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FrozenGate by Avery

Switching regulator that regulates current instead of voltage?

Joined
Nov 7, 2013
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Hello,


I was scouting Linear Technology's ICs to find a switching regulator which regulates the output current, as required for laser diode drivers.
(alternative: Maxim Integrated)

Preferably I want a regulator which doesn't need an inductor to operate. I read that charge pumps might be what I'm looking for.
However the charge pumps I found mostly have low input voltage requirements which is a problem since the power supply will be 9VDC.
Those which DO accept over 9V input voltage, can deliver only 250mA output, which is not enough.

If this is Really not possible, the best alternative will be a Buck regulator.

The problem is that all these ICs that I find regulate output voltage? But I need to regulate the output current, as I read in the sticky post on this board.
Are there ICs out there that regulate output current? Or am I supposed to add some things to the IC to trick it into regulating current?

I have built a lot of linear current regulators which dissipate a lot of heat in the power converter.
9V input voltage, 400mA output @ 2V is sadly a lot of wasted energy since the other 7V needs to be dissipated.
This is why I finally want to try and build one which uses a switched regulator.

List of my requirements:
- No inductors
- 9-15V input voltage
- Current regulation
- ~0mA to >700mA Output current

Not many requirements, but sadly I can't find any regulator which matches all of this.



I see people here on the forums building epically small buck drivers, and I can't help myself going crazy about understanding how it's possible that they can make it so small, and with so less components..
Are those drivers current regulators?

Now I don't need one of those, what I want is to be able to build a driver myself, and most importantly, to understand how this can be done.
There is no need for the driver to be this small. I'm always happy to use T7 or T5 packages because I can't work with DFN, SO, MS, .. packages. I don't have the skills to solder them.


Any ideas?
 
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A charge pump is basically a switched capacitor converter. A capacitor carries charge from the input to the output. For step up charge pumps, a cap is basically charged in parallel with the input and discharged in series and vice versa for step down. It is quite efficient for low currents and if the output is a multiple of the input. There are charge pumps IIRC that are designed for driving LED backlights but still have the low current limitation. It's inherent for charge pumps since you are limited by the switch on resistances and the size and ESR of the flying capacitor.

If you want higher currents, there's no way but to use inductors. Why are you afraid of them anyways? you can always find one in scrap boards and rewind them to get other values.
 
Switching regulators rely on inductance for their operation, so requiring no inductors is silly.
You can get ICs with internal inductors, but their current capacity is very low.

Remove the no-inductors requirement and search again.
 
There is no way to avoid using an inductor. It isn't possible to regulate current with a charge pump
because the instantaneous current through a capacitor tends to go toward infinity. In a real capacitor, it
is only limited by the ESR of the capacitor and resistances in whatever is connected to it (actually stray
inductances as well, but they are insignificant. So technically a charge pump actually contains
inductors(!) and avoiding them is completely impossible(!!)) Anyway, that resistance will convert the
power directly into heat and take out the weakest link. So many poor defenseless little electronic
components have been sacrificed to gain this knowledge. :cryyy:
 
Alright, forget the 'no inductor' requirement.
I've also found a lot of ICs with built in inductors, which do go up to high currents such as 1-2A! Fantastic ^^

Now the problem still remains: they are voltage regulators?!

About the different packages: It seems like the larger packages such as T5, T7, N8,.. always have a much larger 'minimum output voltage' (~4V) in contrary to the ~0.1-0.5V which I find for the smaller packages, why is this?
It's a big problem for me because I can't solder these small packages, but the large packages become useless if they output a minimum of 4V to the diode.
 
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Maybe searching for buck / boost LED drivers IC's may give you some results? You just need to scope the output and check if theres smooth turn on, hows regulation (current out error in %). I didn't build any switch mode drivers, but I've always wanted to.

Subscribing to thread ;)
 
A 2A buck regulator IC with internal inductor? This I gotta see.

Regulators are only trying to get a specific voltage on a feedback pin. You can vary the voltage or the current by varying the ratio that the output has to this expected feedback voltage. Google the LM317 for a simple example. Compare the voltage and current modes of operation.

You can convert the current to a voltage using a shunt. You can then amplify that voltage to match the feedback requirements of the IC using many different methods.
 
Maybe searching for buck / boost LED drivers IC's may give you some results? You just need to scope the output and check if theres smooth turn on, hows regulation (current out error in %). I didn't build any switch mode drivers, but I've always wanted to.

Subscribing to thread ;)
http://www.linear.com/product/LT3517
This is probably something I'm looking for, but it does not have an internal switch.
Also, why are these packages so tiny? 4mm wide?! how am I supposed to solder this, my soldering iron tip is like 5mm wide.



A 2A buck regulator IC with internal inductor? This I gotta see.
Linear Technology - Parametric Search for Step-Down (Buck) Regulators with Internal Inductor
They even go up to 36A

Regulators are only trying to get a specific voltage on a feedback pin. You can vary the voltage or the current by varying the ratio that the output has to this expected feedback voltage. Google the LM317 for a simple example. Compare the voltage and current modes of operation.

You can convert the current to a voltage using a shunt. You can then amplify that voltage to match the feedback requirements of the IC using many different methods.
Thank you for explaining how to make the current regulation!
Sadly the packages with internal inductors don't have the feedback pin because that's also internal :/
 
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Sure they do. Take a look at the LTM8050, for instance.

FB (Pin K7): The LTM8050 regulates its FB pin to 0.79V.
Connect the adjust resistor from this pin to ground.
The value of R FB is given by the equation
RFB = 394.21/(VOUT– 0.79), where RFB is in kΩ.
So I need to amplify the current signal from a sense resistor, into a resistive value? It would be a lot easier if the fb was an analog voltage input.
 
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Ooooh so the FB resistor that you must add, creates a voltage divider together with the 499k resistor?

Is it still possible then to have the output of the OpAmp connected to this pin?
Because that would mean that the OpAmp output is connected to Vout trough the 499k resistor. Isn't that bad?
 
No no, you would connect the sensing resistor to ground in series with the load. Then FB would connect
at the node between the resistor and load. It would require some experimentation. These modules aren't
cheap either, like $45 apiece, but I don't know what your budget is.
 
Oh like that!
So to choose the output current, I simply change the sensing resistor into a POT?
(however I'm not sure there are like 0.1ohm pots)

EDIT: I calculated the sensing pot, and would have to be a 5or10ohm pot.

And indeed.. That is way too expensive. I was thinking about €5 or something ^^



How about we go back a step. This µmodule stuff actually isn't even what I'm looking for.
let's go back to larger components such as the TO220 packages ^^ It's no problem if the driver gets larger.
This one for example:
http://www.linear.com/product/LT1959
(found in this list, and sorting by 'minimum output voltage) because that needs to be low, right?)

The feedback pin on this component:
FB:
The feedback pin is used to set output voltage using an
external voltage divider that generates 1.21V at the pin
with the desired output voltage. Three additional functions
are performed by the FB pin. When the pin voltage drops
below 0.8V, switch current limit is reduced. Below 0.7V
the external sync function is disabled and switching fre-
quency is reduced. See Feedback Pin Function section in
Applications Information for details
hmm isn't that "regulation" bad? it'll only react when it goes below a certain value.
 
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Oh like that!
So to choose the output current, I simply change the sensing resistor into a POT?
(however I'm not sure there are like 0.1ohm pots)

You would need a high power pot. Instead use a regular sense resistor and put a 10k pot
across that. Adjusting the pot down would increase the current. You just have to be
careful not to bottom it out, or else it will max out and probably blow something. The thing
to do would be to connect another resistor to ground in series with the pot to set a fixed
minimum below which it cannot go.


Now you're talking.

(found in this list, and sorting by 'minimum output voltage) because that needs to be low, right?)

We're actually looking for a low reference voltage, but logic would dictate that a low output
voltage would equate to a low reference voltage. Reference voltage isn't usually on a
product selector anyway.

The feedback pin on this component:

hmm isn't that "regulation" bad? it'll only react when it goes below a certain value.

Not necessarily, but it might be undesirable. We are after all using this for purposes other
than what it was designed for. The only time the reference voltage will be in that range will
be during startup (and shutdown, but that don't count) and open load circuit conditions.If
you can find a "current-mode" regulator, that will work much better. CC switching
regulators need a load at all times, or else bad things happen.
 


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