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In text books some controlled or dependent voltage and current sources are shown. What are they really? Can someone explain them with real examples?

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  • \$\begingroup\$ What do you already know? Have you learned about op-amps? Have you learned about transistors? \$\endgroup\$
    – The Photon
    Dec 2, 2023 at 6:31
  • \$\begingroup\$ Yes I know about transistors where where voltage across the collector resistor depends on base current. \$\endgroup\$
    – Alex
    Dec 2, 2023 at 6:54
  • \$\begingroup\$ Alex, I have added CircuitLab simulations to my answer. Later, I will expand them with explanations. It would be helpful for you to try to grasp the ideas in these configurations before my explanations. \$\endgroup\$ Dec 2, 2023 at 11:53

2 Answers 2

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Short answer

The difference between the ordinary constant sources and dependent sources is that the former have only output (the input is implicit, constant) while the latter have both input and output (actually they are converters). So:

  • VCVS is a voltage-to-voltage converter or voltage amplifier, follower or attenuator.

  • VCCS is a voltage-to-current converter or transconductance amplifier.

  • CCCS is a current-to-current converter or current amplifier.

  • CCVS is a current-to-voltage converter or transimpedance amplifier.

So the independent sources can be thought as "dependent sources with constant input quantity", and dependent sources as "independent sources with variable input quantity".

How to present circuits

In text books some controlled or dependent voltage and current sources are shown...

The problem of these textbooks is that they only show these circuit building blocks but do not explain what is inside them; so let's fill that gap. The best way to do this is by building the circuit step by step showing what the idea is at each step. Let's illustrate it by reinventing some of these famous circuits - the basic current-to-voltage converter, the dual "voltage-to-current converter", and the composite voltage-to-voltage converter.

Building a current-to-voltage converter

The main idea underlying the building of these active circuits is that they are improved passive circuits. So let’s start building the current-to-voltage converter with its simplest passive version - the resistor.

Non-inverting configuration

Passive current-to-voltage converter unloaded: In this arrangement, the input current source Iin passes its current through the resistor R thus creating, according to Ohm's law, the output voltage Vout = Iin.R.

schematic

simulate this circuit – Schematic created using CircuitLab

As you can see from the graph below, the relationship between the current and voltage is linear.

STEP 1.1

Passive current-to-voltage converter loaded: The problem with the passive I-to-V converter, as with any passive circuit, is that the input (current) source supplies the load. Let's test it by loading the circuit with the same 1 kΩ resistance as the resistor R; hence the figurative name RL1k. The easiest way to do this is to go into the voltmeter parameters window and reduce its resistance to 1 kΩ. Now it will serve as a “voltage-visualized load”.

schematic

simulate this circuit

The result of this intervention is that the total resistance R||RL1k and accordingly the output voltage drop twice. Let’s investigate it graphically for three load resistances - 100 Ω, 1 kΩ and 10 kΩ. As you can see, the impact of the load is significant.

STEP 1.2

Active current-to-voltage converter: The trick we can use to solve this problem is known as negative feedback (I prefer to call it active copying). It means to supply the load RL1k from another voltage source Vout that copies the voltage VR across the resistor R. We can do it by comparing VR and Vout with a floating null indicator NI (a sensitive voltmeter) and adjusting Vout so that to keep a zero voltage difference across NI.

schematic

simulate this circuit

Now the RL1k load is supplied by the Vout voltage source instead of Iin; so it can be low resistive enough.

Op-amp buffered current-to-voltage converter: Of course, an op-amp would do this tedious job much better than we do; so let us entrust it to the op-amp. Now it will compare its output voltage with the input one and will adjust it so that to zero the voltage difference. The name of this circuit is op-amp follower.

schematic

simulate this circuit

STEP 1.4

Inverting configuration

In the circuit of the non-inverting I-to-V converter, the input current source was perfect ("ideal"), and we have solved the problem caused by the load.

Imperfect current source short circuited: But the input source can be imperfect (made by a voltage source Vin and a resistor Rin in series as in the schematic below). It is of course expected to be short-circuited by an "ideal current load" ("piece of wire" or "ideal" ammeter). Then the current would be exactly Iout = Vin/Rin.

schematic

simulate this circuit

Imperfect current source R loaded: But we have connected a resistor R = 1 kΩ, so the current has decreased twice - Iout = Vin/(Rin+R), and the output voltage is also decreased twice.

schematic

simulate this circuit

STEP 2.2

Active I-to-V converter without negative feedback: Now we can take one more trick out of our "bag of circuit tricks" :-) In fact, it is not only electrical, but widely used in life, and it says the following: If a harmful quantity appears, we destroy it with a useful quantity of equal value. In our case, the voltage drop across the resistor is harmful; therefore we should destroy it with an equal value voltage. This means connecting an additional source -VR in series so that its voltage is added to the input voltage Vin (i.e, it must be negative to ground). In this CircuitLab simulation, this source is implemented as "behavioral" with voltage -VR.

schematic

simulate this circuit

As a result, the "undesired" voltage drop is neutralized, and the imperfect input voltage is virtually short circuited. The upper rail has zero voltage; that is why it is called virtual ground.

STEP 2.3a

STEP 2.3b

Active I-to-V converter with negative feedback: In life and circuits, we prefer to produce the compensating quantity by the negative feedback principle. We can do it as above, by comparing Vout = -VR and VR with a grounded null indicator NI (a sensitive voltmeter) and adjusting Vout so that to keep a zero voltage difference across NI. An interesting fact is that we use the compensating voltage -VR as an output voltage.

schematic

simulate this circuit

Op-amp I-to-V converter with negative feedback: As above, we make an op-amp to do this boring job. It "observes" the voltage of its inverting input (virtual ground), and adjusts its (negative) output voltage to keep the virtual ground.

schematic

simulate this circuit

Perfect current source: Of course, we can drive the I-to-V converter with a perfect current source Iin.

schematic

simulate this circuit

The result is the same.

STEP 2.6a

STEP 2.6b

Voltage-to-current converter

Passive voltage-to-current converter

We have seen above in the Schematic 1.1 how the humble resistor R can convert, according to Ohm's law, the current Iin flowing through it to a voltage drop Vout = Iin.R across it. Conversely, the same resistor can convert, according to Ohm's law, the voltage Vin applied across it to a current Iout = Vin/Iin flowing through it (see Schematic 2.1).

Active voltage-to-current converter

To eliminate the harmful voltage drop across the load (imperfect ammeter) connected in series to the resistor, we can apply the same clever trick as above - adding the same voltage in series. When we change the load resistance RL, this voltage will vary keeping up a constant current. Lets investigate it for three load resistances.

RL = 0 Ω: If the ammeter is "ideal", there is no disturbing voltage drop across it; so there is nothing to compensate and the op-amp output voltage VOA is zero. The output current is Iout = Vin/R = 1 mA.

schematic

simulate this circuit

RL = 1 kΩ: Now let's set the ammeter's resistance to 1 kΩ. A 1 V disturbing voltage drop appears across it, and to compensate it, the op-amp output voltage becomes -1 V. The output current is Iout = (Vin - VRL + VRL)/Rin = 1 - 1 + 1 = 1 mA.

schematic

simulate this circuit

RL = 10 kΩ: Finally, let's increase the disturbing ammeter's resistance to 10 kΩ. A 10 V disturbing voltage drop appears across it, and to compensate it, the op-amp output voltage becomes -10 V. So the output current is Iout = (Vin - VRL + VRL)/Rin = 1 - 10 + 10 = 1 mA again.

schematic

simulate this circuit

Voltage-to-voltage converter

Passive voltage-to-voltage converter

When we load the passive voltage-to-current converter (a resistor R1) with another resistor R2 acting as a current-to-voltage converter (see Schematic 2.2), a simple voltage-to-voltage converter (aka voltage divider) is formed.

Active voltage-to-voltage converter

As above, we can remove the voltage drop across R2, and use the compensating voltage as an output (see Schematic 2.5). Thus we obtain the famous inverting configuration. Depending on the R2 value, three versions can be obtained.

R2 = 1 kΩ: If R2 = R1, this inverting configuration acts as an inverter (inverting amplifier with a gain of -1).

schematic

simulate this circuit

STEP 4.1

R2 = 10 kΩ: If R2 > R1, this voltage-to-voltage converter acts as an inverting amplifier.

schematic

simulate this circuit

STEP 4.2

R2 = 100 Ω: And when R2 < R1, the circuit acts as an inverting attenuator.

schematic

simulate this circuit

STEP 4.3

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    \$\begingroup\$ I would just say WOW \$\endgroup\$
    – Alex
    Dec 2, 2023 at 18:28
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Constant:

In integrated circuits, people build current mirrors. Current mirrors are designed to output a fixed current. one value; not supposed to change. So, when other circuit designers use this current mirror, they abstract away all the stuff needed to make a current mirror and they will draw it all as one symbol: a constant current source.

Dependent:

In integrated circuits, people use the transistor a lot. In analog circuits, transconductance is a big concept for the transistor. Like Circuit Fantasist wrote about converters, "transconductance is a converter from voltage to current and the small signal current output by the transistor is proportional to the small signal gate-source which means the output current of the small signal transistor is dependent on the small signal gate-source voltage".

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    \$\begingroup\$ Welcome to SE EE! Helpful explanations... just wondering if the capital letters at the beginning of the sentences disappeared from English grammar :-) \$\endgroup\$ Dec 2, 2023 at 11:57

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