when a real current source is connected to a load...

[Heidi]

When a realistic current source is connected in series with a load, a resistor for example, the current flowing through the load will be less than the value the current source provides?

If the answer is positive, is that why we put a parallel resistor with the ideal current source to account for the current lost in the load?

 

[steveB]

That seems like a trick question to me. You are asking about a "realistic current source", but you as "is this why we put a parallel resistor with the ideal current source".

Well, a realistic current source is not as simple as an ideal current source with a resistor, and the ideal current source is not part of the real device. Hence, the question begins to lose its meaning.

Still, if I understand the core of what you are asking, I'll answer "yes" and "yes".

 

[Heidi]

That seems like a trick question to me. You are asking about a "realistic current source", but you as "is this why we put a parallel resistor with the ideal current source".

Well, a realistic current source is not as simple as an ideal current source with a resistor, and the ideal current source is not part of the real device. Hence, the question begins to lose its meaning.

Still, if I understand the core of what you are asking, I'll answer "yes" and "yes".

Thank you, Steve.

Is your answer still 'yes' if I ask my question this way? "Is it true that the reason we use an ideal current source and a resistor in parallel with that ideal current source to model a real current source is because when a real current source is connected in series with a load, the current flowing in the load will be less than the value the real current source provides?"

 

[steveB]

Yes, That sounds correct to me.

 

[KeepItSimpleStupid]

"Is it true that the reason we use an ideal current source and a resistor in parallel with that ideal current source to model a real current source is because when a real current source is connected in series with a load, the current flowing in the load will be less than the value the real current source provides?"

I'm not sure I'll agree. Why? Because just like a voltage source, a current source has a characteristic impedance just as an audio amplifier has a characteristic impedance, but they hide it in a different specification "Damping factor".

I've spent a good part of my life measuring currents on the order of pA and stuff gets wierd in that arena.

Even the "current through a device" can get wierd too. There are leakage currents to take into account and or minimize. So, now it's time to introduce Triax cables with some graphite between the insulated wires. Why?

The graphite, dampens the piezoelectric effect. The friction between the conductors generates current. Also, don't move the cables. In the Earth's magnetic field, that generates a current.

A voltage and a dielectric is basically an unwanted resistance, so if this "unwanted" resistance shows up, then what?

So, now we have a cable with two shields, called a Triax cable. The outer shield is a Earth, but the inner shield is at the potential of the inner conductor. it's driven to be that. Now if the shield is at the potential, then there is no voltage difference and therefore no unwanted current. This driven conductor is known as a guard.

So, when measuring "stuff", you have to put it in a shielded box and put a plate in there driven at the potential (voltage) to minimize leakage.

it's a different world where even moisture can change the readings.

Sometimes, the effect of turning on the current source will introduce a current spike. In that case it makes more sense to measure coloumbs and divide by the time, You can zero the count and start counting for say 30 sec and divide Coulombs/time and get Amps or 1E-12 Amps in my case. I had to do this in the 1 to 10 pA range. The instrumentation I had didn't measure 1e-15 Amps.