0
\$\begingroup\$

Having solved the mystery of "pulling" and "pushing" devices, it would be interesting to do the same with "sourcing" and "sinking" devices.

Background

A transistor with a constant base-emitter voltage represents a constant current source (to be more precise, we should also include the power source).

For example, in the CircuitLab schematics below, I have adjusted the base-emitter voltage so that to pass 1 mA current through the load. If you want to change it, open Vin's parameters window, hover the mouse over RL, and adjust Vin while looking at the reading. If the parameters window covers RL, then move the schematic to the right edge of the screen, open and click on parameters to pin it, and return the schematic to the left side of the screen.

Note something interesting in this arrangement - the collector and emitter currents are equal because the base current is not added to the emitter one.

Varieties

Depending on the place of insertion, we call this device by different names.

"Current source"

If we connect the transistor Q with its emitter to the positive supply rail, and the load RL to ground, we call it "current source".

schematic

simulate this circuit – Schematic created using CircuitLab

"Current sink"

If we connect an NPN transistor Q with its emitter to ground, and the load RL to Vcc, we call it "current sink".

schematic

simulate this circuit

Both?

But what should we call (what is) the transistor when it is "floating" between two loads RL1 and RL2?

schematic

simulate this circuit

\$\endgroup\$
3
  • 1
    \$\begingroup\$ This might be opinion-based. It is anyway just a constant current load no matter if it is on high side, low side, or middle. \$\endgroup\$
    – Justme
    Commented Apr 26, 2023 at 20:05
  • 1
    \$\begingroup\$ I would call it a current regulator, or current limiting circuit. \$\endgroup\$ Commented Apr 27, 2023 at 15:44
  • \$\begingroup\$ @Math Keeps Me Busy, Or maybe "constant-current non-linear resistor", or "current-stabilizing resistor"(device)", or simply "current stabilizer"... \$\endgroup\$ Commented Apr 27, 2023 at 15:52

3 Answers 3

4
\$\begingroup\$

Q's collector is a current sink.

Q's emitter is a current source.

You can call Q whatever you desire.


The contents of the magenta box is neither a source nor a sink. There are two sources and two sinks within the box.1 source and 1 sink are mentioned above. The positive terminal of the 10V supply is a source. The negative terminal of the 10V supply is a sink.


The contents of the yellow box is not a source. The transistor collector is. As is the positive terminal of the 10V supply. The negative terminal of the 10V supply is a sink.

The contents of the green box is not a sink. The transistor collector is. As is the negative terminal of the 10V supply. The positive terminal of the 10V supply is a source.


Of course if you consider electron flow everything changes. Point of view is important.

\$\endgroup\$
3
  • \$\begingroup\$ RussellH, I am delighted to have the opportunity to name this circuit. Don't you think "Q's emitter is a current source" can be a little misleading? \$\endgroup\$ Commented Apr 26, 2023 at 20:43
  • 3
    \$\begingroup\$ @Circuitfantasist:...Nope \$\endgroup\$
    – user319836
    Commented Apr 26, 2023 at 20:46
  • 1
    \$\begingroup\$ @RussellH LOL! :D \$\endgroup\$ Commented Apr 26, 2023 at 20:54
3
\$\begingroup\$

It is merely a floating current source.
Also, whether it is pushing/sourcing the current or pulling/sinking the current, it is still a constant current source.

schematic

simulate this circuit – Schematic created using CircuitLab

UPDATE: Adding the same schematic with a transistor:

schematic

simulate this circuit

\$\endgroup\$
18
  • 2
    \$\begingroup\$ @Circuitfantasist I would say it is a current source regardless of the current direction, as it is sourcing either positive or negative current. \$\endgroup\$ Commented Apr 26, 2023 at 20:56
  • 2
    \$\begingroup\$ @EdinFifić: LOL, Now there's a point. Is sourcing a negative current the same a sinking a positive current? :) \$\endgroup\$
    – user319836
    Commented Apr 26, 2023 at 21:03
  • 1
    \$\begingroup\$ @EdinFifić: OK I'll stop now. :D Cheers \$\endgroup\$
    – user319836
    Commented Apr 26, 2023 at 21:21
  • 1
    \$\begingroup\$ @RussellH Cheerios! :D \$\endgroup\$ Commented Apr 26, 2023 at 21:26
  • 1
    \$\begingroup\$ @Circuitfantasist (Added your schematic) It matches it completely. There is about 6.5uA more current coming out of the emitter than what is going into the collector because the voltage source adds it through the base, but if the V source is isolated from the circuit (just like the current source in my schematic) it will also take that current back into the V source (it would circulate between base and emitter). If the V source is galvanically part of the whole circuit, it still matches my schematic 99.354% (1mA into the collector, 1.0065mA out of the emitter), which is good for most cases. \$\endgroup\$ Commented Apr 27, 2023 at 9:05
0
\$\begingroup\$

First I would like to thank the authors of the two answers for their responsiveness and benevolence. They gave me valuable ideas that enriched my understanding of this phenomenon, which I want to share with you in my answer.

As usual, I have accompanied my step-by-step explanations with CircuitLab simulations. In order not to clutter schematics with measuring devices, I completely rely on the attractive DC live simulation. For this purpose, just hover the mouse over the circuit and see the local voltages and currents.

These are "man-controlled" experiments where you adjust some parameter to get the desired value of the current (this is the best way to understand the phenomenon). To do this, open CircuitLab parameters window and start changing the quantity magnitude by observing the current through a DC live simulation.

Because these are conceptual schematics, where possible I have used convenient decimal values of the quantities - 1 or 10 [V, mA, kohm]. This simplifies the calculations since [mA].[kohm] = [V].

Similarity to a voltage source

The new insight for me was that these three current source configurations are analogous to the three voltage source configurations obtained by choosing a different reference point.

Floating sources

Both voltage and current sources are two-terminal devices and, in the general case, we see (work with) both ends of them. Thus, a voltage source has a positive and negative terminal, and as they say, the load will "see" both positive and negative voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

Similarly, a current source has an output and input terminal, and the load "sees" both.

schematic

simulate this circuit

A clarification: Since CircuitLab needs a ground, I have split the RL load into two parts (RL1 and RL2) and grounded the middle point.

Grounded negative terminal

However, we prefer to use only one terminal and connect the other to a reference point (e.g., "ground"). In most cases (e.g. in cars) we prefer to ground the negative terminal and then we get a positive voltage. Figuratively speaking, the load "sees" a positive voltage V+.

schematic

simulate this circuit

Similarly, if we ground the input terminal of a current source, the load will "see" its output terminal and will consider this device actually as a "source".

schematic

simulate this circuit

Grounded positive terminal

In some cases, we ground the positive terminal and thus we get a negative voltage (the load "sees" V-).

schematic

simulate this circuit

In the same way, we can ground the output terminal of the current source so the load "sees" its input terminal and will consider this device actually as a "sink".

schematic

simulate this circuit

What is in the middle?

In the second part of my answer, I will consider the arrangement of interest where a floating current source is stretched through two resistors (RL1 and RL2) between the supply rails.

Ideal current source?

Let's first put an "ideal" current source from the CircuitLab library in this place.

schematic

simulate this circuit

The problem is that we do not know what is inside that circle with an arrow - is it a "true" current source that "produces" the current or an element that only keeps the current constant.

A source indeed

We can check it very easily by replacing the voltage source with a piece of wire. As we can see, the current continues flowing and the voltage across it reverses its polarity. So, we conclude, the CircuitLab current source is a true current source that contains energy inside.

schematic

simulate this circuit

Implementations

But in reality this is not always (more precisely, almost never) the case. Let's see what are these possible things that can be put in the circle with the arrow.

The schematics below are conceptual; their purpose is to show the basic idea. But I have mentioned the possible specific implementations (see also my answer about how to create current sources with "static" but colorful schematics).

Dynamic resistor

The most popular thing to put in the circle is a "varying resistor" which has the property of changing its resistance so as to keep the current constant. To imitate it, open Rvar parameters and begin changing the resistance so that to obtain the desired 1 mA current.

If now I come and try, for example, to increase RL1 or RL2, you will feel this and decrease Rvar accordingly so that the total resistance RL1 + Rvar + RL2 remains unchanged and hence the current.

schematic

simulate this circuit

In real circuits, a transistor with constant base-emitter voltage does this boring work (see the schematic at the end of the question body). Also the so-called "constant-current diodes" made with JFET are a typical example ("diode" in the sense of a 2-terminal device, not PN junction).

Note that this is not a source in the literal sense of the word ("producing" current); it is a current regulating device. The current is produced by the voltage source V. The dynamic resistor is a passive device; so the voltage of its upper end is positive. As we saw above, the CircuitLab current source is not that.

Constant voltage source and dynamic resistor...

If we want to make a "true" current source, we need to include a constant voltage source in series with the dynamic resistor. Or, if you prefer, we need to move the external voltage source V inside the circle.

schematic

simulate this circuit

Note that now the "curcle" is an active device (in the sense producung power); so the voltage of its lower end is positive.

... dusturbed by resistance...

If I now come and insert a 1 k disturbing resistor Rdist, you will sense this and reduce Rvar to 7 k as above so that the total resistance remains unchanged (10k).

schematic

simulate this circuit

... or voltage

If I insert a disturbing 1 V voltage Vdist, you will sense this and increase Rvar up to 9 k so that the total current of 1 mA remains unchanged. And v.v., if I reverse Vdist, you will decrease Rvar to 7 k.

schematic

simulate this circuit

Dynamic voltage source and constant resistor...

We can swap the role of the voltage source and the resistor making V varying and R constant. For example, the op-amp inverting current source acts this way.

schematic

simulate this circuit

... disturbed by resistance...

Now if I insert 1 k disturbance, you will increase V to 11 V to compensate the disturbance and keep 1 mA current.

schematic

simulate this circuit

... or voltage

If I insert 1 V disturbance, you will decrease V to 9 V to keep 1 mA current.

schematic

simulate this circuit

What should we call it?

Let's finally discuss the names extracted from the answers... although this is the least important issue:

  • Current source, current sink, floating current source

  • Current source, current sink, current source and sink

  • High side, low side and middle (floating) current source

  • Pinned to Vcc, pinned to ground, floating current source

There is some taftology because once "current source" is a generic name and then it is used as a specific name. Example: Current sources are divided into current sources and current sinks.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.