amplifiers input
newer low drop out LDO regulators could help but will still have a couple of volts dropped.

FWIW not all circuitry will appreciate constant current ( bad idea unless you know what the load is exactly)
common stumbling block for lamp or tube filament retro fits > not accounting for drop out V and variable resistance loads

 

[Chris Daly]

but the constant current source will have a voltage burden associated with that design reducing the voltage for the amp. 12V -1.24 - 2.5V (at 2A ) leaving ~ 8V at the amplifiers input
newer low drop out LDO regulators could help but will still have a couple of volts dropped.

FWIW not all circuitry will appreciate constant current ( bad idea unless you know what the load is exactly)

Hi
No the voltage burden is very minimal, so small, it normally is not a factor
when a LM338 is used as a current regulator.

I will take some measurements and report back.

Cheers / Chris

 

[infinia]

Hi
No the voltage burden is very minimal, so small, it normally is not a factor
when a LM338 is used as a current regulator.

I will take some measurements and report back.

Cheers / Chris

its on the data sheet ( drop out voltage chart )

its the minimum voltage Vin-Vo the IC needs for linear operation

 

[Chris Daly]

Yes drop out figures published in datasheet are accurate, so in a 13.8v environment
expect 11.6v @ 2 amp current draw, with a nice level of protection for the amplifier
being used, and at fairly low cost in parts.

Cheers / Chris

 

[t-head]

well we haven't progressed from post 24 so...
my main concern is when you equalize the battery monthly... you could / should disconnect all loads. FWIW inverters usually send alarms past 15V which is my problem.

I have no inverter...all loads are disconnected at panel during the daylight when panel/controller is operational...equalization phase is weekly if battery sits at float and is 14.4v (for AGM battery) and lasts 2 hours. Amp will be used at night after panel stops producing current. I have a DMM and can measure things when amp is operational (at least one month away...

 

[t-head]

'This is a fully isolated 12V to 12V DC/DC converter module, or "half-brick." It has 3000 volts isolation from input to output, has a standard footprint, and a wide voltage input of 10-20vDC. The 12 volts output is at 4166 mA, giving 50 watts.

The short circuit protection is done by an automatic recovery "hiccough mode," meaning that when it detects an overload it tries to come on, then if the fault is still there it will shut off. This happens continually until the fault is corrected.

This is perfect for regulating 12 volts from battery or automobile voltage sources.
These are high quality, brand name USA converters'

In stock Vicor VI-J01-CY 50 Watt DC/DC 12 to 12 volt isolated converter and regulator modules in stock at 1/2 the price from PowerStream.

Something like this should work, no?

 

[infinia]

you don't need that. I checked your amp schematic it has a DC/DC inside, the only other thing is tube filaments are running straight across the DC input.
tube filaments should be OK with +/- 10% variation, I would just add a few diode drops to convert the battery 13.6V down to 12V after the added fuse.