I know that the two poles of a battery connected in a closed circuit would act as a voltage source. This is because the +ve terminal is at a higher potential compared to the -ve terminal. This is what drives the current through the circuit.

But for a current source, how does it work. Doesn't the -ve terminal acts like a current source?


Doesn't the -ve terminal acts like a current source?

The term "current source" doesn't mean something like "the terminal that electrons come out off" though I can understand why you might be thinking of it that way.

A current source is the dual of a voltage source. Unlike a battery that produces an (approximately) fixed voltage for a wide range of currents, a current source produces an (approximately) fixed current for a wide range of voltages.

Here's a simple example. If you place a resistor of resistance R across a voltage source of voltage \$V_S\$, the current \$I\$ is given by:

\$I = \dfrac{V_S}{R}\$

In other words, the current \$I\$ through the voltage source is determined by the external circuit which, in this case, is just a resistor.

In contrast, if you place a resistor R across a current source of current \$I_S\$, the voltage \$V\$ is given by:

\$V = I_SR\$

In other words, the voltage \$V\$across the current source is determined by the external circuit. The current source produces a fixed current that is (ideally) independent of the load attached to the source.

Whereas "good" voltage sources have very low internal resistance, "good" current sources have very high internal resistance.

In fact, one can approximate a current source with a high-voltage source in series with a large resistance.

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A ideal voltage source produces whatever current is necessary to maintain the voltage. A ideal current source produces whatever voltage is necessary to maintain the current. Yes, it really is that simple.

Let's say you have a ideal 10 mA current source. If you short it, 10 mA will flow and the voltage will be 0. If you put a 1 kΩ resistor accross it, 10 mA will flow and the voltage will be 10 V. If you put a 10 kΩ resistor accross it, 10 mA will flow and the voltage will be 100 V. If you leave it open circuit, the voltage will go as high as it needs to to make 10 mA flow. That could be a few 10s of kV initially to break down and ionize the air between the two contacts. Whatever it takes. Once the air is ionized (a spark has formed), the voltage will go down again since ionized air is somewhat of a conductor and much less voltage is needed to sustain the 10 mA than to break down the air initially.

Real current sources have real limitations, of course. One obvious one is the maximum voltage it can produce. Another is how well the current is regulated. You might measure a real current source, for example, putting out 10.0 mA at 0 V and 9.9 mA at 10 V. Note that this drop can be expressed as a resistance. 10 V / 100 µA = 100 kΩ. At first approximation, you can model this real current source as a ideal current source with 100 kΩ accross it internally that you can't do anything about. By Norton to Thevenin (look these up if you don't know them) equivalence, you can also think of this as a 1 kV voltage source with 100 kΩ in series with it.

A perfect current source has infinite resistance and a perfect voltage source 0 resistance.

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