MOSFET linear regulator circuit

[VEGETA]

The part I meant is LT6106 which is here: http://www.linear.com/product/LT6106  - Not the one you spoke of, and it doesn't show bandwidth in the specs.

__

You mean by "The current sensing amp only works if the voltage at the sensing shunt is higher than the level for the output resistor." is that the output resistor is the 0.1R shunt or the load itself?

I tested your circuit with the input voltage source set to 25v (constant) and it was good, except for low currents. It doesn't regulate anything less than 50mA or so, I don't remember.

I didn't get your idea about floating supply but if you mean making the CC op-amp connected to the negative supply instead of ground, then what is the gained benefit of this?

If omitting the op-amp will cause stability issues then it might not be the case. All I suggested is the use of high side current sense amplifier instead of the previous way... not to mention that the previous way had issues with <80mA CC mode.

I am confused now, I want anyone of these 2 to work especially with low currents. If I have to use LT1678 for the 2 CC op-amps (and the circuit works for low current sets) then I am forced to = no option. However, the problem now is that the circuit doesn't limit under 100mA. Also, it is not very accurate, like putting 300mA limit (1R load) will give around 299mA through that load (between out+ and out1) while giving like 330mA through the current shunt to ground.

[VEGETA]

one strange thing I forgot to mention, which is the input voltage of the mosfet must not get bigger than certain values. I mean, 22v is ok but 25v is not! the circuit doesn't operate well when it is big. No reason for that!

Sorry for confusing you even more, but stuff are getting complicated for me. I thought it is over but this low current thing made it very bad!

[salbayeng]

Kleinstein, a slip of the keypad there, the current sense amp is the LT6106.
Similar parts are the ZXCT1009 and ZXCT1010. Check your distributor catalog for "current sense amplifier" and "difference amplifier"
(Always check the min and max operating voltage of the part is suitable)

[VEGETA]

Kleinstein, a slip of the keypad there, the current sense amp is the LT6106.
Similar parts are the ZXCT1009 and ZXCT1010. Check your distributor catalog for "current sense amplifier" and "difference amplifier"
(Always check the min and max operating voltage of the part is suitable)

I tried the correct part but as I told you, simulation gave wrong results when I put it in our circuit. Now either we work on the circuit with high-side sensing using LT6106 making it stable or ditch it and return to the previous one and solve low current issue. Both are monsters, like choosing between 2 devils xD.

[VEGETA]

Does solving the problem of low currents lies in ac analysis only? I managed to do ac analysis but didn't know exactly what to probe. If I want to optimize say the CC loop, what should I probe? is it the output voltage of the cc regulator divided by it's feedback? and the same for the diff amp for the current?

Next step is determining a solution. From what I saw in this thread, adding caps and resistors seems the only way to solve these problems as it adds zeros to the loop. So I believe I should determine the frequency at which a certain loop gets unstable (say CC one) then use the formula to determine values for the C and R needed. Also, I know from my knowledge that a series resistor after the op-amp output does increase stability but determining its value is something I don't know yet... besides trial and error noobish technique

That leaves us with driving high capacitance loads and to check the cross over between CC and CV. I believe solving the previous stuff is the priority.

I hope someone knows about this, I'd like to learn how to solve these issues. The whole design is on-hold waiting for this thing to get stable... although other designs like the eez supply uses exactly the same configuration xD.

[Kleinstein]

The classical way of doing the AC analysis for loop stability, is to add an extra voltage source in the loop and look at the loop gain.

If one has a reasonable starting point, to have the regulator stable at least with an easy load, I personally prefer a different way. Looking at the output impedance will show the regulator performance and will also show if there is instability with any load impedance (especiall capacitive). If the regulator goes unstable or come close to it,  the output impedance will show a sharp resonance at that frequency. Besides a resonance directly visible in the curve, there is also the possibility to get a resonance if the external load is changes to a low low capacitor in the output impedance is close to a perfect inductance (or with an external inductor if the regulator haves like a low loss cap). True instability (undamped oscillation) can occur if the regulator circuit has more than 90 deg phase shift in any direction. So looking at the output impedance (and especially the phase shift) one can check for problems at any possible load impedance for just a single curve. One can also see in which range critical load capacitance can be.

The output impedance curve one aims for has usually a more or less fixed shape:
At high frequencies it is the capacitance at the out put that sets the impedance. There is a maximum in impedance than where the regulator sets in - for the usual lab supply this is somewhere in the 1 Ohms and 100 kHz range. A rather peaked maximum in impedance is indicating ringing and should be avoided. To lower frequencies the impedance goes down like 1/f for the very simple regulators. However just this single slow down to very low frequencies would mean a near perfect inductive behavior and thus ringing and possible instability with capacitive loads. To counteracts, usually a kind of horizontal step is added to the impedance curve, somewhere at about 1/10 - 1/100 the maximum value and about a decade wide.
Knowing this shape to aim for, it is usually not hard to adjust the caps for compensation - even if this means a little try and error.

Especially with the fet output stage one still has to check with different DC currents, as the output stage characteristics depend on the current level. At lower current the output gets slower, and the open loop output impedance gets higher - both makes regulation more difficult, especially with a MOSFET output stage. So one has to find a kind of compromise to work well at low and high currents - I don't know of there is a way better than (a guided) try and error here.

For the CC loop, the corresponding property to look at is output conductance: so have a voltage source as a load and look at the AC output current (e.g. current at the source). For the CC mode, such a regulator is not very good anyway. So much of the output conductance is set by the output capacitance anyway. So there is no need to optimize to much faster than that. The CC compensation might be more important for the CV/CC transition than steady state CC mode.
If CC mode has stability problems this could be trouble with the output stage in first place.

For the full program, the first step is usually looking at just the output stage by it's own. Here two properties are important first if the open loop output impedance. This should have less than 90 degree pase shift as well. If needed for higher frequencies an RC snubber can be used, making it part of the output stage.

The second, more important test of the output stage is using the output stage to drive a near perfect short (e.g. current sink + large cap in F range), and look at the trans-conductance. This should be well behaved (not too much phase shift) and sets the speed limit for the regulator. This gives a rough idea where later loop gain must be below 1 in CC mode, and where the output capacitance should take over.

With having a floating current regulator, I meant having the OP(s) for that powered from a supply between MOSFET source and something like 3-4 V below that. As the MOSFET output stage does not work that good at very low currents, it is a good idea to have a constant current sink as a minimum load anyway. This constant current could also provide power to the CC circuit, with just a zener to give them a stable supply. This needs the constant current to work form something like -4 V instead of -0.5 V. In a first stage I would not care so much about how to transfer the measured current and the set-point - this adds to the circuit but is not critical or really hard. This type of circuit is not very common - likely because in old times there where not that many OPs to work with a low supply like 3 V. With low accuracy, as a simple current limit it is common to just have the transistor at the emitter resistor to deviate base current. So the only new this is to combine precision and speed.

For the OPs choice there is no absolute need to be faster than the LM358. Current regulation is slow in original circuit anyway and thus there will be current overshoot in CV/CC transition and this might very well need the extra protection. Faster OPs help here only a little - the type of circuit with the GND reference current limiter is what makes current limit slow. The floating current regulator is a way to speed up current limiting, even without using fast OPs - it just needs extra circuitry and the negative supply. With a negative supply one also would not need single supply anymore.

[VEGETA]

Hello again,

You keep speaking about output impedance and its curve using ac analysis but I didn't find a way to see that, let alone learn about it and check it.

How to look at output stage alone? it is just the mosfet, do you mean cc and cv are not with it? you mentioned that the problems with CC can exist in the output stage itself, but cc is slow by default and I am not too fond with keeping this design as is if I can get it faster.

can you show or do you know an ltspice example for that? I mean make some adjustments to solve one problem. I will study Liv's design more to get some ideas... I might do the effort and mimic eez supply in ltspice to check for it.

does these designs use what you call floating regulator? as they use negative supply.

[Kleinstein]

To see the output impedance of the circuit, one has an AC current source at the load (usually together with the DC load current and with AC amplitude = 1 in the settings). Than in the AC voltage for the output gives the AC voltage for a simulated 1 A AC current - so this is directly the output impedance in Ohms, just with the little unusual dB scale. So 0 dB is 1 Ohms, -20 dB in 0.1 Ohms and so on.

To test the output stage alone, this is without the OPs, so a reduced circuit. I attach such a spice file for the trans-conductance test. Performance of the shown stage is acceptable above about 5 mA. Below might need some care, or sufficient current to a sink.

The second file attached is the principle with a floating current regulator, but ground base voltage regulator. The inductance in series to the shunt is more or less the typical parasitic inductance - it helps to prevent current overshoot. The current sources used in the circuit still need to be replaced with actual circuit. The the current limit is set to 1000 times I4. That circuit should work with relatively slow OPs (e.g. LT1013 is similar speed as LM358). The OP for CC mode could be something like an MCP6002 (though poor precision).

[VEGETA]

I've spent sometime checking the principle of your new circuit, and I think I finally understood it. Here is my explanation:

CC Loop:

you have an op-amp which has Vout+ as its positive terminal and it is always above the negative terminal by a voltage that equals exactly the voltage difference around the current shunt resistor. For 1A output current -> 100mV drop voltage which means the positive terminal of the CC op-amp is 100mV more than the negative one at all times.

Now to set a constant current limit, you need to inject this 100mV in the negative input to make it equal to the positive one (or slightly above) to get the diff amp to put the negative rail as its output and thus drive the MOSFET gate down to the negative rail (-5V) which is a CC operation.

To do so, you put a current source of 1mA --> 1mA*100R = 100mV which makes the negative terminal of the op-amp equals (V_I+) - 100mV at all times. Meaning, you only allow a 100mV drop voltage to exist on the current shunt (=1A) and if it exceeds this then it won't be allowed.

Now that is a brilliant idea! and it is so damn accurate even in very tiny currents!

Aside from that, I didn't understand the purpose of the other current sources (to make minimum current or stability) but since they are fixed currents I can make them using the famous LM334! No worries!


However, the problem is making the current source I4 which is the critical one! how to make it adjustable and software controlled? LM334 won't do it here plus I need something like 1uA resolution...

I thought of making an op-amp comparator (or something like a dummy load constant current sink) but I didn't get a good idea. I want something to be controlled via MCU as well as reading the output current to the MCU. Do you have any idea? I tried just adding a voltage below the 100R but it didn't work properly.

Is this the only thing left in the circuit? I hope xD

____

now please see this image of your circuits: http://imgur.com/a/AK6Pk

I ran the ac analysis and probed the points in the image. Where and what are the points that you spoke of? I mean how do you probe to see what you seek in the bode plot?

You spoke about the output impedance being 1v AC/1A AC but what is the point that you drew on the plot? and the other schematic as well. This is the only thing I am still totally confused with until now. You've been cooperating nicely with me and I am doing much effort in learning and trying but this portion is still not understandable for me.

To further explain the core of my questions: how did you know that the circuit behaves well above 5mA?



______

Domo Arigatou gozaimasu Kleinstein-sensei!

[VEGETA]

Wait... the MCP6002 doesn't accept higher voltage than 6v as its power rail, so how is it accepted here? aren't there any better Linear tech part?

in this design there are 2 op-amps, i would really like to pick a dual op-amp for both of them if possible... will LT1013 be good enough for both? as for the rest of the circuits which needs something like buffering and stuff like that, the lm324 will be good enough in my opinion.

[Kleinstein]

An important part of the version with floating regulator is, that the two OP have a different supply voltage: the one for the voltage regulation uses GND and 30 V so this could be an LT1006 (single to LT1013) or LM358. If the negative supply is used too one could also non single supply OPs like 741 / OP07.

The current regulating OP has a different supply, relative to the output (before the shunt). This is only something like 3 to 5 V and the OP needs to work at it's upper supply limit. So a low voltage Rail-Rail OP like the MCP6001 is correct here. On can't use a dual OP for both.

The current sources for the supply part don't need to be that accurate. More like a 2 transistor version or current mirror to minimize the voltage lost.

The current source I4 to set the current limit is also used to make is possible to transfer the signal from GND referenced to the output referenced part of the circuit. So this would be more like an OP + N-MOSFET (e.g. 2N7000) based one. So the second half of an LM358 for voltage regulation could be used here.

One might still want to have a current monitor, to transfer the measurend current down to GND.

[VEGETA]

But if you put MCP6001, its positive rail will be I+ which can go to 20v while it has a negative rail of fixed -5v. This will be 25v difference which is bigger than the 6v right? that was what I talked about.

I might see what to do for the current sources but I didn't get their goal yet, aren't they for minimum load or stability?

I tried doing the I4 thing as you said but didn't work properly for me. I put a nmosfet with an op-amp and a 1k resistor to ground. However, I retried and it seemed to work. Circuit in the attachments.

I still need to make it software compatible buy adding another op-amp before the positive input of the opamp. I originally wanted to make this not by current sink method of yours but with an op-amp that does an offset voltage, adds it to the I+ voltage to make the drop voltage necessary. However, why is it a must to have a current sink?


"One might still want to have a current monitor, to transfer the measurend current down to GND. "

what current to ground? you mean the output current? diff amp will do the job if it doesn't affect the loop.

[Kleinstein]

The current sink / source method is a way to get very good common mode suppression, just like the current sensing amplifiers do. One could alternatively use a differential amplifier of cause, if accuracy is not that important. For current sensing this is very real option, especially with a floating amplification before that. For transferring the set-point, I would still prefer the current source, as a common mode error could result in negative output resistance.

The current controlling OP has it's own voltage regulation with the zener diode - so the OPs voltage is limited to 4.7 V in the simulated circuit (the lowest voltage zener I found in standard parts list). It is only the current sink that is using the -5 V.  So it is only the current sink circuit hat will see the high voltage (e.g. up to 25 V volts or so). The current regulating OP is working between V_out and V_out-4.7 V.
Having the supply of that OP moving with the output voltage removes the need for a high slew rate and bandwidth.

The current source circuit in the spice file will work ok with a lower controlling voltage V3 (e.g. < 8 V, or better < 5 V). In the final circuit it one would need to have R1 going to the -5 V supply and use only something like 4 V (relative to the -5V)  max, as the output side could go slightly below zero too.

[VEGETA]

Here in the attachments a new version of the current sink for setting the CC stuff... I guess it is good enough (except that I picked a mosfet randomly). The set in the example is 100mV (from MCU DAC or filtered PWM) which means 1A output in the final circuit. However, since there are op-amps in this circuit, will it be stable or will it return the previous nightmare?


Now I saw the voltage on the CC op-amp, both rails are positive voltage! is that the "floating" supply? But why is that? isn't just connecting the negative rail is enough?

I didn't quite get the last paragraph but I think you meant relying on -5v rail instead of ground because the output (?) can go to negative slightly. i modified the circuit so you can put your additions as the circuit is so small, won't take time.

what is there left to do?


THANKS!

[VEGETA]

Now as for the negative supply, I found some charge pumps which are cheap around 0.75$ maxim part on digikey but they are only 50mA. How much current should it be to tolerate our circuit?

[Kleinstein]

The circuit would need something like 10 mA from the negative supply. It very much depends how much current is needed / wanted to make the MOSFET power stage behave good. So it can depend on the used MOSFET. It may be a little more if additional OPs use the negative supply, e.g. use TL07x instead of LT1013 for voltage regulation.
 
I don't like charge pumps very much, as they produce quite some noise on the input side - though there is switched mode regulator anyway. Depending on how the +30 V is generated, one might be able to combine this with the negative supply and maybe the supply for a controlling µC and display. One could even end up running the µC/ display from GND and -5V.

For the CC OP it helps to have it's supply floating (or better being relative to the output). Having only one side moving with the output (and have the positive side fixed to the 30 V) would cause more trouble, as the OPs supply could than change fast and OPs are not well specified for that. Also the supply current of the OP is still of some use if goint through the MOSFET - so why waste it.

[VEGETA]

I will generate the 30v by using a boost converter directly from the battery pack (8.4v -> 30v) which doesn't have to provide much current because it is for op-amps only.

I didn't find a cheap charge pump, here is the digikey search: http://www.digikey.com/products/en/integrated-circuits-ics/pmic-voltage-regulators-dc-dc-switching-regulators/739?FV=15c0002%2C112801ce%2C1f140000%2Cffe002e3&mnonly=0&newproducts=0&ColumnSort=1000011&page=1&stock=1&pbfree=0&rohs=0&quantity=1&ptm=0&fid=0&pageSize=25

The only part I found reasonable is MAX1720 which supplies -Vin and 50mA which is enough as you explained. It is 12KHz frequency which might be good low noise. I already have a 5v linear regulator to provide 5v for MCU and other stuff... I will use it as an input to this charge pump. both works fine.


The other current sources/sinks can be replaced by transistor... though I didn't get why they exist there.

[Kleinstein]

The boost converter to get 30 V might be able to also provide a negative supply, from an additional winding (making a transformer from the inductor). The 30 V also still need to supply about 5 mA for the MOSFET gate driver.

I3 is there as a minimum load for the MOSFET and to power the floating OP for the CC mode.

Using I2 instead of the 10 K resistor that was there is earlier circuits saves a little on the current from 30 V. One might get away with even less current than 1 mA. The 10 K resistor would draw 3 mA worst case and this current would also be missing at the FET - I3 would need to be higher too. Just a resistor might also work, if it can be more than 10 K.

[VEGETA]

Cheapest reliable boost IC on digikey is MIC2288 which can output up to 1.2A which is far far more than we want. You say we need 5mA for MOSFET gate and it can get it when I make the battery pack its input.

This part is straightforward, originally I wanted it to be more complicated by having one big boost followed by a buck as a pre-regulator. However, say boost is 80% efficient, and buck is 80% also then what makes the total efficiency? it is a bad idea... originally I didn't know a good SEPIC converter especially how to control its voltage directly by MCU. Then I figured out the way of using a digital potentiometer (I2C or SPI one) to control it as its feedback resistor. Then I found a suitable sepic converter which by all means a better choice here for various reasons.


Now getting a 30v rail to power the op-amps is not a big deal, one small boost is enough. I think this MIC2288 and other similar ICs like it is good enough. For the negative rail, I am gonna use MAX1720 or similar part taking the battery pack as its input, which is the simplest solution. I saw your solution of an inductor in the EEZ supply but it will require an extra linear regulator which won't be economical as well. What is your suggestion?

Using LT1013 for voltage op-amp and the current source controller is a good choice but I don't know if it is the same as LM358, also the op-amp that compares I+ voltage with the current set voltage (0-200mV = 0-2A) from the MCU is, I think, must be a little accurate since it is gonna deal with 1mV of accuracy so will LM324/LM358/LT1013 be good enough?

I searched Linear.com for suitable op-amps for suitable op-amps and found these: LT1800, LTC6261, LTC6220. You can check this page sorted by the cheapest: http://www.linear.com/parametric/Precision_Amplifiers#!cols_1006,1030,1075,1047,1021,2275,1004,1005!s_2275,0!gtd_!1047_yes!1021_%3E=2!1075_yes

Do you find a suitable cheap part in there? I tried to match your original suggestion.

[Kleinstein]

The OP used for current regulation needs to be a low voltage RR type. So I think the cheap MCP6002 should work, if we accept the rather high possible offset, that might limit the lowest set current. There are a few similar, more accurate OPs available (could be even Az types (e.g. MCP6V2x), but this is kind of overkill). The second OP could be of good use to amplify the measured current signal (e.g. 10-50 times). Alternatively the second OP could be used for a current sense amp with an extra p-FET.

The OP for voltage control does not need to be accurate, so the LM358 is good enough - the LT1013 is about the same speed, but much more accurate and expensive. The LT1006 is the single version of the LT1013. As we have a negative supply one could also use non single supply OPs - so many more to choose from. A low supply current might be important here, as these are powerd from the 30 V.

One could use a second OP as a differential amplifier to get the measured current signal relative to GND - here the negative supply might be needed. With a current sense amplifier one would likely need it to get the signal from relative to the negative supply back up. The OP for the I4 source (set -current) could be powered from something like the µC supply and the negative one - no need for speed and high accuracy.

[VEGETA]

Searching Linear parts for the CC op-amp that is similar to MCP6002 resulted in failure xD. I tried LTC6088 in LTSPICE and it showed some little oscillation near the end of the on-time (2A) around the 1A limit (30mA peak-to-peak) but LT1013 didn't work too. The other op-amps mentioned by me in the last post didn't work too despite being better quality than MCP6002.

I tried the op-amps and these are the ones which worked: LTC6085 (the quad version of LTC6084 -> not the same drawing = hard to re-wire xD) and LT6106 which is a current sense amp. LTC6084 is 2$ from digikey which is acceptable. I could get cheaper parts but dunno if they work since ltspice doesn't support them. I am not a Linear.com fanboy xD


You mean using the 2nd op-amp to actually measure the output current with extra p-fet? how is that assuming we need a diff amp for current sense, not to mention that it will be using the same power supply of I+ and -5v (4.7v zener). Speaking of this negative rail supply... it is only -5v but in simulation I find the negative rail of the op-amp goes to even beyond -5v. How is that?

The op-amp for I4 is supplied from +5v and ground, why there is a need for negative rail which is only a -5v charge pump? Also, for the differential amplifier for current sensing, it can be powered from +30v and ground without the need of negative rail... won't it be like +-35v as a total voltage difference between the rails?

[Kleinstein]

For the current sense amplifier the LTC6084 should be OK - though the small case might be inconvenient. The CC OP needs to work at the upper rail - so usually an Rail to Rail OP, this is why the LT1013 does not work.
The main downside of the MCP6002 is the large offset - one could get around it with adjustment and 2 extra resistors (in case the offset is negative).

To sense the current, I see 3 options:
1) Use a special current sense amplifier, a resistor to the -5 V and than an differential amplifier (powered from batt (e.g. 7.2 V) and -5 V) to bring the signal back to GND level.
2) Build the current sense amplifier from an OP (e.g. 2nd half of CC) and P-Channel MOSFET (optional PNP darlington) and rest as before.
3) A third, more low cost option would be to have an OP (e.g. 2nd half of CC) to amplify the current signal (e.g. x 20) and than have a differential amplifier (powered from +30 and GND) to bring the Signal to GND level.

For the options 1 and 2 the differential amplifier should need a negative supply as the inputs would be at a negative voltage. For the option 3 the amplifier does not need the negative supply.

For the I4 current source one would normally need the -5 V supply for the OP too. The point is that the current sink does need to go all the way to zero volts and maybe even a few mV negative.  One might get away with a less accurate modified circuit, without the negative supply for the OP.

One could use a modified gate drive circuit, that can get away with less current (e.g. 1 mA instead of 5 mA) there.

[VEGETA]

though the small case might be inconvenient.

what is that?

Price for the LT6084 is ok so a good device with reasonable price. No need for MCP6002 here I guess. Still you didn't give me the answer of the power supply for this op-amp (2nd op-amp in the IC rather than current set), if it is a floating with mostly positive supply for both + and - rail... So how will it suite the other op-amp in the same IC which is meant to do something else?

1) Use a special current sense amplifier, a resistor to the -5 V and than an differential amplifier (powered from batt (e.g. 7.2 V) and -5 V) to bring the signal back to GND level.

why the need of negative supply? it is gonna be I+ and I- difference so it won't go negative, right?

2) Build the current sense amplifier from an OP (e.g. 2nd half of CC) and P-Channel MOSFET (optional PNP darlington) and rest as before.

Here is what I asked about, using the 2nd CC OP which has the same floating supply... how can it work? And, if I am gonna use a diff-amp why the need for p-mosfet?

3) A third, more low cost option would be to have an OP (e.g. 2nd half of CC) to amplify the current signal (e.g. x 20) and than have a differential amplifier (powered from +30 and GND) to bring the Signal to GND level.

Well, this is weird. Where is the "current signal"? it is mainly the I+ voltage and I- voltage but it is not like this in our loop. What is the "bringing signal back to ground using a diff-amp"?


My first thought was the following: I already have a /10 voltage divider (0.1% resistors) for Vout (or I-) node to get the output voltage... so it would be good to make yet another voltage difference on the I+ node by a factor of /10. Now get a simple diff-amp between the 2 which, most importantly, not a part of the control loop so I guess it won't affect stability. Let us say it is 2A and the voltage is 20v out this means something like 20.2-20 = 0.2v now a x10 diff-amp = 2v so 2V = 2A as the original simple design was. I guess the other LM358 can do the job or any op-amp really, so we can use the other op-amp in the IC that contains the CV op-amp to do it. It is powered from anything like 30v or 5v to ground. If it won't affect the CV op-amp, I can even hook it to -5v supply.

^ What do you think about that?

For the I4 current source one would normally need the -5 V supply for the OP too. The point is that the current sink does need to go all the way to zero volts and maybe even a few mV negative.  One might get away with a less accurate modified circuit, without the negative supply for the OP.

That was not my question, it was that the negative rail of the op-amp went below -5v while we don't have such a capability! you know the -5v is all the limit.

My guessed answer is: because we used a current source model instead of a true realistic circuit.

[Kleinstein]

The LT6084 comes only in rather small form factors, like TSSOP8 or similar - no DIP8 or SO-8. Otherwise it is a better replacement for the MCP6002 (especially much tighter offset specs).

The power supply of this OP is relative to I+. To a certain degree we can chose something like 3 to 5 V with the zener diode. So we have something like I+ and I+ -4V as there supply. The power comes through the current source (I3).  A second OP with the same power can be used to amplify the voltage over the shunt or to build a current sense amplifier by hand (OP+P-FET). It is classical high side current sense amplifier circuit that needs a P_FET or with lower accuracy a PNP darlington. The only difference to the normal current sense circuit is, that the OP is powered from the supply relative to the shunt and not to GND - this is not a problem at all, but more of an advantage for the OP.

Because I+ and I-  can go essentially all the way to GND level, the output current of current sense circuit can not flow through a resistors at ground level - one will need a more negative level here. To bring the signal from that more negative level to the ADC would require something like a difference amplifier, but other options are there. Still this makes the first two options rather complicated.

The 3rd. option is using a difference amplifier to measure (for display) the current. To reduce the required common mode rejection of the differential amplifier it is a good idea to amplify the voltage over the shunt first. The is relatively easy, as we already have the supply relative to the shunt for the CC loop. It is just easier to have an extra x 10 stage here than to go for 0.1% or better resistors at the difference amplifier. So this is relatively close to your suggestion, just in not using a divide by 10 first, but the extra times 10 at the shunt. The difference amplifier could than be something like times 1 (so 4 equal resistors) and still be powered from 30V/GND (15 V would be sufficient).

The possible lower than -5 V voltage in the simulations could be due to the current source model used. I did not see it, and operations should not need it. How much negative supply is needed, depends on the voltage for the 2 floating OPs - we need something like 0.5-1 V more than the zener voltage.

As a 4th option one could also power the current display from the "floating" 3-5 V supply - there is a current of some 5-10 mA available anyway.

[VEGETA]

So your idea is to use the other LTC6084 (powered from the I+ and zener which is referenced to -5) to amplify the I+ and I- signal by 10? this is true because max current of 2A will output 0.2v drop over the shunt -> x10 = 2v so that 1v = 1A. However, you mentioned a diff-amp powered from 30v to ground after this... what is it? My guess is the floating supply from I+ and zener is always gonna give +5 or so voltage difference which is enough to output 0-2v signal, why do we need an extra diff amp op-amp?

The power comes through the current source (I3)

To bring the signal from that more negative level to the ADC would require something like a difference amplifier, but other options are there

I just didn't get this.

As a 4th option one could also power the current display from the "floating" 3-5 V supply - there is a current of some 5-10 mA available anyway.

I guess this is related to the first suggestion of yours. I got the idea of getting the difference of I+ and I- then amplify it by 10 to get 2v max output when 2A max current occurs. Now what is the 4-10mA that is already available?

BTW, can I3 be made of a cheap op-amp and a 2n2222 transistor (or nmosfet like I4)? or should I use transistors for it? it is referenced to -5v no 0v.