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Need Help Current Source for 5A and 0,1V,,,Is It Possible?

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oneitusatu

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Hi,

i am designing Low Resistance Measurement,my problem here is i need power supply that provide current source 5A and very low voltage almost zero(if it possible).

the resistance i would like to measure is in uOhm,,

1)i was planning using ic regulator for current www.micrel.com/_PDF/mic69502.pdf ,, anyone have experienced using it?and wwhat is the problem?

2)i was planning using the dischargin of capacitor,so when caps is full then we have microseconds time at that discharging of the caps for measurement,ive tried a simple experiment with it and it could provide very low voltage and because of very low resistance in caps it could provide high current too?but my problem here is in measuring the voltage & ampere at that microseconds using microcontroller?have anyone tried the same thing?

3)what is the side effect if we have high drop out voltage?because my output would be at least 0,1volt?

4)is there any other/simpler way to get my specification current source?


Rgrds
 

oneitusatu said:
Hi,

i am designing Low Resistance Measurement,my problem here is i need power supply that provide current source 5A and very low voltage almost zero(if it possible).

the resistance i would like to measure is in uOhm,,

1)i was planning using ic regulator for current www.micrel.com/_PDF/mic69502.pdf ,, anyone have experienced using it?and wwhat is the problem?

2)i was planning using the dischargin of capacitor,so when caps is full then we have microseconds time at that discharging of the caps for measurement,ive tried a simple experiment with it and it could provide very low voltage and because of very low resistance in caps it could provide high current too?but my problem here is in measuring the voltage & ampere at that microseconds using microcontroller?have anyone tried the same thing?

3)what is the side effect if we have high drop out voltage?because my output would be at least 0,1volt?

4)is there any other/simpler way to get my specification current source?
Click to expand...

It is possible to use MIC69502 as a current regulator as when you look at its internal diagram it consists of 3 blocks: a 0.5V reference, an Opamp (as error amplifier) and a PNP power transistor. The selection of the current sensing resistor will be similar to the LM3157 – see picture below – ref.=0.5V.
If you don’t have MIC69502 you can replicate it just by using those three blocks mentioned above …
Be aware though that all these options will require a supply voltage, which in the case of MIC6905 is around 5-6Vdc. Without it no sensible regulation can be achieved ..

I don’t think you can relay on a charge of a capacitor to precisely measure resistance. I’d rather add a dc switch to the circuit mentioned above, turn the circuit on, allow it to stabilise, take a measurement (read voltage at constant current) and turn the switch off ..

:wink:
IanP
 

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Sure it's possible, but just keep in mind that the "efficiency" of the supply is going to be extremely low with extremely low resistance loads. It can be done both with linear regulators or SMPS (SMPS will be much less lossy than linear regulators, but more complicated).

The capacitor idea may work in theory for measuring resistance, but in reality it won't work for uOhm loads because the variation in the ESR and capacitance (with temperature and aging) will completely swamp out the effect of the actual load resistance.
 
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    IanP

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Maybe you could employ a different measurement strategy? Maybe it would be possible to create some resonant L/R/C circuit with the measured resistance. When excited with ac signal the changing resistance would effect in changing complex impedance of the circuit. Maybe you could measure the resistance by locating resonant peak? I don't know if this is feasible (component-value-wise) but it's a different idea.
 
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    IanP

    Points: 2
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IanP said:
It is possible to use MIC69502 as a current regulator as when you look at its internal diagram it consists of 3 blocks: a 0.5V reference, an Opamp (as error amplifier) and a PNP power transistor. The selection of the current sensing resistor will be similar to the LM3157 – see picture below – ref.=0.5V.
If you don’t have MIC69502 you can replicate it just by using those three blocks mentioned above …
Be aware though that all these options will require a supply voltage, which in the case of MIC6905 is around 5-6Vdc. Without it no sensible regulation can be achieved ..

I don’t think you can relay on a charge of a capacitor to precisely measure resistance. I’d rather add a dc switch to the circuit mentioned above, turn the circuit on, allow it to stabilise, take a measurement (read voltage at constant current) and turn the switch off ..

:wink:
IanP
Click to expand...

for the supply voltage 5-6 Vdc,do i need battery or another power supply for it?is there any effect of regulating regulated voltage?but if i use batteries then there'll be lot of batteries or charging the batteries since the high ampere produced,am i right?


mtwieg said:
Sure it's possible, but just keep in mind that the "efficiency" of the supply is going to be extremely low with extremely low resistance loads. It can be done both with linear regulators or SMPS (SMPS will be much less lossy than linear regulators, but more complicated).

The capacitor idea may work in theory for measuring resistance, but in reality it won't work for uOhm loads because the variation in the ESR and capacitance (with temperature and aging) will completely swamp out the effect of the actual load resistance.
Click to expand...

what is exactly ESR?i didnt find more information on that,can you explain it more?and what is the correlation with it

poorchava said:
Maybe you could employ a different measurement strategy? Maybe it would be possible to create some resonant L/R/C circuit with the measured resistance. When excited with ac signal the changing resistance would effect in changing complex impedance of the circuit. Maybe you could measure the resistance by locating resonant peak? I don't know if this is feasible (component-value-wise) but it's a different idea.
Click to expand...

wow nice idea but i still dont get it and dont have any experience in playing with impedance thing,would you please post some articel and circuit example for any project have done with it?
 

oneitusatu said:
for the supply voltage 5-6 Vdc,do i need battery or another power supply for it?is there any effect of regulating regulated voltage?but if i use batteries then there'll be lot of batteries or charging the batteries since the high ampere produced,am i right??
Click to expand...

In any case you do need a power source, a battery or a ps ..
If your circuit is "smart" and takes high-current measurement over short period of time, say, 1ms-5ms, then a battery should last for some time ..
:wink:
IanP
 

IanP said:
In any case you do need a power source, a battery or a ps ..
If your circuit is "smart" and takes high-current measurement over short period of time, say, 1ms-5ms, then a battery should last for some time ..
:wink:
IanP
Click to expand...

you said that on previous post about the replicate regulator that i could make it by myself,so how i pick the right transistor for what my requirement?
 


IanP said:
I'd consider p-channel MOSFET, eg.
www.irf.com/product-info/datasheets/data/irfh5053pbf.pdf
and increase the supply voltage to >=9V (Vgson) ..
:wink:
IanP
Click to expand...

how about the schematic for this transistor?because i dont see much typical application for this mosfet in datasheet
especially for current source,would you give me the hint how to make it,please?
btw why do you suggesting using p-MOSFET?

what for the precision op-amp i should use?and how to find the right component for this?

thank you
 
Last edited:

oneitusatu said:
what is exactly ESR?i didnt find more information on that,can you explain it more?and what is the correlation with it
Click to expand...

This may be worth looking into, to help with what you're doing.

ESR stands for Equivalent Series Resistance. It has to do with a capacitor's internal resistance when charging and discharging. ESR is not the same as leakage resistance.

Even a very small resistance in a cap (a tenth of an ohm) can reduce its performance.

There are ESR meters available, and there are credible messages of praise from technicians who claim the meter lets them quickly diagnosis bad caps that they would ordinarily have called good, and they can repair equipment that would previously defy their best efforts.

An ESR meter works by injecting a pulsed waveform under 300 mV amplitude. This is coming through a low ohm resistor (maybe 5 ohms).

Because the applied voltage is so low, it does not turn on any diodes in neighboring wires.

The probes are applied to the terminals of the cap. A good capacitor draws away the entire signal because it has very low ESR. An oscilloscope reads just about zero.

A capacitor which has any amount of ESR causes the oscilloscope to show a reading.
Example, 0.1 ohm ESR will cause a reading of 300mV times .1/5, equals 6 mV.

That's the concept for how to make a poor man's ESR meter. It is not far from being a low ohms meter as you plan to make.
 
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To refer to the problem specification in the original post, are there any reasons to do a µs & high current measurement? This would be the case when measuring non-linear components, e.g. transistor or diode saturation voltages. For linear resistors, you get most likely a more accurate measurement with less circuit effort by performing a moderate time duration (10 ms up to s) measurement at a more convenient current current level.
 

sorry ive been busy all this time,and i still cant figure it out how to make this PS,

BradtheRad said:
This may be worth looking into, to help with what you're doing.

ESR stands for Equivalent Series Resistance. It has to do with a capacitor's internal resistance when charging and discharging. ESR is not the same as leakage resistance.

Even a very small resistance in a cap (a tenth of an ohm) can reduce its performance.

There are ESR meters available, and there are credible messages of praise from technicians who claim the meter lets them quickly diagnosis bad caps that they would ordinarily have called good, and they can repair equipment that would previously defy their best efforts.

An ESR meter works by injecting a pulsed waveform under 300 mV amplitude. This is coming through a low ohm resistor (maybe 5 ohms).

Because the applied voltage is so low, it does not turn on any diodes in neighboring wires.

The probes are applied to the terminals of the cap. A good capacitor draws away the entire signal because it has very low ESR. An oscilloscope reads just about zero.

A capacitor which has any amount of ESR causes the oscilloscope to show a reading.
Example, 0.1 ohm ESR will cause a reading of 300mV times .1/5, equals 6 mV.

That's the concept for how to make a poor man's ESR meter. It is not far from being a low ohms meter as you plan to make.
Click to expand...
yes ive been doing research for this ESR and i didnt found ESR exmple that suitable with my specification,i need an accurate microOhm measurement

FvM said:
To refer to the problem specification in the original post, are there any reasons to do a µs & high current measurement? This would be the case when measuring non-linear components, e.g. transistor or diode saturation voltages. For linear resistors, you get most likely a more accurate measurement with less circuit effort by performing a moderate time duration (10 ms up to s) measurement at a more convenient current current level.
Click to expand...

i was going to measure resistance in a contact,and the ordo its on microOhm

---------- Post added at 17:21 ---------- Previous post was at 17:18 ----------

Please if anyone know the information of current source,and how to determine the right component ?
what transistor should i use for current source 5a and 0,1 v other than http://www.irf.com/product-info/data...rfh5053pbf.pdf??and please tell me how to configure it

Thanks
 

i was going to measure resistance in a contact,and the ordo its on microOhm
Click to expand...
Yes you told about microOhm before. But there's a trade-off between measurement duration and voltage measurement accuracy, due to different mechanisms: e.g. amplifier noise, current source settling time, circuit inductance.

At the end of the day, it may be easier to achieve the intended resistance resolution with lower current and longer measurement time. Looking at the key parameters of industry standard µOhm meters can give insights.

As I already mentioned, a pulse measurement can't be avoided if a non-linear resistance is involved, e.g. with power semiconductors. As far as I'm aware of, this isn't the case with a contact resistance measurement.
 

Perhaps I'm missing the point but why does the current have to be so well regulated?

When measuring very low resistance in the past, admittedly in the low mOhm rather than uOhm range, the technique I used was four wire differential measurement. The current was passed through a known very low value resistor before the part under test, the voltage across that resistor was then compared with the voltage across the measurement points. Screened wires were used to carry the voltages to a remote amplifier. Using Ohms law the difference in voltages could be used to calculate the unknown resistance regardless of the current as long as it was high enough to provide a reasonable measurable PD. It also eliminated cable resistance which was many times that of the devices I was measuring. Because the current isn't as important, switching it on and off is much easier.

Brian.
 
FvM said:
Yes you told about microOhm before. But there's a trade-off between measurement duration and voltage measurement accuracy, due to different mechanisms: e.g. amplifier noise, current source settling time, circuit inductance.

At the end of the day, it may be easier to achieve the intended resistance resolution with lower current and longer measurement time. Looking at the key parameters of industry standard µOhm meters can give insights.

As I already mentioned, a pulse measurement can't be avoided if a non-linear resistance is involved, e.g. with power semiconductors. As far as I'm aware of, this isn't the case with a contact resistance measurement.
Click to expand...

how do you say measuring low resistance with the lower current?according to basic ohm's law (V = I x R) you have provide lowest voltage and highest current as possible to get lowest resistance value?

and yes you are right,the idea about measuring with capacitance is not my concern anymore so i dont need to measure in uSec,
 

The simple point is that voltage resolution is also variable to some degree, e.g. a function of measurement time. Considering only resistor noise, you get a tenfold resolution improvement with a factor 100 measurement time. In addition, you have 1/f noise and offset voltages and offset drift, in so far there's of course a lower limit.

I also agree with betwixt, that the measurement problem should be understood as a ratiometric one.
 

betwixt said:
When measuring very low resistance in the past, admittedly in the low mOhm rather than uOhm range, the technique I used was four wire differential measurement. The current was passed through a known very low value resistor before the part under test, the voltage across that resistor was then compared with the voltage across the measurement points. Screened wires were used to carry the voltages to a remote amplifier. Using Ohms law the difference in voltages could be used to calculate the unknown resistance regardless of the current as long as it was high enough to provide a reasonable measurable PD. It also eliminated cable resistance which was many times that of the devices I was measuring. Because the current isn't as important, switching it on and off is much easier.

Brian.
Click to expand...

wahh,then i have misconcept before,so you're trying to say that we dont need high current and high voltage as my requirement in uOhm resistance measurement?
and i was planning using four wiring measurement too

then if i only put adc of microcontroller for measuring voltage before and after resistance i could get the resistance?its so that simple??do i need more circuit for that?

betwixt said:
Perhaps I'm missing the point but why does the current have to be so well regulated?
Click to expand...
because i need to generate a constant current to make the calculation ohms law easy , or i have misconcept again?
and afaik the current flows in closed circuit is same as source,then do i still need a current source for that?
i am confuse here,
 
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I'm still not sure about the constant current. I agree it makes the math simple but you can still do the calculation the way I suggested and it works at ANY current, it doesn't even have to be constant. The only requirement is that is should be high enough to allow reasonable voltage to be measured. You are correct that the current at source is the same as at the measurement point, ignoring input currents to the measurement amplifiers.

Four wire measurement is essential, the voltage drop in the wiring would otherwise be magnitudes higher than that across the measurement point. There may be advantage in placing the amplifier(s) close to the measurement location to further reduce noise in the system.

Brian.
 

betwixt said:
I'm still not sure about the constant current. I agree it makes the math simple but you can still do the calculation the way I suggested and it works at ANY current, it doesn't even have to be constant. The only requirement is that is should be high enough to allow reasonable voltage to be measured. You are correct that the current at source is the same as at the measurement point, ignoring input currents to the measurement amplifiers.

Four wire measurement is essential, the voltage drop in the wiring would otherwise be magnitudes higher than that across the measurement point. There may be advantage in placing the amplifier(s) close to the measurement location to further reduce noise in the system.

Brian.
Click to expand...

how do i pick the right amplifier for that??
i still confusing about picking the right component based on my requirement,any links for each component?

and my other question
why would we still need a current sense or voltage sense circuit/ic if we already have the adc of microcontroller?is it just a noise or what?anyone please explain it further
 

I think you are underestimating the problem or not fully explaining it. Before we go further, can you tell us what exactly it is you are measuring that has such low resistance?
Give us some idea of the range of resistance you want to measure, it make a big difference to the way the amplifier is specified.

To show you worst case scenario, for a normal 10-bit ADC with 5V reference to show full scale while measuring 1 uOhm would need a current of 5 million Amps !

Brian.
 

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