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2A LED Driver for 4-17v range

ACharnley

Member
Hi,

I've not looked at LED drivers before but now need to source one for the above voltage range. 4.5v min input can be at a push.

After hours trawling through Google, Ti's website etc and feeling like I'm missing a trick, it seems most are lower current drivers for LED series whereas the CREE XP-L I'm intending to drive is ~3.2v at up to 3A.

Teach me!

Cheers!

Andrew
 

ronsimpson

Well-Known Member
Most Helpful Member
The TPS61500 only shows "boost" circuits in the data sheet. It will not do what you want. I thing I could modify the circuit to make it work "buck/boost".

I have used parts from diodes.com many times. I know this only 1.5A but it does: vin=4 to 18
Here is an example of boost/buck.
1570838529183.png
ST Micro: LED2001, LED2000, ST1CC40
Linear Technology/Analog Devices: LT3477
Monolithic Power Systems Inc.: MP24833
ON Semiconductor: NCL30060
Diodes Incorporated, AL8843S
TI, TPS92512
Here is some parts that might work. Happy reading.

With Vin min=4.5 you might have troubles with a buck only PWM. There might not be enough head room.
I think you meed to think about buck and boost/buck. Do not use boost.
----edited----
It was pointed out that you might not know what a "boost" PWM is.
examples of what a boost can do: Vin 4.5 to 17 with a Vout of 20V.
example of what a buck can do: Vin 4.5 to 17 with a Vout of 3.0V.
example of what a boost/buck can do: Vin 4.5 to 17 with a Vout of 10V.
 
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ACharnley

Member
Erm yes I'd linked to the wrong one. The actual one was TPS54201. Texas have done a fine job hiding all their high current drivers in the parametric search. This one is 4.5v. I can ensure Vin is 5.5v but as you do say, 4v would be better.

MP24833 is low efficiency @ lower voltage. I really like the spec of MP2410A but it doesn't appear widely available.
 

ACharnley

Member
Could this be a candidate for DIY?

I'm using the PWM of the MCU to control the PWM of a sync buck and in this case the load isn't capacitive and so the minefield of DCM inefficiency considerations for the lower FET don't apply. I'm already reading the current of the LED so it appears to have duplication.

One problem though is based on the 24MHz MCU I'd only have a 240 step resolution at 100KHz, which for a 3A load is going to require a larger inductor.
 

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ronsimpson

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Most Helpful Member
I don't know what the voltage level is for mpptD5. The two transistor buffer looses 0.7 volts. Is there enough signal to turn on a MOSFET?
When the supply drops to 4.5V will there be enough gate drive for the PFET?
Because of the slowness of software, If the supply voltage changes, the duty cycle can not be changed fast enough. (If supply changes from 10V to 15V very fast the current will clime)
1570884497809.png
 

ronsimpson

Well-Known Member
Most Helpful Member
Here is a little 8 pin IC that has the two MOSFETs inside.
Soc. frequency is 850khz so the inductor is small.
If the current gets too high there is two ways the current is under controle. Loop one responds in a couple of nano-seconds and the second in mili-seconds.
If you want to limit the brightness you connect your computer to the "INH" pin. PWM that pin at 100 to 1000hz. I use a small 8 or 6 pin u-computer to do that and it also watches buttons and conditions. Often the micro computer also watch the LED's temperature.
1570885151370.png
I have not used this part, I have used parts like this to make the entire PCB and LED fit inside a light bulb. (flash light bulb)
 

ronsimpson

Well-Known Member
Most Helpful Member
Why I do not like software PWM.
Using the example in this post: Vin=6 volts and Vout=3V (LED voltage). 100khz 50% duty cycle.
The inductor has one end on Vout at 3V.
The inductor is switched from two points. 6V and 0V. 50% the LC filters this to 3V.
When the switch is on the inductor has 3V across it. (6V-3V=3V) and the current ramps up from 2.5A to 3.5A (averate to 3A Iout)
When the switch is off the inductor has -3V across it. The current ramps from 3.5A to 2.5A.
Every thing is stable and happy.
--------
Now the Vin increases to 9V. The software does not have time to respond.
Vin=9V. The charge up voltage on the inductor is 9v-3v=6V.
Current charges up:
2.5A to 4.5A in the first cycle.(the slope is 2x because the voltage is 2x) Then 4.5Aback down to 3.5 of the off time of the first cycle.
In the second cycle: 3.5 goes to 5.5 and back down to 4.5A.
In the third cycle: 4.5 to 6.5 and back to 5.5A.
In the forth cycle: 5.5A to 7.5A and back to 6.5A.
This trend continues until the software changes the duty cycle or something barkes.
-------
If you have a IC controling the LED current:
There is probably a 4.5A over current sense that will open up the switch with in 5nano-second. This will shorten the "on time" and thus the duty cycle so current can not run away.
Next: with in 1mS the error amp will see the current is too high and change the duty cycle so the current is back to 3.0A that you want. And that will fix the current limit condition.
------
If you "PWM the IC to dim the light" this happens at a much slower rate. At 1khz your u-computer can make more levels of brightness.
One of the reasons we often turn on/of/on/offf the IC to dim the light is that white LEDs change color when run at a low current. So we run the LED at 100% current 1/4 the time to get dim. This leaves the LED running at full current and making the right color. Your eye is slow and does not know the light is not there all the time.
 

ACharnley

Member
I don't know what the voltage level is for mpptD5. The two transistor buffer looses 0.7 volts. Is there enough signal to turn on a MOSFET?
When the supply drops to 4.5V will there be enough gate drive for the PFET?
Because of the slowness of software, If the supply voltage changes, the duty cycle can not be changed fast enough. (If supply changes from 10V to 15V very fast the current will clime)
View attachment 121061
I have to assume the PIC32 will be fast enough to catch the supply change, plus there'll be an appropriate amount of capacitance to help. The 4.5v will be ok for the FETs.

I need to find out though whether the current through the LED is linear with duty increase but I expect it is not, that the duty increases voltage linearly not current which for the LED will be logarithmic. This will be a problem as a 100Khz equivalent (using a 10uH inductor) with a 24MHz CPU is only 240 steps. If the window between the LED being dim and full is minimal then this won't be enough.
 

ACharnley

Member
Here is a little 8 pin IC that has the two MOSFETs inside.
Soc. frequency is 850khz so the inductor is small.
If the current gets too high there is two ways the current is under controle. Loop one responds in a couple of nano-seconds and the second in mili-seconds.
If you want to limit the brightness you connect your computer to the "INH" pin. PWM that pin at 100 to 1000hz. I use a small 8 or 6 pin u-computer to do that and it also watches buttons and conditions. Often the micro computer also watch the LED's temperature.
View attachment 121064
I have not used this part, I have used parts like this to make the entire PCB and LED fit inside a light bulb. (flash light bulb)
There is only one loop, it isn't current mode as there's no tap off the end of the inductor. Instead there's a sense resistor and it's using that point, which makes it voltage mode. Normally there is another input which takes a resistor to set the max current regardless of the input PWM but this chip doesn't feature that protection (neither would software PWM unless I stuck a comparator on the output and auto shutdown the PWM - worth considering for sure if the software PWM will work).
 

ACharnley

Member
As I thought the voltage is linear D = Vout/Vin.

Assuming a max input voltage of 17v = 17/240 = 70.8mV

The LED is dim at 2.5v and at full brightness at 3.3v.

Disregarding input voltage sag =

3.3v/2.5v = 0.8v, 800/70.8 = 11 control steps.
 

ACharnley

Member
I better understand what these LED circuits are doing now. The maximum current is set (the 910 ohm resistor here) and then the PWM signal is literally turning on/off a FET on the output. I can't easily replicate the maximum current draw so I will need a off the shelf IC.

Thanks for point out ST1CC40, I hadn't come across that one before and it looks perfect. Can I ask how you found it?
 

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ronsimpson

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Most Helpful Member
The maximum current is set (the 910 ohm resistor here)
No. The current is set by 0.133 ohms and 0.1 volts. The regulator is not regulating voltage (as in 3V) but current. (OK it is regulation voltage as in 0.1 volts on the FB pin) (I did not look at the FB voltage on the TPS54200 but it is usually 0.1 to 0.2V)
1570907314182.png
 

ronsimpson

Well-Known Member
Most Helpful Member
As I thought the voltage is linear D = Vout/Vin.
A LED driver is not a voltage to voltage PWM. It is voltage to current. At about zero current the output voltage is 2.5V, using your numbers. At full current the voltage moves up to 3.3V. (That is very dependent on temperature and what LED you get) Typical voltage = 3.3 but Max might be 4 and min is about 2.8. Just guessing.
There is only one loop, it isn't current mode as there's no tap off the end of the inductor.
I do not understand. These ICs measure the current ramping up in the inductor. That is how they know when to shut off the switch. current mode.
 

ACharnley

Member
Will not cause any measurable current in the LED. Same thing for 140mV.
It will once it gets to 2.5v though, that's what I mean by there being only 11 steps before 3.3v is reached. It's not enough. Reading the TI document I understand this is called analog dimming and a digital CPU doesn't have the granularity to do it this way, hence they use "pwm dimming" where the output is regulated to the full amperage by the buck and then turned on/off rapidly to dim the LED.

A normal current mode buck measures voltage across the inductor but for the LED driver you referred to you can see there's no tap. Instead it's using voltage mode (but the voltage is derived by the current sense resistor). So if the controlling PWM input duty is 100% the current is limited by the voltage on the fb pin, derived by the sense resistor (for which current converts to voltage). In other words the sense resistor limits the maximum current and has to be calculated.
 

ACharnley

Member
No. The current is set by 0.133 ohms and 0.1 volts. The regulator is not regulating voltage (as in 3V) but current. (OK it is regulation voltage as in 0.1 volts on the FB pin) (I did not look at the FB voltage on the TPS54200 but it is usually 0.1 to 0.2V)
View attachment 121073
Yes sorry, the 910 + cap are a filter. We're saying the same thing, but there is only one feedback loop.

Thank you again for pointing out ST1CC40. It only uses a 50mOhm resistor @ 3A so is more efficient than the TI one by some 5% at high load. Lower voltage point as well, a perfect fit!
 

ronsimpson

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Most Helpful Member
A normal current mode buck measures voltage across the inductor but for the LED driver you referred to you can see there's no tap.
Please show me an example.

Here is a typical current mode boost PWM. The MOSFET current is measured by the voltage across the circled resistor. The inductor/MOSFET current is the ramp used to set the point when the MOSFET is turned off.
1570911244757.png

1570911564277.png
 

ronsimpson

Well-Known Member
Most Helpful Member
Maybe we are not communicating.
The first PWMs years ago are "voltage mode". The PWM Comparator does not look at current but looks at the voltage made by the Oscillator. The Osc makes a sawtooth or ramp. The output of the Error Amp and the Osc ramp sets the duty cycle. In the calculations there is no place for the PWM to see current. A over load is hard to see and it takes many cycles for the IC to understand there is trouble.

Current mode is much faster to respond to short circuits and inductor saturation. Because the "ramp" comes from current in the inductor or transistor. If the current ramps up too fast the power transistor is turned off in nS. A current mode PWM sees the current and responds well.

It does not matter if the output is voltage or current.
In current mode the error amp sets the "Ve" level which is the max current for this cycle.
I drew some red lines on the picture below. If the input voltage doubled the current ramps up 2x faster. The Ve point is reached in 1/2 the time and the power transistor is turned off in 1/2 the time. If you change the input voltage greatly the Ve voltage will not need to respond. This makes the current mode PWM very good at responding to Vin changes.
1570912218413.png
 

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crutschow

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Most Helpful Member
If you want to try a roll-your-own circuit without using a specialized controller IC, below is the LTspice simulation of a fairly simple constant-current hysteretic control SMPS using common IC components..
A hysteretic control SMPS has the advantage of being unconditional stable with no feedback compensation required.
The simulation shows the output current does not change (expect for the ripple) with an input step voltage change from 4V to 17V.

Note that, as with most SMPS circuits, it will not work well (or at all) if you build it on a plug-in breadboard.
It should be built on a vector board (preferably with a copper ground plane) due to the high frequency, high current signals in such a circuit.
All signal and ground paths carrying the high currents should be built with as short a connection as possible.

1570911402163.png
 

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