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Led driver of constant current - is it really constant ?

kellogs

Jan 7, 2014
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I don't know what the application for your circuit is (what is it being used for) but maybe look at circuits used in solar garden lights, like these (there are 2 different circuits described on this site): http://www.talkingelectronics.com/projects/SolarLight/SolarLight.html

These circuits include a solar charging circuit for a 1.2V cell, an oscillator that boosts the voltage to drive a white LED, and a circuit to cut off the LED during daylight. If you're looking for continuous operation (day or night) the solar cell and photocell circuits could be taken out.

These solar light circuits don't have a current limiting resistor on the LED since it's driven with pulses, making it more efficient, and it's a low power circuit running from a single AA cell.

That's quite an old school piece of work! Unfortunately it won't easily fit into my head. For the second circuit:

The inductor produces a voltage (in the form of a spike) that is higher than 2.1v to illuminate the high-bright yellow LED - in fact it is much higher but the LED converts this energy to light.
You cannot measure this voltage with a multimeter as the voltage is produced in the form of spikes.
If you remove the LED, a CRO will show the spikes are higher than 40v.

Something tells me that if I replaced that battery with an alkaline AA cell, the LED will light up from brighter to dimmer as the battery depletes (1.6 -> 0.8 V). Is this the case ? What sort of dimming percentage should I be looking at for the whole battery life ?
 

Sunnysky

Jul 15, 2016
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Constant Current source using 200nA diode @ 376mV from Vbe=Vce driving an open-collector Comparator with an 18 Ohm Current sensor for 20mA = 376mV/18Ω . Unregulated R limited LED compared on the right runs from 28 to 12 mA while Vbat ranges from 3.2 to 2.2V... My Simulation here http://tinyurl.com/y683znyl with Vbat slider on the right.

upload_2020-8-7_22-14-21.png

LED current drops to 10mA when this circuit Vbat drops to 2V for a typical high bright Red or Yellow LED.
 

kellogs

Jan 7, 2014
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Thanks a lot Tony!
Relearned a few things today (if I ever learned them from school, that is :) )... just one thing I can not grasp on the circuit - what does the 470 pF do ? In my simulation, removing it doesn't cause much difference.
And one more - can I replace the first transistor with a diode of any sort and tweak the resistors ?
 
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Harald Kapp

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driving an open-collector Comparator
@kellogs : Note that "open collector here refers to the assembly of opamp, 470 Ω base resistor and transistor. Using a true open collector opamp (or comparator) IC will not work as this could not deliver the base drive for the transistor.

can I replace the first transistor with a diode of any sort and tweak the resistors ?
The BJT is used in a configuration called "diode connected transistor". This setup has a much steeper I-V characteristic than a diode. Depending on your stability requirements a diode may work, too.

what does the 470 pF do ?
The capacitor is probably there to suppress noise. The circuit will possibly work without the 470 pF, but be a bit more noisy.
But: as the capcitor is in th
e feedback loop of the comparator it will influence the frequency dependent loop gain and therefore may be required to suppress oscillations. A thorough analysis would be in order taking into account the characteristics of the comparator to get a definitive answer. Or build the circuit on a breadboard and change the capacitor's value to see how the circuit behaves.
 

Sunnysky

Jul 15, 2016
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My faux pas for calling it an OC comparator, rather it is any Op Amp with Q2 as a common emitter current sensed (voltage controlled) current sink. Rb/Re=30~50 for ~any Q2. Any silicon diode will work instead of Q1 , C to reduce startup current of diode voltage source. Otherwise no change in steady state.

Q1 or diode has a NTC thermal coefficient. -4mV/‘C (?)
 
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