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Car power to constant current 3.6A 50v boost converter

CommanderLake

Oct 2, 2012
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I would like to power one of these: http://uk.farnell.com/jsp/search/productdetail.jsp?sku=2816596&CMP=i-bf9f-00001000
from a car current regulated at up to 3.6A 50v, I've spent days searching for the most efficient solution but I cant find a boost converter controller in an easy to use package such as a PDIP with which I can adjust the current on the fly.
The LED does in fact consume 3.6A at 50v.
What would be the best solution other than using an inverter and my extremely heavy and inefficient linear bench supply?
 

Harald Kapp

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Look for "LED boost driver" chips. Available from many manufacturers.
 

Harald Kapp

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See e.g. this selection table.
This table doesn't accomodate "single" LEDs with VF=45 V. Instead select 1 string of 16 LEDs with an LED voltage between 2.5 V and 3.2 V (totals to 40 V ... 51 V).
 

Harald Kapp

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You're unlikely to find a PDIP or similarly sized case for this kind of driver. These are rather modern chips employing modern housings to minimize size and parasitic losses.
 

Harald Kapp

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The 400 mA are probably for the application circuit with the 0.47 Ω sense resistor in figure 1.
Current setting on page 10 explains how to chose a sense resistor for other currents.
 

Harald Kapp

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is there an easier way to design the circuit?
I don't know of any. Designing a switch mode conveter involves a few steps of calculation and some design decisions. Unless you buy an off the shelf converter/driver for your application, I'm afraid you're stuck with doing those calculations.
 

CommanderLake

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I'm stuck on page 14 of the MP4008 datasheet...
Assuming a 13.5v input, an inductor value of 10uH, an input current of about 14.2A and an inductor P-P ripple of 5.4A
also assuming Rcs1 is the resistor from CS to the output of M1 and Rcs2 to ground (see page 16)
I get an Rcs1 value of 0.014 and Rcs2 value of (less than or equal to) 0.0255

It then states "The CS resistance must be less than RCS1 and RCS2.", what on earth does that mean, it doesn't state a CS resistance and the resistors cant be less than them selves?
 

Harald Kapp

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Have a look at the typical application circuit. There is only one Rcs. You should have stumbled a bit earlier over the 2 equations for 2 different Rcs1 and Rcs1. The meaning of this is: There are two equations for Rcs that need to be satisfied. One leads to Rcs1, the other to RCS2. In order to fulfill both conditions Rcs must be less than both Rcs1 and Rcs2. In your case a value of Rcs = 0.01 Ω would fulfill this condition.
 

CommanderLake

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Thanks Harald, that makes sense, but what is the other resistor without a label connected to the CS pin for?

Edit: the datasheet for the MP4008 evaluation board (EV4008) shows the unlabelled resistor connected to the CS pin is 100 ohms.
 
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Harald Kapp

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This resistor is probably not critical (couldn't find any details on this resistor either). My guess is that it is meant to limit input current into CS in case of a current spike and resulting voltage spike across Rcs. You're probably good with 100 Ω.
 

CommanderLake

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The light isn't for the car itself I just want to be able to power it from a car efficiently and it's a great opportunity to learn more about switch mode power supplies, see this post: www.electronicspoint.com/forums/threads/create-controlled-50v-pulse.290795/page-2#post-1783396

I'm almost done calculating the compensation network on page 15 of the MP4008 datasheet: www.monolithicpower.com/en/documentview/productdocument/index/doc_url/L01QNDAwOF9yMS4wMS5wZGY%3D/release_date/MjAxNS0wOC0wNiAwMDowMDowMA%3D%3D/
I'm stuck at the end of equation 19, specifically GCS, I know conductance is the reciprocal of resistance but what's the resistance of the CS circuit, is it the unlabelled resistor that's 100 ohm on the evaluation board?

RCOMP = ((14 + 0.078) / 0.078) * (2500 * 0.000164 * 2 * 3.1415926535897932384626433832795) / (440 * (1 - 0.75) * 0.01) = 422.68554696193973838092327168815
Does that look right?
 
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Harald Kapp

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What does the compensation network do and do I really need to implement it?
This network compensates the frequency dependend characteristics of the circuit to ensure stable operation to avoid e.g. uncontrolled oscillations. You need to implement it.

RCOMP = ((14 + 0.078) / 0.078) * (2500 * 0.000164 * 2 * 3.1415926535897932384626433832795) / (440 * (1 - 0.75) * 0.01) = 422.68554696193973838092327168815
Does that look right?
I think you got the 23rd digit of Pi wrong.
Just kidding. Your components will be 1 % accurate at best, so no need to calculate beyond that resolution.

what's the resistance of the CS circuit, is it the unlabelled resistor that's 100 ohm on the evaluation board?
No, it is not. It is, as the text states, the conductance of the compensation network consisting of the series and parallel connection of Rcomp, CZ and CP.
 

dave9

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Please explain the need for this.

I am in favor of DIY, controlling the circuit and making sure it is Done Right.

In this case, what are you hoping to achieve? I assume you know you will need a massive heatsink that costs more than some already made light bar from (China?). I assume you know your alternator may not be able to sustain this.

I assume you are an expert in designing reflectors so this LED has a useful beam?

I might be an a55. I don't like to kill enthusiasm for a project but why??

I can't imagine an outcome that is worth the effort, but i am always learning if you have some niche application that needs this.
 

CommanderLake

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Please explain the need for this.

I am in favor of DIY, controlling the circuit and making sure it is Done Right.

In this case, what are you hoping to achieve? I assume you know you will need a massive heatsink that costs more than some already made light bar from (China?). I assume you know your alternator may not be able to sustain this.

I assume you are an expert in designing reflectors so this LED has a useful beam?

I might be an a55. I don't like to kill enthusiasm for a project but why??

I can't imagine an outcome that is worth the effort, but i am always learning if you have some niche application that needs this.
As I mentioned above, see this post: https://www.electronicspoint.com/fo...ntrolled-50v-pulse.290795/page-2#post-1783396

I'm using a 500mm diameter 250mm focal length(f0.5) precision Fresnel lens with an aluminium frame I made to hold the 172w led at just the right distance to focus at infinity, as you can see in the pictures it lights up the clouds so I want to take it somewhere with a vast landscape with a friend of mine and see how far it shines and take more pictures.

But since the led already uses 172w I need an efficient power supply that can output 50v at 3.6A so I've decided to make it another project, I have a big heavy heatsink already that should be sufficient for the boost converter.

As I'm also into audio and making speakers and subwoofers I know a car can output about 800-1000w before the voltage drops significantly so 200ish at about 14 amps is nothing.

Its all just my idea of fun.

No, it is not. It is, as the text states, the conductance of the compensation network consisting of the series and parallel connection of Rcomp, CZ and CP.

So I need the value of Rcomp to calculate the value of Rcomp? That's chicken and egg stuff!
 
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