Load modelling: Using Resistor or current sink as a load?

I am confusing what is the difference in modeling a load (unknown load). Some simulations model load as a pure resistor while the others use a current sink. I would like to know what is the difference between them?

P/S: For example, Moura (in his famous LDO book) recommended that in order to check the stability of the LDO: use both of resistor and current sink as the load for the simulation as 2 extremes.

I am wondering why?

Appreciate your comments.

Sincerely,

A simulated current source has an infinite resistance so it causes a different phase-shift in the feedback loop as compared to a low resistance load, thus both are used for simulation purposes to check the feedback loop stability at the extremes of output load resistance.

crutschow said:

A simulated current source has an infinite resistance so it causes a different phase-shift in the feedback loop as compared to a low resistance load, thus both are used for simulation purposes to check the feedback loop stability at the extremes of output load resistance.

Click to expand...

Thank for your quick reply. But I am still wondering why.

For example for checking stability of my LDO (targeted Iout is: 1mA to 1A, Vout_LDO = 2V), this means the Rload is equivalent to Rload(min) = 2Ohms to Rload(max) = 2kOhms as 2 extremes. But when checking with 2 current sinks (1mA and 1A, respectively) as 2 extremes, the results from R and current sink are totally different.

Could you give me more inside, please?

Thank you!

I am still wondering why, for example, if we use R and current sink as 2 extremes, why don't we just use R with min and max values for checking stability?

Also, should I understand that with iload min, and max, we have to use both of R and curent sink for each case respectively?
That means:
Case 1: iLoad min: Rmax and current sink = iLoad_min
Case 2: iLoad max: Rmin and current sink = iLoad_max

Thank you

Hi,

It seems you think in DC only.

For sure a constant current sink of 100mA connected to 5V - in terms of DC - is the same as connecting a 50 Ohms load.

But "stability" means AC stability.
And here is the difference: A current sink/source has infinite AC impedance. Although it is drawing DC current.

Imagine you have 5V DC and a ringing of +/-100mV (ringing cuased bay anyy error influence). So the output voltage is somewhere between 4.9V and 5.1V.
With a resistive load of 50Ohms you get an output current range of 98mA ... 102mA (it changes by 4mA, and these 4mA are damping the ringing)
But with a current sink the current is constant... no current here to damp the ringing.

To avoid such situations some circuits use a "zobel" circuit. Basically an RC series combination. Here just imagine a 100R / 1uF combination. It draws no DC current (no power loss) but it generates AC current to dampen oscillations.

Klaus

Dear Mr. KlausST,

Great! I understand the problem now. Really appreciate!

Thank you very much!