Swend
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Hi
What is your range of current to measure?
Hi,
it will do something. And it won´t explode.
Whether it will do what you expect - nobody knows... because you don´t say what you expect.
For every 1A of current through R_Sense I expect 5mV of output voltage.... if it doesn´t start to oscillate because of the missing power supply capacitors.
Klaus
Hi,
I assume the AD8221 is slower than the LT´s internal INA. Maybe slower than you expect.
My recommendation: First decide your requirements/specifications, then look for suitable parts.
The performance will also depend on PCB layout. --> I recommend Kelvin wiring to the shunt.
Klaus
* I can´t validate it a "appropriate use" ....The thing I was uncertain about was the R1/R2 and if was an appropriate use of same considering it's low voltages and placed directly on the power supply output..
It always confuses me when the shown circuit shows something different to the text.
So - even with the new informations - I can only repead what I´ve written in post#2: "It will do something".
I´ve still a lot of doubts:
* does the circuit make sense? (Usually one wants to amplify a shunt signal, but you just attenuate it.)
* I can´t see a match to your 0...50A. If you turn R2 pot down to zero it surely will not limit the current within the desired 0..50A range. Instead it won´t limit the current at all ... and then it is likely that something explodes.
--> I can only recommend to avoid this situation.
Also a drawback in this topology is that you will not know exactly WHAT current you are setting it for ! Whether you use a 270deg pot or a 20T multiturn. Surely you would want to know this in your setup.
Since this is quite a simplistic circuit, why not hook it up and see what you get ?
I agree. No need for showing the whole SMPS schematic.I personally prefer to have a partial circuit where the object of interest is in focus, instead of e.g. a full A3 size circuit diagram.
Brad:
Since V_BE drifts by about 2mV/°C and the shunt is 2mOhm...
I expect an output drift equal to about 1A/°C.
Can you verify this?
You can detect and amplify sense voltage via common-base operation of a transistor. My simulation has 2 milliOhm sense resistance. Even that low amount is subject to several Watts when carrying 50A. If you're willing to experiment, you can tap across a span of several inches of cable, to obtain a fraction of a volt differential for your circuit to measure.
If you turn the pot down - a.k.a. reduce R2 you will see less current than there really is - do you want to do this?
better to adjust the demand or reference up and down and see the corresponding change in the actual shunt - rather than step down the shunt voltage with your resistive divider ...
I just feedback what is not that clear to me when I read the posts.
Your headline says "variable" and twice in the first post you mentioned the same. For sure we assume this is what you want to focus on.
Now you write it doesn't have to be variable.
Now I'm not sure if you want to use a pot at all. But if you want to use a pot .... since it has three terminals, there are several ways to connect it instead of R2. You may connect it completely wrong.
And there are several connection schemes that will work ... but have different benefits and drawbacks.
Thus if you want to use a pot, then I recommend to show us how you intend to connect it.
After all I'm still not sure if you will be satisfied with the solution.
Especially, because I don't know why the fine tuning is that important for you.
If you want to build a power supply with adjustable output current limit, then you may be disappointed, because
* by turning the pot down you decrease the feedback, thus you increase the coil current
* by turning down the current feedback too much you risk the SMPS to get damaged
now you mention "average current".
But by modifying the input signal from the shunt you also adjust for peak current. (You see the OPAMP marked "IC1". This is a part of the whole control loop - even voltage regulation.)
In case you just want to adjust "average current", then I recommend a constant current circuit at I_AVG pin.
True.I have the term from the datasheet of the LT1339, they claim that the LT1339 current limiting feature works by measuring the "average current"
The controller does involve cycle by cycle current limiting in it´s regulation loop... but in opposite to other controllers the inverting input of the comparator IC1 (pin VC) is influenced by the value of the average current circuit.and not like most other controllers who measure the current on a pulse by pulse basis.
I see.Likewise I have the term "integrator" from the datasheet.
You are correct.Is that not how a current limiting usually works, by lowering the output voltage?
The capacitor on pin5 is essential for the average current limiter to work. Don´t disconnect it.Does that mean that I keep the capacitor on pin 5 together with a constant current circuit, or completely discard the cap? And for what range of current does the constant current circuit have to supply? How to calculate it?
What they call an integrator has a flat, gain of 1 from 0Hz to fc, then gain falls with -3db/octave. fc is the wanted corner frequency of 1/(2 x Pi x R x C). --> These are the characteristics of a 1st order LPF.
A true integrator has no (wanted) corner frequency and the gain at zero frequency goes in direction infinity.
To simplify things you have two options:
* either pull down the current at the node to increase the average load current
* or pull up the current at the node to reduce the average load current (safe way)
Let´s discuss this way.
There is a 50k resistor. With every uA at the node IAVG you "add" 50mV.
With a current of 50uA you add 2.5V and the average current limiter will reduce ouput voltage.
--> with 0...50uA you reduce the nominal average current linearily from 100% down to 0%.
Note: I expect that the precise current regulation at low load currents is impossible, because the V_OL limit of the "Current Sense Amp".