Hi,

I am a noob and I did search the forum but could not find a hit on my project or questions but got some very close answers (I did only look at 4 pages of search results though).

I am trying to replace a 9v 2 amp DC out (110VAC in) power supply with a 12v 7.5 AH sealed lead acid battery. This is to solve the problem of a digital display picture frame device that must sit on a coffee table in the center of the room with no VAC outlets under the table. I built a LM388T linear voltage regulator circuit and it seems to work fine until I tested my regulated 9v output voltage with loads. Had significant drop in 9V output voltage under small loads drawing < 2 amps.

I realized that the 12v battery even when fully charged will drop voltage below 12v when a large current draw is applied. Therefore if V(in) (battery voltage) is not stable and goes down under load then V(out) will go down and be below my desired 9v. Did not test the battery and linear voltage regulator on the digital picture frame display device for fear of harming it.

(1) Are there any incorrect statements or important feedbacks needed for what has been said so far?

(2) Do I really need to build a switching voltage regulator (like at http://www.dimensionengineering.com/DE-SWADJ3.htm) and my circuit becomes just this device and input/output leads? I read these switching regulators are what are needed in battery powered robots were servo motors cause big current draws and therefore voltage drops in the battery that the digital circuits don't like to have - is this true?

(3) If a switching regulator is the way to go is the DE-SWADJ3 the best device? Seems a little pricey at $25/ea but that was about what the entire LM388T circuit cost me when all was said and done.

(4) Should I go with a DE-SWADJ3 regulator over a LM388T regulator just from a power conversion (better efficiency) point of view?

All feedback is welcome.

tia


moonie1

 

A 12v lead-acid battery (actually, around 12.8v-12.9v @77°F, 25°C) will be considered fully discharged at around 11.4v. If you discharge a lead-acid, SLA or AGM battery that deeply, it will have a short lifespan indeed.

With such a small margin between the battery voltage and the desired voltage, you don't have much to work with. You really need a DC-DC converter that will shut off when the battery is discharged to the desired threshold. This isn't a simple task.

 

Even a fully charged SLA 12v under a load will drop a volt or so to <11volts when under a significant load. I have put a fully charged new SLA 12v on a volt meter added a load and have seen the voltage drop.

That is why I surmised and asked for conformation in my Q#1 above is this what is causing my battery powered linear regulated (LM388T) circuit to drop well below 9v when I put a load on it?

 

Even a fully charged SLA 12v under a load will drop a volt or so to <11volts when under a significant load. I have put a fully charged new SLA 12v on a volt meter added a load and have seen the voltage drop.

Small SLA's don't handle heavy loads very well. That's a fact of life.

That is why I surmised and asked for conformation in my Q#1 above is this what is causing my battery powered linear regulated (LM388T) circuit to drop well below 9v when I put a load on it?

If the battery itself doesn't handle loads well, how do you think that adding a regulator will improve that situation?

A buck/boost regulator might, but that will only spell an even earlier demise of your battery.

You need to re-assess your power requirements vs the battery capacity.

 

Battery capacity it seems not to be an issue. I have access to allot of batteries and even a SLA 12v 12AH did not do the trick. Do I need an 8D on the coffee table to make this little digital display work (joke intended).

 

A switching regulator will give you the best efficiency but for ease of use I would go for a low dropout linear regulator such as a LM1084 or LM1085. Both have a dropout voltage of about 1.3 volt so they should work down to a battery voltage of approx 10.7 volt (well and truly flat).

 

Let me see if I can understand (I know I don't but I am trying). I found this definition.

Drop-out voltage is the minimum voltage across the regulator that's required to maintain the output voltage at the correct level.

So with a "normal" operation a ~12V battery (Vin) and ~9V device needing power (Vout) I should do well with any linear regulator that has a drop-out voltage <3V.

Maybe my circuit I built with the LM338 is wrong? I can tune the potentiometer so my output voltage is exactly 9V. However when a load is applied I get <9V and I see that the battery has also dipped below 12V (~11.2V) which is normal. Should these linear regulators keep a constant Vout even if Vin changes and you don't vary the potentiometer at all? If so what is the parameter called for a linear regulator that lets you give a Vin +/- X% and still get a stable unchanged Vout?

 

As long as you maintain the drop-out voltage (or greater) across the regulator it will keep the output constant. If the voltage across the regulator falls below the dropout voltage you loose regulation and your output voltage will drop.

The fact that your battery dips down to 11.2V under load would indicate that it is either flat or has gone high resistance (a common fault with sealed lead/acid batteries). Either way a input 11.2V is too low to let a LM338 produce a regulated 9V output. This is why I suggested a low dropout regulator instead of the LM338.

As long as you follow the design equations in the data sheet there is not much that can go wrong but it might be an idea to attach a copy of your circuit so we can check it out.

 

Thanks so much, I think I understand now. As long as Vin-Vout > drop-out voltage of the regulator I can maintain Vout for any current draw < regulators max current.

If I put two 12v SLAs in series and get 24v then I should be way above the ~2-3v drop-out voltage for the LM338T. This way I should maintain my 9v Vout. This is a way to "test" my circuit.

The circuit came off the internet as a tried and true linear voltage regulation circuit. I may have messed up the connections though and the cap values maybe in question too. However, I do get a nice change in Vout when I turn the potentiometer in the circuit as promised. As I increase the potentiometer's resistence Vout goes up.

 

If you put two SLA's in series and try to use a linear regulator to get the output down to 9v, the linear regulator will be dissipating a great deal of power.

To keep things simple, let's say that the batteries can output 24v @ 1A.
You have a load that will sink 1A @ 9v.
Power dissipated in the load will be E x I (Voltage x Current) = 9v x 1a = 9 Watts.
The linear regulator will have to dissipate the remaining power; (24v-9v) x 1a = 15v x 1a = 15 Watts. This is very wasteful, as 62.5% of the power consumption will be in the regulator itself.

 

Yes the 24v p/s was just to test the circuit and see if the 9v output is maintained by the circuit when I put it under a load.

Also are you saying the switched mode power supply, DE-SWADJ3, only has a 80% efficiency and other SMPSs can have efficiency's ~90%? If so what is a good price SMPS with 90% or better efficiency that will regulate down to 9v a 12v battery and do so with 2A at the 9v?

 

Well it turned out that the linear voltage regulator circuit I made with the LM388T must have had a build mistake or one of the components was bad or mis-spec'ed (capacitors?). Never could get it to maintain a Vout when I put a big current through it even though the LM388T was spec'ed a 3A max. I had Vin-Vout > Vdrop during the test and Vtest still dropped to zero under large loads. Anyone can gues what went wrong ??

<snip>

Also the device is idiot proof as you just need to connect Vin, Vout and Vground. Not cheap at $25 though but very efficient for 12Vin and 9Vout at ~95%.

Thank you everyone for all your input and help. Much appreciated.