Current Amplification.

[sdmays]

I am designing a traditional buck dc-dc converter to power a one watt LED at approximately 3.1VDC @ 350mA from a 12VDC source. The circuit works fine, but I have to 'hand' regulate the output current. In other words, I adjust a potentiometer up and down to keep IOut close to the desired 350mA. My goal is to amplify the voltage accross a 100mOhm resistor in series with the LED to ground in order to monitor/regulate the current through it and the LED.

To amplify the voltage, I have used an LM339 (as a low frequency op-amp without any capacitors), an LM324, and even a ZTX851 as a rigged common-emitter amplifier (all three amplifier configurations were powered by a single source, i.e. 12VDC to ground). The 339 will amplify the 0 to 35mV signal, but the output has a 2.6 VDC offset(?), and the gain is not linear when it is amplifying the signal. The 324 doesn't really do anything. It has an output of 750mVDC, and that is it.

Does anybody have a SIMPLE circuit that will amplify a 0 to 35mV signal by a factor of 100? This is a personal project, therefore I do not want to spend good money on a rail-to-rail op-amp unless necessary - someone is bound to have used some good old-fashioned ingenuity to design something as simple as this.

One last note, my buck converter is not made from a fancy, dedicated buck converter IC. I don't want to let any cats out of the bag other than to say the operation of the converter is handled by an LM339. I'm saying this to let people know I am asking this because I plan to do everything by hand, instead of allowing someone else at MegaCorp to do my thinking for me (other than to help me with this circuit (ha-ha)).

Thanks!!!

 

[m3]

Hi,

LM339 is a open collector comparator, try use a CA3140.

Regards

M3

 

[nbucska]

yes. You have 12V to play with -- use 100X larger sense R.

<nbucska@pcperipherals.com>

 

[sdmays]

Thanks for replying!

nbucska, it would certainly make it easier if I used a larger value sense resistor, however, I don't choose to dissipate the extra power in the larger value resistor. The maximum power the 100mOhm resistor would dissipate is 12.25mW (plus a few additional mW dissipated by the amplifier) as opposed to 1.225W in a resistor 100 times greater (10 Ohms). I am building a flashlight with the LED as the light source, so I am trying to make the driver as efficient as possible.

 

[sdmays]

Use a high side current shunt monitor such as the Burr-Brown INA138 or the Zetex ZXCT1009. The data sheet for the Zetex part shows the current monitor being used in a buck converter. Both of these parts are very easy to use and, although the desired 12.25mW may not necessarily be feasible, the power dissipation can and will be MUCH lower than using a large sense resistor.

For example: use a 0.6Ohm resistor in series with the load and connect a several kOhm resistor (I actually calculated the exact value a few weeks ago, but I forgot what it was) to the output of the Zetex current monitor and the output should be around 1.23V nominally.

 

[rodar]

On Semiconductor make a chip part # NCP1200 that
incorporates a fixed frequency switch mode supply that
can easily do current control. Just use an inductor in
series with your LED and a fast recovery diode around both
LED and inductor. I have used this chip to make a direct
line powered LED driver using this method.
Rodar

 

[lcsjk]

Not sure if you are using the 339 for an amp or a switching comparator. However, the 339 is not internally compensated for linear operation so you need to put a capacitor in the feedback loop (output to (-) input. Remember that you are not current regulating but are regulating the voltage across the 0.1 ohm resistor. (By the way, why not use a 1.0 ohm resistor which will only dissipate 0.122 watts, easily handled by a .25 watt resistor.) I suggest putting a cap across the .1 resistor and the feedback cap. Without more topology, I can't help much.

 

[logbook]

It might be more “interesting” to use an open-loop system. Consider this: if you just chop the 12V at a defined duty cycle through an inductor into the LED you will get the required efficiency and power, without having to go to the intermediate step of generating a stable voltage. Let the current in the LED be a pulsing ramp at some frequency like 100kHz and you will not burn out the LED or see the flicker. This should be more efficient, cheaper and simpler than your present scheme.

 

[nbucska]

Hi logbook:
The 12V is 12V only nominally -- when you charge it, it can
be significantly higher, in discharged condition lower.

I would use some current feedback based control.


<nbucska@pcperipherals.com>

 

[TheSpeakerGuy]

There's a problem with adding a gain element within the control loop. You would be adding additional gain and phase to what was a compensated system.

If you're not concerned with efficiency, put the LED in series as the upper feedback element, and a resistor from the LED to ground. The current would then be VFB / R, where VFB is usually 1.25V.

There's many design ideas in the past couple years addressing this. I did this back in 1999

http://www.maxim-ic.com/appnotes.cfm/appnote_number/940/ln/en

 

[sdmays]

TheSpeakerGuy: I am trying to increase my efficiency as much as possible, therefore, I will stick with what I have written above. By using a 0.3Ohm sense resistor and approximately a 1.21kOhm output resistor fed by a Zetex ZXCT1009, I should be able to control the current through the LED. I have chosen the National Semiconductor LM2676ADJ, which requires 1.21VDC for feedback, to control the switching, etc.

I was originally trying to use the LM339 to handle the switching. It is very easy to create a PWM using the LM339. All I would have to do is control the duty cycle. I still plan to one day use the LM339 to completely control a buck converter or a boost converter.

 

[jimkirk]

sdmays,

Did you configure the 324 opamp for non-inverting gain? This should work. Also, the input common mode includes zero, but the output range does not, that could explain the 750 mV you're seeing.

An alternate method (not sure, as I don't know details of your topology) is to use a section of your 339 comparator with the voltage across the current sense resistor to one input, a voltage reference to the other input, to directly control the peak current through the LED. As long as the overall sawtooth wave shape doesn't change too much with supply voltage, this may be acceptable. You'll be dealing with small voltages and offset erors must be considered (the commercial grade 339 has up to 5 mV input offset voltage, the 339A version has only 2 mV), but with some slight tweaks, once you've set a peak current, it should be pretty stable over temperature/voltage supply.