A Non-Linear Load with the relation of i=(v^3)/R

[Poyan100]

Hello Everybody
I need to design a load with the following relation between voltage and current:

v=R*(i)^(.33)

In fact, I need to have a system that its current is a function of its voltage

i= (1/R)*v^3.

Does anybody have an idea how I can design this circuit?

An idea that I have is to design a current source that its current is a function of its voltage based on the given equation. But I do not know how to do that.

I really appreciate your helps


Regards

Poyan

 

[KlausST]

Hi,

working voltage unipolar (DC) or bipolar (AC)?

What voltage range?


for unipolar:
I´d build a DAC controlled current source. And a ucontroller with ADC.

Measure the voltage with ADC - calculate current with micro - output value to DAC.

Did i understand it right?
0V --> 0A
0.1V --> 1mA
0.2V --> 8mA
0.5V --> 125mA
1V --> 1A
2V --> 8A
5V --> 125 A
...

Klaus

 

[Poyan100]

Hi,

working voltage unipolar (DC) or bipolar (AC)?

What voltage range?


for unipolar:
I´d build a DAC controlled current source. And a ucontroller with ADC.

Measure the voltage with ADC - calculate current with micro - output value to DAC.

Did i understand it right?
0V --> 0A
0.1V --> 1mA
0.2V --> 8mA
0.5V --> 125mA
1V --> 1A
2V --> 8A
5V --> 125 A
...

Klaus

Hi Klaus
It is for an AC system. You understanding about the relation between current and voltage are correct.
Also, thank you for your suggestion. Do you know how I can make it as a integrated component in a circuit. Imagine that I want to take out a resistance form my system and replace it with a circuit that has this physical behaviour.

 

[KlausST]

Hi,

again:

What voltage range?
Or
What current range?

******************
AC...
can you use a rectifier?
What frequency range?

Klaus

 

[Poyan100]

Hi,

again:

What voltage range?
Or
What current range?

******************
AC...
can you use a rectifier?
What frequency range?

Klaus

Hi
The voltage is sinewave with amplitude of 2 volt and frequency of 25 Hz. The current then is i(t)=R*v(t)^3. R is a constant value and can be selected by ourselves. So, the current range is tunable based on our constraints.

Cheers

 

[KlausST]

Hi,

current is not fixed.. then what max. current do you need?

What precision do you need?
(for example with max 10A a in this application a precision of 1mA is hardly possible)

Can you use fixed resistors that can be switched ON by a simple voltage comparator logic?

Klaus

 

[Poyan100]

Hi,

current is not fixed.. then what max. current do you need?

What precision do you need?
(for example with max 10A a in this application a precision of 1mA is hardly possible)

Can you use fixed resistors that can be switched ON by a simple voltage comparator logic?

Klaus

Hello
Yes. That is why I am saying we can select the R to have current in desirable range. For instance lets assume R is 10 , then the current would be
0V --> 0A
0.1V --> 0.1mA
0.2V --> 0.8mA
0.5V --> 12.5mA
1V --> .1A
2V --> 0.8A

I hope it helps. Now we need to think about the precision. Maybe we cannot have this profile for all ranges, for example we can have that relation only if our current is above 12.5 mA, it means that only for voltages above 0.5V we can have this relation i mean i(t)=v^3/10.
Regards
Poyan

 

[KlausST]

Hi,

If you select R=0.001, then you will see 8000A.
Do you want this?


Klaus

 

[chuckey]

How about using a LED bargraph driver (LM3914/5/6?) but instead of driving LEDs it drives a series of switching transistors each one switching in a different load resistor. These each give ten steps with a DC input. So the circuit would be an AC amp + rectifier, then a DC amp and off set adjustment, then LM39XX, then 10 (or 20) switching transistors with their special resistive loads. Without doing the maths I would have thought that a reasonable dynamic range/ accuracy would be had.
Frank

 

[Poyan100]

Hi,

If you select R=0.001, then you will see 8000A.
Do you want this?


Klaus

Hello
You know it is not practical to think about having 8000A, that is why I am saying we can select R to have a reasonable range for current. For instance, in my case I am saying having R=10 is good and resulted:
0V --> 0A
0.1V --> 0.1mA
0.2V --> 0.8mA
0.5V --> 12.5mA
1V --> .1A
2V --> 0.8A

How about using a LED bargraph driver (LM3914/5/6?) but instead of driving LEDs it drives a series of switching transistors each one switching in a different load resistor. These each give ten steps with a DC input. So the circuit would be an AC amp + rectifier, then a DC amp and off set adjustment, then LM39XX, then 10 (or 20) switching transistors with their special resistive loads. Without doing the maths I would have thought that a reasonable dynamic range/ accuracy would be had.
Frank

Thank You Frank. It is a good idea and in fact I will have a variable resistor that change the impedance to mimic the relation between current and voltage. It is good but always the system will be linear not non-linear.

 

[KlausST]

Hi,

You know it is not practical to think about having 8000A,

I´m working for a silicon producing company, doing current regulations with peak of more than 1000 V RMS and currents up to 6000A RMS, so peak is more than 8000A.

It is definitely practical...

To Franks solution:
I think he wants a "voltage controlled" paralleling of decreasing resitors.
So that your load adapts to voltage:
0.1V --> 1kOhm
0.2V --> 250 Ohm
0.5V --> 40 Ohm
...


Klaus

 

[Poyan100]

Hi,


I´m working for a silicon producing company, doing current regulations with peak of more than 1000 V RMS and currents up to 6000A RMS, so peak is more than 8000A.

It is definitely practical...

To Franks solution:
I think he wants a "voltage controlled" paralleling of decreasing resitors.
So that your load adapts to voltage:
0.1V --> 1kOhm
0.2V --> 250 Ohm
0.5V --> 40 Ohm
...


Klaus

Thank you Klaus. In terms of practical I meant for my case, The source of power that I have is a pendulum arm attached to a generator through a gearbox and typicaly it can not to produce more than 10W for the speed that I am running it.