Buck Converter as constant current source

Thread Starter

imraneesa

Joined Dec 18, 2014
222
Dear Sir,

I am designing a buck converter using to light laser diode which has forward voltage of 4.7v. It have feedback voltage reference as 0.25v, which i used to set the constant current. i have used two 0.33ohm resistors in parallel to get a constant current of 1.5A.


Vin = 8.4v
Vout = 4.7v (laser diode forward voltage)
Iled = 1.5A
Frequency = 314Khz
I ripple = 20% of Iled = 0.3
L= (VIN – VOUT ) × D × T / Iripple
=((8.4-4.7) x (4.7/8.4) x (1/314000) / 0.3 ) x 1000000 uH
= 21.977uH

I have selected 22uH. but when i finished the board and tried to check if i am getting 1.5A constant current. I failed, I am only getting 1.1A~1.2A. What am i doing wrong. i have selected 10uf capacitor for input and output. and C3 as 0.1uf.
 
Last edited:

Kjeldgaard

Joined Apr 7, 2016
379
How does the layout look like?

For at the required current, it need not much additional copper to the path before it becomes the measured errors.
 

crutschow

Joined Mar 14, 2008
23,379
Any resistance between the GND connection of FP7103 and the GND connection of R1 and R2 will cause the observed error.
You may have to reduce the value of R1 and/or R2 to compensate.
 

Kjeldgaard

Joined Apr 7, 2016
379
At the uploaded image I can't see the detail I wanted.
But you must be aware of is to keep the measuring current and the voltage measuring points on the shunt resistor separated, it's often called kelvin connections.
As Crutschow writes, it is the connection between the GND terminals on FP7103 and "bottom" of R1 / R2, and just as important FB on FP7103 connection to the "top" of R1 / R2.
 

Kjeldgaard

Joined Apr 7, 2016
379
Right now I do not have anything installed, that can open a BRD file.

So I would I like to have screen dumps of the copper layers, preferably with silk screen so it is to find the individual components.
 

Thread Starter

imraneesa

Joined Dec 18, 2014
222
Right now I do not have anything installed, that can open a BRD file.

So I would I like to have screen dumps of the copper layers, preferably with silk screen so it is to find the individual components.

Dear Sir,
I am attaching this image if it is any useful. i dont know how to look in dump files. Please look into this image if it will make any sense. Thank you sir,
 

Attachments

Kjeldgaard

Joined Apr 7, 2016
379
It looks like there is a "long thin" liner from the L- pad two R1/R2 junction, and FB are connected to the "wrong" end of that liner.

What happens if you solder the laser minus wire directly to the R1/R2 junction pads?
 

drvlas

Joined Oct 16, 2015
28
Hi everybody!

I've no experience in LED lighting, and not much in DC-current design. So my idea may be some kind of stupid. But here it is.
We have to cope with big current / small voltage issue. And we have to feed a load at some distance from our circuitry. Voltage drops and such stuff. Right?
Now, let's shorten the points L+ and L-. We'll have a good 1.5 A current in R1/R2. Okay. And what is the current into the VCC terminal of FP7103? It equals 1.5 A plus 4 mA into the FP7103 itself.
The load (laser diode) is not connected to a GND anyway, eh? Connect it between the voltage source and FP7103 - and enjoy 1.504 mA of load current.
Will it work?
 

Thread Starter

imraneesa

Joined Dec 18, 2014
222
It looks like there is a "long thin" liner from the L- pad two R1/R2 junction, and FB are connected to the "wrong" end of that liner.

What happens if you solder the laser minus wire directly to the R1/R2 junction pads?
when i checked the path from fb to the junction it is not touching any where. and when i connect the -ve of laser to the direct junction of the R1 and R2 i am getting the same amperage like before, no change.
 

Kjeldgaard

Joined Apr 7, 2016
379
Now I'm about to run out of ideas.
But things to limit current, could be the inductors DC resistance or if there is used a unsuitable type of ceramics for the input and output capacitors.
What's the data on these three components?
 

Thread Starter

imraneesa

Joined Dec 18, 2014
222
Now I'm about to run out of ideas.
But things to limit current, could be the inductors DC resistance or if there is used a unsuitable type of ceramics for the input and output capacitors.
What's the data on these three components?

Dear Sir,
This is inductor
https://www.digikey.com/product-detail/en/bourns-inc/SRN6045-100M/SRN6045-100MCT-ND/2756164

This is 10uf capacitor
https://www.digikey.com/product-detail/en/murata-electronics-north-america/GRM21BR61C106KE15K/490-6473-1-ND/3845670

This is 0.1uf capacitor
https://www.digikey.com/product-detail/en/samsung-electro-mechanics-america-inc/CL21F104ZBCNNNC/1276-1007-1-ND/3889093

So capacitors also involve in limiting current?

I increased the inductor ripple from 20 to 30% so i end up using 10uH.
 

Kjeldgaard

Joined Apr 7, 2016
379
10uH, 2.5A, 60mOhm: OK
10uF, 16V, X5R: OK
100nF, 50V, Y5V: Here I would also choose an X7R or X5R type, but at the low operating voltage it's probably not a problem.

Certain ceramics lose a lot of the capacity when you increase the operating voltage.

I had, some years ago, a 10uF, 16V, Y5V input capacitor in a power supply and my measurements showed a wild big ripple, corresponding to well below 1uF effective capacity.

A search for X5R Y5V difference provides a lot of reading material, and the first hit is from Maxim: Why Your 4.7µF Capacitor Becomes a 0.33µF Capacitor

How much have you measured on the circuit?
And what measurement equipment you have access to?
 

Thread Starter

imraneesa

Joined Dec 18, 2014
222
Any resistance between the GND connection of FP7103 and the GND connection of R1 and R2 will cause the observed error.
You may have to reduce the value of R1 and/or R2 to compensate.
10uH, 2.5A, 60mOhm: OK
10uF, 16V, X5R: OK
100nF, 50V, Y5V: Here I would also choose an X7R or X5R type, but at the low operating voltage it's probably not a problem.

Certain ceramics lose a lot of the capacity when you increase the operating voltage.

I had, some years ago, a 10uF, 16V, Y5V input capacitor in a power supply and my measurements showed a wild big ripple, corresponding to well below 1uF effective capacity.

A search for X5R Y5V difference provides a lot of reading material, and the first hit is from Maxim: Why Your 4.7µF Capacitor Becomes a 0.33µF Capacitor

How much have you measured on the circuit?
And what measurement equipment you have access to?
Right now i am using R1=0.33ohm and R2=0.27ohm. Since the reference voltage is 0.24v the current should be equal to 0.24/0.1485 = 1.6A.

But when i measure it on circuit i am only getting 1.34A. i have two digital multimeter and oscilloscope. i am measuring it near positive input. The source i am using is two 3.7v lithium 16340 batteries. when i checked both batteries in series it shows 7.9v.
 

dannyf

Joined Sep 13, 2015
2,196
But when i measure it on circuit i am only getting 1.34A.
Without knowing where you measured it and how you measured it, such measurements have no meaning.

If I were you, I would make sure that wiring is correct first. Then I would mark on the schematic key voltage measurements, like input, output, fb voltage, etc.
 

Thread Starter

imraneesa

Joined Dec 18, 2014
222
Dear Sir,

I measured it at the input. I just found the culprit.
Kjeldgaard was right. the problem was the capacitors that i was using as 10uf. i dont know why. when i changed them i get all correct amperage that suppose to be present. I request you to suggest me a good capacitor of 10uf so that i can use them on my project. Thank you everybody for taking time to help me.
 

Kjeldgaard

Joined Apr 7, 2016
379
I am pleased to hear that you are heading towards a functioning circuit.

You do not write much about the capacitors you had in the PCB when the current was too low, and what types that gave the correct result.
 

Thread Starter

imraneesa

Joined Dec 18, 2014
222
I am pleased to hear that you are heading towards a functioning circuit.

You do not write much about the capacitors you had in the PCB when the current was too low, and what types that gave the correct result.
Dear Sir,
I just used some capacitors from other board but the package i was using was 805 and these new ones are 1206.
But all my happiness gone again. something is going wrong with my parts i think. again i am getting trouble with the different capacitors also. again the current not matching with the resistors. same old problem. i feel something is not correct. i have used sample board which came from the company and even in that i am not getting the current as required. i dont know where is the problem. if it is inductor? i am getting good 320kHz from the chip going to the inductor. and reference voltage near the current sense resistors is 260 to 290mv.
 

Thread Starter

imraneesa

Joined Dec 18, 2014
222
Any resistance between the GND connection of FP7103 and the GND connection of R1 and R2 will cause the observed error.
You may have to reduce the value of R1 and/or R2 to compensate.
Dear Sir,
I want to use 16340 lithium which is 3.7v battery. but when full charge it, i will get 4.2v. so for my project i am using 2 of them. that is voltage 8.4v. in designing inductor for the buck converter as constant source current. which voltage i should consider. should i use 8.4 or 7.4v?

I have got the sample board for the same project. which use to give me correct current according to the resistors i choose. but now i changed inductor. since then i am not getting correct current at all. it had 10uh 3.6a inductor and now i am using https://www.digikey.com/product-detail/en/bourns-inc/SRN6045-100M/SRN6045-100MCT-ND/2756164
My buck converter has 320kHz fixed frequency.
My observations
When i connect 8.4v battery i am getting 1.4A current where as i suppose to get 1.8A.
When i connected 12V battery I am getting 1.1A current.
I know it will change according to the Duty cycle. But how do i select Inductor which will give me constant current under the voltage rage of 7.4v to 8.4v.
 

ian field

Joined Oct 27, 2012
6,543
Dear Sir,

I am designing a buck converter using to light laser diode which has forward voltage of 4.7v. It have feedback voltage reference as 0.25v, which i used to set the constant current. i have used two 0.33ohm resistors in parallel to get a constant current of 1.5A.


Vin = 8.4v
Vout = 4.7v (laser diode forward voltage)
Iled = 1.5A
Frequency = 314Khz
I ripple = 20% of Iled = 0.3
L= (VIN – VOUT ) × D × T / Iripple
=((8.4-4.7) x (4.7/8.4) x (1/314000) / 0.3 ) x 1000000 uH
= 21.977uH

I have selected 22uH. but when i finished the board and tried to check if i am getting 1.5A constant current. I failed, I am only getting 1.1A~1.2A. What am i doing wrong. i have selected 10uf capacitor for input and output. and C3 as 0.1uf.
The simplest buck is the hysteretic type, you need to retain the voltage regulation as a failsafe - but you can also sense current to terminate the conduction cycle.

The downside of the hysteretic is high ripple, so it needs good filtering between it and the LD.
 

Benengineer

Joined Feb 6, 2016
133
Dear Sir,

I am designing a buck converter using to light laser diode which has forward voltage of 4.7v. It have feedback voltage reference as 0.25v, which i used to set the constant current. i have used two 0.33ohm resistors in parallel to get a constant current of 1.5A.


Vin = 8.4v
Vout = 4.7v (laser diode forward voltage)
Iled = 1.5A
Frequency = 314Khz
I ripple = 20% of Iled = 0.3
L= (VIN – VOUT ) × D × T / Iripple
=((8.4-4.7) x (4.7/8.4) x (1/314000) / 0.3 ) x 1000000 uH
= 21.977uH

I have selected 22uH. but when i finished the board and tried to check if i am getting 1.5A constant current. I failed, I am only getting 1.1A~1.2A. What am i doing wrong. i have selected 10uf capacitor for input and output. and C3 as 0.1uf.
The equation you are using for inductor is not right. You should use ∆I = Vin*D*D’*T/2L. Based on your spec ripple current, you can find L value. Also, you need to know what output vlotage max ripple you want. Then you can find out capacitor value with equation ∆v = ∆I*T/8*C.
Try it and let me know your result.

Ben
 
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