Author Topic: Another counstant current load for critics  (Read 1813 times)

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Offline 001Topic starter

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Another counstant current load for critics
« on: November 03, 2019, 05:49:25 pm »
Hi

Is it good or bad project?  :-//
https://diyodemag.com/projects/dummy_load
 

Offline floobydust

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Re: Another counstant current load for critics
« Reply #1 on: November 03, 2019, 08:42:57 pm »
Bad.
 

Offline Whales

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Re: Another counstant current load for critics
« Reply #2 on: November 03, 2019, 09:55:16 pm »
From the article:



(Viewing this image required javascript + a lightbox.  :'( )

Offline Whales

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Re: Another counstant current load for critics
« Reply #3 on: November 03, 2019, 10:01:54 pm »
Wait, they consider this a switching device?  But this looks like a linear regulator.

Hmm, they have a scope shot of something that looks switching.  There are no grid squares, so I can't tell what the 100mV scale means, but I suspect they have just zoomed into the ripple.

Also I am not sure you really need the 9V regulation stage.  Ditch it and use your 12V directly unless your 12V source is unstable.  I suspect it was born of them changing it from the "5V" design they mention.

Overall: it's a simple design built with jellybean parts.  The IRF540 most people will use will be a greymarket fake (it works fine but won't meet datasheet specs, be wary of their remarks regarding the 175degC die limit).  My only thought is to whether or not this device is stable and under what conditions.
« Last Edit: November 03, 2019, 10:10:46 pm by Whales »
 

Offline Whales

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Re: Another counstant current load for critics
« Reply #4 on: November 03, 2019, 10:08:29 pm »
Some specific quotes about this design:

Quote
Whilst this isn’t a perfect system, due to inefficiencies in both the MOSFET and the operational amplifier

"inefficiencies in the MOSFET" isn't the right wording or line of thought here.  That's too easy to construe in many incorrect ways, this class of device always has the same "efficiency" regardless of the MOSFET used.

Quote
the op-amp will attempt to rectify this difference by increasing the voltage at the output. This, in turn, saturates the gate of the MOSFET, which allows current to flow from the MOSFETs Drain to the source.

I would have thought this device operated the FET mostly in the linear region, not the saturation region.



I once bought or obtained an issue of Diyode after being recommended it.  I didn't end up liking it, but I can't remember the full details.

It looks like they're focusing on the whole round trip of making a device from breadboard to box.  That's nice.  It's made from jellybean parts, it's performance won't be anything fancy, and there are a lot of tweaks I would do to the PCB for better perf; but I think that's outside their aims and misses the point.  It's not intended to be an amazing spec dummy load, but instead shows you how you can start making your own (simpler) alternative to the off-the-shelf products the mention in the opening.
« Last Edit: November 03, 2019, 10:12:22 pm by Whales »
 

Offline bson

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Re: Another counstant current load for critics
« Reply #5 on: November 04, 2019, 08:42:27 pm »
I would have thought this device operated the FET mostly in the linear region, not the saturation region.
For a FET the saturation region is the linear/ohmic region.
 
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Offline Dave

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Re: Another counstant current load for critics
« Reply #6 on: November 05, 2019, 09:09:41 am »
For a FET the saturation region is the linear/ohmic region.
:palm: :palm: :palm:



Source
<fellbuendel> it's arduino, you're not supposed to know anything about what you're doing
<fellbuendel> if you knew, you wouldn't be using it
 
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Offline mikerj

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Re: Another counstant current load for critics
« Reply #7 on: November 05, 2019, 10:50:23 am »
Quote
the op-amp will attempt to rectify this difference by increasing the voltage at the output. This, in turn, saturates the gate of the MOSFET, which allows current to flow from the MOSFETs Drain to the source.

I would have thought this device operated the FET mostly in the linear region, not the saturation region.


The FET in a linear regulator has to operate in the linear/ohmic region.  If the FET becomes saturated then the current/voltage is no longer under control.
 

Online magic

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Re: Another counstant current load for critics
« Reply #8 on: November 05, 2019, 05:58:30 pm »
No, when the channel saturates then current stays nearly constant regardless of minor variations in drain voltage, controlled solely by gate voltage. Quite a desirable condition in a constant current sink, actually ;)
 

Offline tooki

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Re: Another counstant current load for critics
« Reply #9 on: November 05, 2019, 06:56:44 pm »
(Viewing this image required javascript + a lightbox.  :'( )
No, as with many lightboxes, the JavaScript merely traps the click and redirects to a lightbox, but the actual link itself is a plain link to the destination image, so that without JavaScript, it’s a functional link. (That’s how the inline thumbnails on eevblog work, too.) You can try this by opening the link in a new tab, which sends you to this URL: https://diyodemag.com/_images/5cbed096c672e05c2cb5a1fc
 

Offline SiliconWizard

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Re: Another counstant current load for critics
« Reply #10 on: November 05, 2019, 07:04:07 pm »
This project is a pretty basic voltage-to-current converter. There's nothing fundamentally "wrong" with it. As is, it has some limitations, but it can work fine for a range of conditions.
(I hate schematics with basic elements like opamps not drawn as actual opamps...)

As it is, one of the opamps is used as a voltage buffer for the "reference" voltage coming from the pots; then the buffered voltage is divided by 2 (to limit the current range). Let's call it Vref = Vpot/2. With R6, the resulting current range would be 0 (there will be an offset though!) to 4.5A: Iload = Vref / R6.

With the LM358 and an IRF540, the probability of it oscillating is small IMO, but not zero. You can add a capacitor in the feedback loop to avoid oscillations.

Due to the topology, the working voltage range is also limited. R6 basically drops the voltage reference (= R6*Iload). Obviously the input voltage must be greater than Vref + Vds.

This basic topology has been discussed many times on here, so you should be able to find a number of threads dealing with it and the possible traps (including oscillation).

As to whether the MOSFET mostly works in the linear or saturation region, I'll leave it as an exercise. :popcorn:
« Last Edit: November 05, 2019, 07:08:31 pm by SiliconWizard »
 

Offline 001Topic starter

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Re: Another counstant current load for critics
« Reply #11 on: November 05, 2019, 07:41:14 pm »
Thanx a lot
Can U link me with better project at these level?
 

Offline SiliconWizard

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Re: Another counstant current load for critics
« Reply #12 on: November 05, 2019, 07:49:00 pm »
 
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