Yes, I realize that the 7812 won't work, but are there any other relative simple devices/circuits which will?

Not really. There are LDO (low drop out) regulators that can narrow the gap between in and out, but not give more than you put in.

 

I should've asked that question before.

You didn't ask for voltage regulation; you asked for current regulation. Very different things.


The 78xx/79xx series of regulators are ancient. You won't get decent response or voltage drop out of them.

 

I don't really *need* voltage regulation because my device has been proven to work across the ranges of 10v - 12.6v. I was just thinking that it would have been nice to give it a steady voltage if that wasn't too hard.

I should've asked that question before.

?????

 

I have been really ambiguous about about what I am powering here. I thought that I might explain it to you guys.

The system in the third picture is called the Headplay Personal Cinema System. The user puts the goggles on his face and plugs the video signal into the black box called the "liberator." The liberator processes the video signal and sends it to the goggles. In the goggles, the video is displayed on small LCD screens and magnified via a set of lenses, giving the user the perception that he is looking at an 80" screen.

In the RC airplane community, there is a discipline called FPV (first person view). This is where the pilot pilots the aircraft through video via a RF downlink to a camera on the airplane. Because this is all done outside, many pilots prefer to use these video goggles over LCD screens because of their portability and usability in the sun.

This portability requires an external battery instead of the AC adapter. One of the most commonly used batteries to power the electric motors on RC aircraft is a 11.1V LiPo (12.6v - 9v). Since many pilots have many of these laying around, they just wire up one to power the video goggles.

However, some people are having problems with blowing the fuses (labeled N) on their goggles (see second picture). This fuse won't blow while goggles are running but will blow during power on (or power off?). This what I am hoping to prevent.

First picture: inside liberator
Second picture: DC plug and fuses in circuit
Third picture: Headplay goggles (liberator on left, goggles on left)

 

Thanks for coming clean! Feel better now?

What power supply do the goggles use normally?

 

Thanks for coming clean! Feel better now?

hahahaha I just figured that was extraneous information and that it would only confuse people if I gave it at the very beginning.

What power supply do the goggles use normally?

AC power adapter rated for 12V - 1.25A

 

Can someone explain to me why I am getting this small current spike at the front?

It's hard to see in the picture, but the current spikes to ~630mA at first and then falls to about 570mA right away.

With my LM358 model (LM358 is the dual version of LM324), I got a 1.2A spike at startup. I fixed it (in the simulation ) by adding a couple of RC networks. See below.
I have included the .ASC file. Below is the LM358 model I used.

	

		
			

		
		
	



	
	
		
		
	


	

		
			

		
		
	



	
	
		
		
	


	

		
			

		
		
	

*////////////////////////////////////////////////////////////////////
* For ordering or technical information on these models, contact:
* National Semiconductor's Customer Response Center
*                 7:00 A.M.--7:00 P.M.  U.S. Central Time
*                                (800) 272-9959
* For Applications support, contact the Internet address:
*  amps-apps@galaxy.nsc.com

*//////////////////////////////////////////////////////////
*LM358 DUAL OPERATIONAL AMPLIFIER MACRO-MODEL
*//////////////////////////////////////////////////////////
*
* connections:      non-inverting input
*                   |   inverting input
*                   |   |   positive power supply
*                   |   |   |   negative power supply
*                   |   |   |   |   output
*                   |   |   |   |   |
*                   |   |   |   |   |
.SUBCKT LM358    1   2  99  50  28
*
*Features:
*Eliminates need for dual supplies
*Large DC voltage gain =             100dB
*High bandwidth =                     1MHz
*Low input offset voltage =            2mV
*Wide supply range =       +-1.5V to +-16V
*
*NOTE: Model is for single device only and simulated
*      supply current is 1/2 of total device current.
*      Output crossover distortion with dual supplies
*      is not modeled.
*
****************INPUT STAGE**************
*
IOS 2 1 5N
*^Input offset current
R1 1 3 500K
R2 3 2 500K
I1 99 4 100U
R3 5 50 517
R4 6 50 517
Q1 5 2 4 QX
Q2 6 7 4 QX
*Fp2=1.2 MHz
C4 5 6 128.27P
*
***********COMMON MODE EFFECT***********
*
I2 99 50 75U
*^Quiescent supply current
EOS 7 1 POLY(1) 16 49 2E-3 1
*Input offset voltage.^
R8 99 49 60K
R9 49 50 60K
*
*********OUTPUT VOLTAGE LIMITING********
V2 99 8 1.63
D1 9 8 DX
D2 10 9 DX
V3 10 50 .635
*
**************SECOND STAGE**************
*
EH 99 98 99 49 1
G1 98 9 POLY(1) 5 6 0 9.8772E-4 0 .3459
*Fp1=7.86 Hz
R5 98 9 101.2433MEG
C3 98 9 200P
*
***************POLE STAGE***************
*
*Fp=2 MHz
G3 98 15 9 49 1E-6
R12 98 15 1MEG
C5 98 15 7.9577E-14
*
*********COMMON-MODE ZERO STAGE*********
*
*Fpcm=10 KHz
G4 98 16 3 49 5.6234E-8               
L2 98 17 15.9M
R13 17 16 1K
*
**************OUTPUT STAGE**************
*
F6 50 99 POLY(1) V6 300U 1
E1 99 23 99 15 1
R16 24 23 17.5
D5 26 24 DX
V6 26 22 .63V
R17 23 25 17.5
D6 25 27 DX
V7 22 27 .63V
V5 22 21 0.27V
D4 21 15 DX
V4 20 22 0.27V
D3 15 20 DX
L3 22 28 500P
RL3 22 28 100K
*
***************MODELS USED**************
*
.MODEL DX D(IS=1E-15)
.MODEL QX PNP(BF=1.111E3)
*
.ENDS
*$

 

AC power adapter rated for 12V - 1.25A

Huh, I guess that's just puny enough to act as a natural current spike limiter, compared to a battery.

 

The easiest way to avoid a spike on start-up is to use a 0.1uF (100nF) cap from Ref to ground. This will help a great deal with avoiding the initial overshoot.

Ron_H,
I used values in my schematic that would be pretty easy to obtain from any Radio Shack, because I was under the impression that our O.P. wanted a reasonably simple circuit that they could obtain parts for locally and immediately. RS doesn't carry a 0.1 Ohm resistor, nor LM358's either; the only suitable opamp they carry would be the LM324.

However, if they have access to an LM358 and an 0.1 Ohm resistor, that would work just fine.

Now that I'm looking on Radio Shacks' site; I see I was in error about the 0.8 Ohm resistor; it's an 8 Ohm resistor. However, they do carry a 0.47 Ohm resistor that would do the trick; with corresponding adustments to the Ref voltage divider.

 

The easiest way to avoid a spike on start-up is to use a 0.1uF (100nF) cap from Ref to ground. This will help a great deal with avoiding the initial overshoot.

Ron_H,
I used values in my schematic that would be pretty easy to obtain from any Radio Shack, because I was under the impression that our O.P. wanted a reasonably simple circuit that they could obtain parts for locally and immediately. RS doesn't carry a 0.1 Ohm resistor, nor LM358's either; the only suitable opamp they carry would be the LM324.

However, if they have access to an LM358 and an 0.1 Ohm resistor, that would work just fine.

Now that I'm looking on Radio Shacks' site; I see I was in error about the 0.8 Ohm resistor; it's an 8 Ohm resistor. However, they do carry a 0.47 Ohm resistor that would do the trick; with corresponding adustments to the Ref voltage divider.

I have a local radio-supply shop which I normally use instead of radio shack. Their parts are highly superior in quality. There is a good chance that they will carry many of those parts.

Quick question about the current-sense resistor. If I used a standard carbon-film resistor, would it be able to dissipate enough power to avoid overheating?

 

You could possibly use a carbon film resistor, if you can find one with a suitable wattage rating. I wouldn't recommend the old cylindrical "brown body" resistors; as their values change significantly over time. Metal film is more stable over temperature. You could even use wire wound, which will have some inductance, so not quite as good as metal film.

 

Here's something even more simple. See the attached.

Rlimit is increased to 1 Ohm. You'll take just a small efficiency hit because of this, as with a 250mA load, 62.5mW will be dissipated in Rlimit, while 2,937.5mW will be dissipated in the load. That's still better than 97.8% efficient.

R1 supplies current to pull the gate of M1 high, tending to turn the MOSFET on.
As current begins to flow through the MOSFET, it also flows through Rlimit; as the current increases past 0.63v, Q1's base starts conducting current, which causes the collector to sink current, pulling the gate of M1 back down, which reduces the current flow through the MOSFET, hence through Rlimit.

When C1 becomes fully charged, the voltage across Rlimit drops enough to turn Q1 completely off, and the gate charges to the battery voltage less the drop across Rlimit.

 

Wow guys, I am really impressed with the help that I am getting here! I thought that this discussion was wrapping to a close, and then discover that SgtWookie has been working to simplify the design.

My dad is an electrical engineer, and although he isn't a power supply guy (he programs FPGAs), he has a well rounded understanding of many other areas and has been helping me also. His comment when I showed him these different circuits was this:

The device which you are powering is not going to be purely resistive. There are many other things to factor in including capacitance and inductance. If the current draw is constantly changing (could be caused by a clock) and the control circuit isn't fast enough to keep up with the changes, it will become unstable and blow up into the rail on either side.

 

Here's something even more simple. See the attached.

Rlimit is increased to 1 Ohm. You'll take just a small efficiency hit because of this, as with a 250mA load, 62.5mW will be dissipated in Rlimit, while 2,937.5mW will be dissipated in the load. That's still better than 97.8% efficient.

R1 supplies current to pull the gate of M1 high, tending to turn the MOSFET on.
As current begins to flow through the MOSFET, it also flows through Rlimit; as the current increases past 0.63v, Q1's base starts conducting current, which causes the collector to sink current, pulling the gate of M1 back down, which reduces the current flow through the MOSFET, hence through Rlimit.

When C1 becomes fully charged, the voltage across Rlimit drops enough to turn Q1 completely off, and the gate charges to the battery voltage less the drop across Rlimit.

lol, isn't this pretty much the same thing that I had suggested in the first post except that the "current limiting" transistor has been replaced by a MOSFET. I guess the load is in a different place also to prevent voltage drop...

 

Here's something even more simple. See the attached.

Rlimit is increased to 1 Ohm. You'll take just a small efficiency hit because of this, as with a 250mA load, 62.5mW will be dissipated in Rlimit, while 2,937.5mW will be dissipated in the load. That's still better than 97.8% efficient.

R1 supplies current to pull the gate of M1 high, tending to turn the MOSFET on.
As current begins to flow through the MOSFET, it also flows through Rlimit; as the current increases past 0.63v, Q1's base starts conducting current, which causes the collector to sink current, pulling the gate of M1 back down, which reduces the current flow through the MOSFET, hence through Rlimit.

When C1 becomes fully charged, the voltage across Rlimit drops enough to turn Q1 completely off, and the gate charges to the battery voltage less the drop across Rlimit.

I was under the mistaken impression that 0.5V was more saturation voltage than he could tolerate. Looking back to post #23, I see that the voltage can apparently be as low as 9V, which means that the tolerated saturation voltage can be as high as 1 Volt. Good catch, Wookie!

BTW, when did the load current change to 250mA? I thought it was about 450mA.

 

I was under the mistaken impression that 0.5V was more saturation voltage than he could tolerate. Looking back to post #23, I see that the voltage can apparently be as low as 9V, which means that the tolerated saturation voltage can be as high as 1 Volt. Good catch, Wookie!

BTW, when did the load current change to 250mA? I thought it was about 450mA.

Here is the deal with the voltage drop. I can have a 0.5V drop, but since my device needs 10V minimum, my battery would only work at 10.5V or higher. This is OK, but is really not preferable since I want to be able to use the full capacity of the battery.

 

Here is the deal with the voltage drop. I can have a 0.5V drop, but since my device needs 10V minimum, my battery would only work at 10.5V or higher. This is OK, but is really not preferable since I want to be able to use the full capacity of the battery.

That is why I posted the circuit with the 0.1Ω sense resistor. It has a saturation voltage of (Rsense+Rds)*Iload ≈ 50 millivolts (with 450mA load).
BTW, what is your steady-state load current? Is it still ≈450mA?

 

That is why I posted the circuit with the 0.1Ω sense resistor. It has a saturation voltage of (Rsense+Rds)*Iload ≈ 50 millivolts (with 450mA load).
BTW, what is your steady-state load current? Is it still ≈450mA?

yes sir, it's still 450mA.

 

Finished!

Sorry, the soldering job was pretty hacked. Next time I will get the proto-boards with the little metal circles around the through-holes.

Here are some pictures. My dad thinks that I should check it out with a dummy load first. He's going to show me how to use an oscilloscope at work, so we can make sure that the control loop is stable.

The MOSFET is mounted funny so that I can attach it to a heatsink.

 

Which version did you build?
You shouldn't need a heatsink unless your load turns out to be more than about 550mA.
Testing with a dummy load is a good idea, just in case you made a wiring error.