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    Variable current sense, will this work?

    Hi friends!

    I'm in doubt if this would work as a variable current sense circuit, especially the R1/R2 voltage divider. I would like to make e.g. R2 variable, would that work? LTSpice says it will, but I intend to place this on a SMPS so I'm not sure, noise and real life working conditions you know.

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    Re: Variable current sense, will this work?

    Hi

    What is your range of current to measure?



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    Re: Variable current sense, will this work?

    Hi,

    it will do something. And it wonīt explode.
    Whether it will do what you expect - nobody knows... because you donīt say what you expect.

    For every 1A of current through R_Sense I expect 5mV of output voltage.... if it doesnīt start to oscillate because of the missing power supply capacitors.

    Klaus
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    Re: Variable current sense, will this work?

    Quote Originally Posted by tyassin View Post
    Hi

    What is your range of current to measure?
    Hi tyassin, between 0-50A

    - - - Updated - - -

    Quote Originally Posted by KlausST View Post
    Hi,

    it will do something. And it wonīt explode.
    Whether it will do what you expect - nobody knows... because you donīt say what you expect.

    For every 1A of current through R_Sense I expect 5mV of output voltage.... if it doesnīt start to oscillate because of the missing power supply capacitors.

    Klaus
    Sorry about that, I expect the same as you do so that's good. The AD8221 is just to simulate the input of the current sense of the LT1339 buck/boost, which also is an instrumentation amp but with G=15, the current limit kicks in at 120mV across R_Sense.



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    Re: Variable current sense, will this work?

    Hi,

    I assume the AD8221 is slower than the LTīs 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
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    Re: Variable current sense, will this work?

    Quote Originally Posted by KlausST View Post
    Hi,

    I assume the AD8221 is slower than the LTīs 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
    Allow me to rephrase "The AD8221 is just to simulate the input of the current sense of the LT1339 buck/boost, i.e. I will not be using the AD8221 but the internal INA of the LT1339". 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..

    So the part's list would be one 100ohm resisitor and one 100ohm potentiometer. The LT1339 INA connects directly across R_Sense without Kelvin wiring as per app note, so do you feel it would be necessary with Kelvin wiring if I add two resistors close to R_Sense and keep the traces sort?



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    Re: Variable current sense, will this work?

    Hi,

    If I understand now, then you donīt want to use the AD8221 at all... In post#1 you didnīt mention this.

    It always confuses me when the shown circuit shows something different to the text.

    It really isnīt much effort to draw a pot instead of the fixed R2. And to draw power supply capacitors... and kelvin wiring (if you have) ... but it prevents confusion.
    --> Please understand that you surely know how your real circuit should look like and how you expect it to work... but we only have the information you give.

    ***
    So - even with the new informations - I can only repead what Iīve written in post#2: "It will do something".

    Iīve still a lot of doubts:
    * does the circuit make sense? (Usually one wants to amplify a shunt signal, but you just attenuate it.)
    * I canīt 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 wonīt limit the current at all ... and then it is likely that something explodes.
    --> I can only recommend to avoid this situation.

    you write:
    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..
    * I canīt validate it a "appropriate use" ....
    * I even canīt find in your posts that "R1/R2 are placed directly to the power supply output".

    --> And - with all low impedance, low voltage sensing circuits - I recommend to use kelvin wiring.

    Klaus
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    Re: Variable current sense, will this work?

    Quote Originally Posted by KlausST View Post
    It always confuses me when the shown circuit shows something different to the text.
    I'm sorry about that, I understand what you are saying, I really try to write it so that I get the least amount of flak, but I'm no communications expert for sure. If I were to attempt a rephrase of thread #1 it would go like this: "Here is a resistivity divider R1/R2, will it work in this depicted environment?" that was the entire gist of my post. 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. I just wanted to know if I could place a a resistive divider where I did and said it would be used in a SMPS, I was in doubt and this is a critical part of the construction. But I understand, you are detail oriented and I wish I could be more like you in this respect.

    You say

    Quote Originally Posted by KlausST View Post
    So - even with the new informations - I can only repead what Iīve written in post#2: "It will do something".

    Iīve still a lot of doubts:
    * does the circuit make sense? (Usually one wants to amplify a shunt signal, but you just attenuate it.)
    * I canīt 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 wonīt limit the current at all ... and then it is likely that something explodes.
    --> I can only recommend to avoid this situation.
    OK, if you could please have in mind that my question is loosely "Here is a resistivity divider R1/R2, will it work in this depicted environment?" and that I would be a happy camper with a one word reply yes/no/maybe, I just wanted to know if I was on the right-ish path.

    Sense: It makes sense to me, I know that low voltage measurements are just as difficult as high voltage measurements and I wanted to avoid even lower voltages across R_Sense and make things even more difficult for myself. So if someone answered my question with a "yes" or "maybe" it would just be a matter of fiddling around with the resistor values and I would get something useful for further development. I have a 10-15Kgs 24VAC transformer (I have no data, it's 40 years old) collecting dust that I would like to put into good use now.

    Explosion: Turning down R2 and disable current limiting completely is a feature that I would like to have in this power supply..

    0...50A: Everything is a compromise for me, I need this power supply and I have limited time and funds - so I know it won't be perfect, I don't need nor want perfection here.

    Noted with the kelvin wiring, thank you.

    In any case I appreciate your input and I'm truly sorry that I'm confusing you with my less than perfect questions, but I'm not perfect.



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    Re: Variable current sense, will this work?

    The circuit as such looks simple and straightforward as is. Clearly the concern is whether it will work as intended especially since voltage levels are really low. Right next to an SMPS would definitely put it in the area to pickup EM due to smps switching and high currents, and the coupled signals are very likely to be of the same order or greater than the signal you are trying to measure.

    So minimising wire loops (inductive pickup), twisted wires, shielding, short wire runs etc will all be required and play a significant part. Capacitive decoupling too might help, but impedances are so low ! Turning down R2 (i.e. make it 0) would make it current insensitive of course. I'm not so sure though what benefit Kelvin wiring would have in your case.

    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 ?



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    Re: Variable current sense, will this work?

    Quote Originally Posted by kripacharya View Post
    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.
    Actually I don't need need variable, I could do with on/off if push comes to shove, but if i can do variable with just two resistors then I don't mind and I can use fingerspitzengefühl for precise adjustment.

    Quote Originally Posted by kripacharya View Post
    Since this is quite a simplistic circuit, why not hook it up and see what you get ?
    That was my intention, right after i design the PCB and order it along with the LT1339.



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    Re: Variable current sense, will this work?

    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.

    Click image for larger version. 

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    Adjust values to obtain desired range of output. It may require extra effort and/or circuitry to see 0V output with 0V input. (This isn't to say an op amp is easier to adjust.) Be on the alert for temperature changes to cause readings to rise or fall.



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    Re: Variable current sense, will this work?

    fingerspitzengefühl ... Brilliant electronic solution!



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    Re: Variable current sense, will this work?

    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 ...



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    Re: Variable current sense, will this work?

    Hi,

    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?

    Swend:
    No need to apologise, nobody is perfect.
    ...and to be that pedantic isn't always a benefit, too.
    I just feedback what is not that clear to me when I read the posts.

    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.
    I agree. No need for showing the whole SMPS schematic.
    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.

    ******
    If I understand right, then you want to fine adjust the current limit of the SMPS IC.
    If so:
    * A resistive divider should generally work....
    * low ohmic resistors are good, but since you work with high frequency, they need to be low impedance, too. Thus don't use wire wound resistors.
    * correct wiring is important

    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. (With the LT it is used as coarse current limiter to prevent the inductor to go into saturation)
    If you want to build a power supply with adjustable output current limit, then you may be disappointed, because
    * I expect the output current will drift with temperature
    * the output current will depend on input voltage and output voltage
    * and you will see difference from part to part
    * by turning the pot down you decrease the feedback, thus you increase the coil current
    (Lets say the shunt is calculated for a 25A current limit...with your solution you adjust the current limit to 50A ... infinte (depending on pot connection). Not in direction zero. No 0A to 50A!
    I'm not sure if you are aware of this.
    * by turning down the current feedback too much you risk the SMPS to get damaged

    Klaus
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    Re: Variable current sense, will this work?

    Quote Originally Posted by KlausST View Post
    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 are correct. Temperature change causes readings to drift. I saw this in my own tests as I squeezed the transistor between my fingers. The reading changed. When I let go, the reading returned to previous.

    The op amp is less sensitive to temperature change.

    It's just that I remember going through contortions when I made a similar current sense circuit (for my homebrew Amp-hour meter). I tried to do it with a single-ended supply and LM324 quad op amp. It has no offset adjustment. So I had to create a resistor network at the inverting input, that created the proper gain (200x). Then I added a resistor going to the positive supply, so that I got 0V output with 0V input. After testing a lot of resistor values I managed to get linear response from 0 to 15A, eventually. But I thought 'There must be an easier way.'

    The task is to accept a low impedance signal, and apply high gain. The output can be high impedance. The op amp fulfills this role, as the output is a half-bridge, one of which transistors turn on to a certain extent, giving us the output voltage.

    So likewise the common-base amplifier has low imput impedance, high gain, and high output impedance. It needs no negative supply, hence can never create a negative reading. It's not an ideal solution but it has a lot going for it. The temperature drift might be solvable with some effort, perhaps with a strategically placed component or two.



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    Re: Variable current sense, will this work?

    Quote Originally Posted by BradtheRad View Post
    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.
    Hi BradtheRad

    I like your solution, but I think I will pass on this one as I know I will be "going through contortions" working with that low voltages. IF I were to consider low voltages, then I think I'd rather make it non-intrusive e.g. I would use a Hall-Effect sensor.

    - - - Updated - - -

    Quote Originally Posted by Easy peasy View Post
    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 ...
    My reasoning was like this; if I settle for e.g. R_Sense=0.02 I get 120mV at 6 amps and current limit kicks in. So if can attenuate V(R_sense) I would get adjustable current limiting from 6 to whatever amps. A compromise.

    I don't know where I would adjust the demand or reference up and down? I see though on the LT1339 the INA output goes into an integrator that gives the average output current which is then compared to ref of 2.5V, the capacitor of this integrator is external, maybe it is possible to adjust current limit here? I don't know if it is a relative simple task or complex.

    I AVG (Pin 5): Average Current Limit Integration. Fre-
    quency response characteristic is set using the 50kΩ
    output impedance and external capacitor to ground.
    Averaging roll-off typically set at 1 to 2 orders of magni-
    tude under switching frequency. (Typical capacitor value
    ~1000pF for f O = 100kHz.) Shorting this pin to SGND will
    disable the average current limit function.

    Click image for larger version. 

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    - - - Updated - - -

    Quote Originally Posted by KlausST View Post
    I just feedback what is not that clear to me when I read the posts.
    I know, maybe it's just me - but I feel your frustration with my unclear questions, and I don't deliberately want to be the source of anyones frustrations.

    Quote Originally Posted by KlausST View Post
    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.
    I think it's just a simple communications problem again. For me the "variable" in this context means R1/R2(pot), because I want to vary the resistance and thus the current limit. And when I say "want" I don't mean I "demand", I mean more in the sense of "wish" i.e. I can wish for something, but there is no guarantee that I will get my wish granted, or maybe the costs in time/money are prohibitive and I concisely decide that I will not pay the associated costs.

    Quote Originally Posted by KlausST View Post
    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.
    On the pot thing, I thought about it, I don't think it would work if I mount the pot on the chassis the wires will probably pick up noise and introduce inductance/capacitance, so if I were to use a pot it would have to be mounted on the PCB, and if it's mounted on the PCB I would have to open the chassis every time I wanted to adjust it, not very practical.

    Quote Originally Posted by KlausST View Post
    After all I'm still not sure if you will be satisfied with the solution.
    I can guarantee you that I will be satisfied with whatever the end result might be, because that is what I will have in my hands and I will have to accept with whatever quirks it has, just like any other piece of equipment I have in everyday life.

    Quote Originally Posted by KlausST View Post
    Especially, because I don't know why the fine tuning is that important for you.
    It's not "that important" it's a "wish", I always wished to own a power supply with variable current limit ever sine I saw one as a kid. Now I'm in the process of making yet another power supply, and maybe this time I could get what I wish for, or maybe not.

    Quote Originally Posted by KlausST View Post
    If you want to build a power supply with adjustable output current limit, then you may be disappointed, because
    I deal with disappointment on a daily basis, I would actually be disappointed if it didn't disappoint me

    Quote Originally Posted by KlausST View Post
    * by turning the pot down you decrease the feedback, thus you increase the coil current
    Please see the reasoning I gave to Easy Peasy

    Quote Originally Posted by KlausST View Post
    * by turning down the current feedback too much you risk the SMPS to get damaged
    Yes, but it's not going into production and the consumer market accompanied by a two year guarantee. This is a personal power supply that I build from the ground up and that I can repair myself the day I blow it up, so I accept that risk knowingly and willingly. Think of it as commie tech e.g. the Wartburg car or the Tomos outboard, you can repair everything on those things with only a screwdriver and a 13mm wrench.



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    Re: Variable current sense, will this work?

    Hi,

    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.)

    Now I donīt know whether you want to manipulate
    * "average current" only
    * or "average current" plus the complete control loop. (which I donīt recommend because of stability reasons)

    In case you just want to adjust "average current", then I recommend a constant current circuit at I_AVG pin.
    Theoretically you could use a simple resistor there, but as soon as you modify the resistor value you modify the cutoff frequency of the LPF, too (which you named "integrator")... While a constant current circuit does not modify the cutoff frequency.
    Good thing here is, that the constant current circuit is uncritical, because the signal at I_AVG is not critical. The frequency is low and the voltage_of_interest is at 2.5V.

    Klaus
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    Re: Variable current sense, will this work?

    Quote Originally Posted by KlausST View Post
    now you mention "average current".
    I have the term from the datasheet of the LT1339, they claim that the LT1339 current limiting feature works by measuring the "average current" and not like most other controllers who measure the current on a pulse by pulse basis. Likewise I have the term "integrator" from the datasheet.

    Quote Originally Posted by KlausST View Post
    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.)
    Is that not how a current limiting usually works, by lowering the output voltage?

    Quote Originally Posted by KlausST View Post
    In case you just want to adjust "average current", then I recommend a constant current circuit at I_AVG pin.
    I would prefer the "average current" only, because of stability reasons and that the constant current circuit is uncritical as you say.

    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?



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    Re: Variable current sense, will this work?

    Hi,

    I have the term from the datasheet of the LT1339, they claim that the LT1339 current limiting feature works by measuring the "average current"
    True.

    and not like most other controllers who measure the current on a pulse by pulse basis.
    The controller does involve cycle by cycle current limiting in itīs 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.

    Likewise I have the term "integrator" from the datasheet.
    I see.
    Two lines of neutral information, in case you are interested in :
    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.

    Is that not how a current limiting usually works, by lowering the output voltage?
    You are correct.
    But what I meant is the different thing:
    Imagine the actual load current is much less than the set "average current limit". Then the average current limiter is obviously OFF. Now the voltage regulation still works via the cycle by cycle current limit (IC1) control loop.
    Indeed it reduces the duty cycle to the correct value to maintain the desired output voltage.

    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?
    The capacitor on pin5 is essential for the average current limiter to work. Donīt disconnect it.

    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)
    Letīs 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".

    Klaus
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    Re: Variable current sense, will this work?

    Quote Originally Posted by KlausST View Post
    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.
    You are right, but I think it is important to differentiate between passive and active and give the guys at Linear/Analog some credit too. A passive LPF is RC and a passive Integrator is also RC and since they are passive there will be attenuation. So the RC in the LT1339 could be a passive integrator they refer to as simply "integrator". The op-amp thing on the right to the integrator must be a voltage comparator only, so it has nothing to do with the integrator as such.

    Quote Originally Posted by KlausST View Post
    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)
    Letīs 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".
    So if I opt for the safe way is this then what you mean?

    Click image for larger version. 

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    (Not an actual circuit. For illustration purposes only)



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