Using an Op-Amp to drive multiple LED's in series.

[Falcon69]

So, here's my question, and I tried a brief search, but everything I found only showed one LED on the Op-Amp.

I have 5v in my circuit.  I'd like to use an op_amp to boost the voltage to around 24 volts to run 6 or 7 LED's in series.  This would lower my power consumption of the circuit from 140mA to just 20mA, plus whatever the op-amp eats up. I have multiple of these circuits on one PCB, so lower the power consumption, would also lower the footprint size of the voltage regulator, as I would not need to get such a powerful one. Right now the one I have puts out 3A, I'd like to get that down to under 1A.

From the data sheet (this is probable not the one I will use, just the first sheet I pulled up)

http://www.ti.com/lit/ds/symlink/lm324a.pdf

It looks like it accepts a voltage input of 3v to 32v, and can put out up to 30mA on the output.

I also have another part of the circuit which requires 14 LED's, so can the output voltage be brought up to around 50volts? (3.2 vF x 14 = 48.8V)

I think the correct term is, Non-Inverting DC Amplifier, that I will be using it as?

Is there anything wrong with doing this?  I ran an online simulator and it seems it works, though, the simulator is not very good.

Thanks in Advance for the advice.

[cowana]

Are you aware that the output of the opamp can never be greater than the supply voltage? While it's easy to set up for any gain you want, the output will always be limited - for the LM324 the maximum output is Vcc - 1.5v.

If you want to power a series string of LEDs with a voltage higher than your supply rail, you'll need a switching boost converter.

[Falcon69]

nope, didn't know that. It won't work then. thank you.

[george.b]

This would lower my power consumption of the circuit from 140mA to just 20mA

Hmm, nope, that would lower your current consumption. Your power consumption (watts, not amps) would remain the same.
Even with a boost converter, your input power would be that of the load, plus conversion losses.

[rstofer]

Look around for an LED driver chip.  It will be organized as a constant current device and you can probably put two LEDs in series and let the driver control the current instead of some very low value resistor.

These devices look a lot like a standard transistor.

http://www.mouser.com/ds/2/268/cl25-587917.pdf

That particular device is going to provide 25 mA of current and that may be a little high.  There are others.

[dacman]

Use the op amp to turn a power MOSFET on and off.

[Falcon69]

You all missing the point of what this thread was for.

cowana....i think you are mistaking an op-amp for a voltage regulator?  an op amp takes a voltage, and multiplies it, based on the resistors selected that are connected to it. For instance, a 1k and 9k resistor, i think, puts a 24volt on the op amps output from a 5v supply to the output.

Now a voltage regulator, yes, some, even most from my understanding, does have a voltage drop across them of around 1.25 volts, some even higher, and they can't output any higher than than the supplie voltage to it.

rstofer.  That's all great and all, IF i had more than 5volts to drive 2 or more LED's in series. I don't, I only have 5v to work with.

dacman.....Why? That's just added components, and I still have the problem with only having 5v to drive multiple LED's in series (which can't happen, maybe RED color I can drive two, but that's it).

I did a search the other day, someone was using a op amp with 5 volt input and controlling 4 white led's that were motion sensors, they turned on and off as he waived his hand. His schematic showed the 4 white led's connected directly to the output of the op amp. But, I can't seem to find the video now or the schematic.

Guess I'll try and see if I have one and see if it works or not.

[Cubdriver]

What is the rail voltage (voltage supplying the op amp's power terminals)?  This is important, as it's a limiting factor - the op amp cannot put out a voltage that is higher than its supply voltage.

-Pat

[Falcon69]

oh, i understand. okay. ya, that would also be 5v as well, as that is the only voltage i have to work with.

I'll have to figure something else out.

[ebclr]

All you need is this



https://www.aliexpress.com/item/0-24V-Top-Mosfet-Button-IRF520-MOS-Driver-Module-For-Arduino-MCU-ARM-Raspberry-pi/32722409551.html?spm=2114.01010208.3.1.ZlOW59&ws_ab_test=searchweb0_0,searchweb201602_3_10056_10065_10055_10068_10054_10069_10059_10073_10017_10070_10060_10061_10052_10062_10053_10050_10051,searchweb201603_6&btsid=4935fb68-a4de-444b-bad2-b68eeafe251c

[Audioguru]

You all missing the point of what this thread was for.

an op amp takes a voltage, and multiplies it, based on the resistors selected that are connected to it. For instance, a 1k and 9k resistor, i think, puts a 24volt on the op amps output from a 5v supply to the output.

Absolutely NOT!
The output voltage of an opamp is always LESS than its power supply voltage. The datasheet for the LM324 quad opamp shows that with a +5V supply its maximum output voltage is +3.8V at low output currents and is +3V at 20mA of output current but some can produce an output current no higher than 10mA.

EXAMPLE: +5V power supply for an LM324 quad opamp, a non-inverting amplifier with a 9k negative feedback resistor and a 1k resistor from the inverting input to ground (a gain of 10 times). With a signal input of +0.3V then the output is +3.0V. With a signal input of +0.38V the maximum output voltage with no load current is +3.8V. With a signal input higher than +0.38V the output will be +3.8V and no higher.

If you have a 0V and +5V signal that needs to turn many LEDs in series on and off then the signal feeds a series resistor feeding the base of a transistor to turn the transistor on and off. The collector output of the transistor connects to the series LEDs in series with a current-limiting resistor from a suitable high voltage power supply. 

[IanB]

I have 5v in my circuit.  I'd like to use an op_amp to boost the voltage to around 24 volts to run 6 or 7 LED's in series.  This would lower my power consumption of the circuit from 140mA to just 20mA...

It is not clear if you understand the conceptual problem with this statement. If you have 6 LEDs each of which is running at 20 mA, then they will consume a certain amount of power. Whether you put them in series, or in parallel, or in some other arrangement, they will always consume the same amount of power and will always draw at least the same current from the 5 V supply. This can never be altered.

One thing that will vary will be the power losses in the circuit. The more complex you make the arrangements to drive the LEDs the greater will be the potential losses, and so even more power will be drained from the 5 V supply.

[Falcon69]

i understand now. I didn't realize that there was a separate supply rail for the opamp. I thought it was a positive and negative input and a single output with the new voltage..    I got it now.

[george.b]

i understand now. I didn't realize that there was a separate supply rail for the opamp. I thought it was a positive and negative input and a single output with the new voltage..    I got it now.

Yup. An opamp can indeed multiply/amplify an input signal by a factor set by external resistors (if you want to understand how, you might want to brush up on Kirchhoff's laws, then it's fairly easy to understand), but the output is always bounded by the supply rails. Some opamps - called "rail-to-rail" - can get their outputs pretty close to the rails, but a typical jellybean part will normally go a couple of volts short of the supply.

Now that's out of the way - if you still want to drive your LEDs in series, you need a boost converter. If you can choose, it seems to me it'd be more painless to drive them in parallel instead.

[dacman]

To multiply the voltage output of the op amp (to higher than the supply voltage without a transformer), it could be turned into an oscillator and connected to a voltage multiplier, which is a series of capacitors and diodes.

[IanB]

i understand now. I didn't realize that there was a separate supply rail for the opamp. I thought it was a positive and negative input and a single output with the new voltage..    I got it now.

It's not because of the op amp, it's because of the power balance. If you have 6 LEDs with a forward voltage of 3 V and a current consumption of 20 mA, then the total power consumption is 6 x 3 V x 20 mA = 360 mW.

Now if you want to obtain 360 mW from a 5 V rail, your current draw will be 360 mW / 5 V = 72 mA.

It doesn't matter whether you use op amps, boost converters, motor-generators, or some other exotic device, this equation will always hold. Even if there was a clever op amp that could produce a higher output voltage than the input voltage, it still would have to obey this equation.

Put another way:

This would lower my power consumption

This statement is physically impossible. There is no circuit that can consume less power than the things it is powering. If ever you find yourself thinking that you have to stop and reconsider.

[JPortici]

What i do in these situations (logic level input, higher voltage/current output) is use a comparator, or an equivalent arrangement.

LM339: quad comparator with open collector output.

It is a comparator so it just needs that V+ > V- to put the output in an ACTIVE STATE.
 - well there is some hysteresis and uncertainity... and switching speed also depends on (V+ - V-) so the higher the faster, or something like that.
 - This means that you can easily use logic levels even at the max voltage supply of 32V

I used the words ACTIVE STATE because in this comparator the output is an open collector type: when V+ > V- a current is applied to the base of a BJT, the output pin is the collector of this BJT. so when it's active it can sink current from the load.

A very basic schematic would be to use the 24V Line as the power supply for the comparator.
The (+) input connected to the logic level thing you use to drive the LEDs through a resistor
The (-) input connected to a bias network, so that there is an appropiate voltage (for example Vlogic/2)
The load, a current limiting resistor and the LEDs in series, connected between 24V and the Comparator output.

Should be able to dissipate the power during switching with no problem, i am doing this exact thing with four led strips and a quad comparator

otherwise there is the equivalent arrangement: comparator/opamp + power bjt/mosfet with suitable VCE/VDS Max and power rating, though at higher power things get tricky