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I have this circuit that should describe a constant current source. From my point of view the pnp-transistor has to work in its saturation mode. Thus the potential on the emitter should be constant. The value of the stable current is then defined by the resistor R1 and that voltage. Is that right so far? My concrete question refers to the variable resistor R2. The current should be constant when R2 is varied. When the resistance increases the voltage drop over R2 also increases by Ohm's Law. Therefore: what happens with "the rest of the voltage"? The voltage across R1 is constant, and the drop over R2 gets bigger, so somewhere the voltage has to decrease.

My idea would be that due to the saturation mode there's a current from the collector to the base and so the increasing voltage is kind of compensated.

enter image description here

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    \$\begingroup\$ You're saying "saturation" where (I'm pretty sure) you mean "active region". For unfortunate historical reasons, saturation in a BJT and saturation in a FET mean more or less opposite things. See this answer: electronics.stackexchange.com/questions/129408/…. \$\endgroup\$
    – TimWescott
    Jan 17 at 15:50
  • \$\begingroup\$ V2 should be between V1+ and R3 and, as shown R1 is either too small or, the repositioned value of V2 is too big. \$\endgroup\$
    – Andy aka
    Jan 17 at 15:50
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    \$\begingroup\$ If that's to be a current source, then R3 can quite a bit smaller -- 220 Ohms would be enough to avoid HF oscillation. (Shorting R3 may still work fine. But with some risk.) In that case, the base is held at +5 and the emitter will be about a diode drop above that, or about 5.8 V. The difference, 12 V - 5.8 V = 6.2 V would be across R1... so about 62 mA. R2 is then way too big to avoid saturation, which you do want to avoid for the collector acting in constant current behavior. You only have 5 V headroom, so R2 can't be larger than 90 Ohm in that case. BE junction is temp dependent. \$\endgroup\$ Jan 17 at 16:10
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    \$\begingroup\$ That can be see here. \$\endgroup\$ Jan 17 at 16:11
  • \$\begingroup\$ @periblepsis do you have a reference for the HF oscillation? \$\endgroup\$
    – internet
    Jan 19 at 23:11

4 Answers 4

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The value of R1 is wrong in the circuit you posted, it should be around 10k Ohms for it to properly operate as a constant current source.

When properly biased as a current source, the transistor operates in the forward active region, not in saturation. The excess voltage is lost across the transistor as the current flows between collector and emitter. In other words, it operates with a large collector-emitter voltage. As a result, the transistor will get pretty warm (current times voltage equals power). Base current will be small (almost negligible).

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  • \$\begingroup\$ Thanks. Ok that clears something for me but there's still a misunderstanding: When R2 increases, the voltage has to be shifted from the emitter-collector to R2. But when acting in active-forward mode that should result in a change of the base current - the base controls the 'resistance' of the transistor. But that would also result in a smaller emitter current as Ic = Ib×B. The current does not change, hence the base current doesn't change and therefore I cannot understand why suddenly the emitter-collector voltage differs. \$\endgroup\$ Jan 17 at 16:09
  • \$\begingroup\$ @olympus_mons No, the base current does not control the "resistance" of the transistor - there is no such resistance. The base current is instead (roughly) proportional to the collector current. So, for a given collector current, you'll always have the same base current regardless of the collector-emitter voltage. (At least approximately.) \$\endgroup\$ Jan 17 at 16:27
  • \$\begingroup\$ Ok, but somehow R2 gains more voltage with an increasing resistance. Thus the voltage over the transistor has to decrease: why does this not affect the base current? \$\endgroup\$ Jan 17 at 16:38
  • \$\begingroup\$ @olympus_mons Not all components obey Ohm's law. Transistors certainly don't. In the forward-active region, the transistor's collector will "push out" as much current as it needs to. For a given collector current, it needs a certain base current. The ratio between collector current and base current is the transistor's current gain (beta), which is given in the datasheet. For example, if the transistor was to deliver 1mA on its collector and has a gain of 100, it needs 10µA base current to do so. (As long as it remains somewhere in the forward-active region, at least.) \$\endgroup\$ Jan 17 at 16:51
  • \$\begingroup\$ Ok, and by varying the resistance of R3 the operating point of the transistor changes and by this the emitter-collector voltage can be determined, referring the output characteristics curves of the transistor? \$\endgroup\$ Jan 17 at 18:32
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A current source should be V1 "independent", so
You need at least 2 diodes between the base and V1 ...
But there are many "schematics" for such a circuit.

Thus the potential on the emitter should be constant.

Like this for example ... And the load (R3) should be < 1.2 kOhm.

enter image description here

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Before I start, let me point out that with the 10kΩ resistor R3 at the transistor's base, this will not work well as a regulated current source. Usually the base in these designs is driven directly by some low impedance voltage source. I'm going to continue assuming that \$R_3 = 0\Omega\$.

The transistor is not saturated while it is properly regulating current.

The "rest of the voltage", as you call it, is between collector and emitter; it is \$V_{CE}\$. That is, by KVL:

$$ V_{R2} + V_{CE} + V_{R1} = 12V $$

We can plug some numbers in there. Assuming \$V_{BE} = 0.7V\$:

$$ V_{R1} = 12V - (5V + 0.7V) = 6.3V $$

$$ \begin{aligned} V_{R2} + V_{CE} + V_{R1} &= 12V \\ \\ V_{R2} + V_{CE} &= 12V - 6.3V \\ \\ V_{R2} + V_{CE} &= 5.7V \end{aligned} $$

You have correctly stated that with a constant current through R2, then as R2 changes resistance the voltage across it must also change. That last equation tells us that a total of 5.7V is available to be shared between R2 and the transistor. Whatever chunk of that 5.7V is "taken up by" R2, the remainder must be across the transistor C-E.

This tells us something about the saturation state of the transistor. As long as there's more than about 0.2V across the transistor, \$V_{CE} > 0.2V\$, it isn't saturated. A transistor is saturated when \$V_{CE}\$ is near zero, at a minimum.

Put another way, as long as there is less than 5.5V across R2, \$V_{R2} < 5.5V\$, the transistor is not saturated, because there must therefore be more than 0.2V between the transistor's collector and emitter.

For the transistor to be on the cusp of saturation, the value of R2 must be such that it develops 5.5V (or perhaps a little more, depending on the transistor, and the current).

Firstly, get an idea of the current in R2. Ohm's law applied to R1 will reveal that. I'm going to ignore base current, assuming it's tiny in comparison, so that collector current and emitter current are equal, \$I = I_C = I_E\$:

$$ I = \frac{V_{R1}}{R_1} = \frac{6.3V}{100\Omega} = 63mA $$

Knowing the current in R2, and the voltage across it when the transistor is just saturated, we can apply Ohm's law to find its resistance:

$$ R_{2(SAT)} = \frac{5.5V}{63mA} = 87\Omega $$

In other words, as long as the resistance of R2 is between 0Ω and 87Ω, the transistor is not saturated (or is on the cusp of saturation), and will regulate current perfectly well. If R2 exceeds 87Ω, the transistor is saturated, and is no longer regulating.

The value you show for R2 in your schematic is much greater than 87Ω, so as shown, that transistor is deeply saturated.

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Introduction

Where "the rest of the voltage" is

Think of the collector-emitter part of the transistor as a “variable resistor” Rce that changes its "resistance" in the opposite direction as the load R2 changes. As a result, the sum of the complementary resistances Rce + R2, the total resistance R1 + Rce + R2, and the current I = Vcc/(R1 + Rce + R2) is constant.

You can simulate this arrangement with a potentiometer connected in series to R1 - one half is Rce and the other is R2. When you move the wiper, R2 will decrease/increase, and Rce will automatically increase/decrease, so their sum and the current does not change (a basic potentiometer property).

Why R3 is not needed

R3 here is a misconception and shows that the circuit author is not aware of the problem. A CE transistor stage with an emitter resistor (current source) must be driven with constant voltage (produced through a voltage divider) and not current (produced through a resistor). This is because the transistor must be able to control its base-emitter voltage from the emitter side, and for this the base must be at fixed voltage. If it is not, it will "move" when the emitter "moves" and nothing will result. Actually, the input of this circuit is its output (the emitter), i.e. this is a common-base configuration. As the load acts as a disturbance, it can be figuratively called "disturbed emitter follower".

R3 would be needed if there were no emitter resistor (grounded emitter). Then the transistor input is almost a short circuit and requires to be driven by current. But to drive an emitter resistor stage through a base resistor is wrong.

Building a current source

These concepts can best be understood if the varieties of current sources above are built step by step and explored through CircuitLab experiments.

Without negative feedback

Conceptual circuit: The simplest current source consists of a resistor Rce connected in series to a constant voltage source Vcc; it drives a resistor load RL. This current source is not perfect because the resistance Rce is constant...

schematic

simulate this circuit – Schematic created using CircuitLab

... and the current will depend strongly if the load changes its resistance RL.

STEP 1.1

The remedy is to change Rce in the opposite direction of the load changes. Let's simulate the behavior of such a dynamic resistor for three values ​​of load resistance (you can set them in the Rce "parameters" window).

RL = 5 kΩ, Rce = 5 kΩ: We need to monitor currents and voltages with measuring instruments but they will clutter the circuit. We can solve this problem by combining the resistors with the meters (i.e. by setting them to the same resistances). So let's represent Rce by a 5 kΩ voltmeter and RL by a 5 kΩ ammeter. As you can see, the total resistance is 10 kΩ, and the current is 1 mA.

schematic

simulate this circuit

RL = 6 kΩ, Rce = 4 kΩ: If the load increases its resistance with 1 kΩ, Rce (i.e. we) will react by decreasing its resistance with 1 kΩ. As a result, the total resistance is again 10 kΩ, and the current, as before, 1 mA.

schematic

simulate this circuit

RL = 4 kΩ, Rce = 6 kΩ Conversely, if the load decreases its resistance with 1 kΩ, Rce will react by increasing its resistance with 1 kΩ. The total resistance will be again 10 kΩ, and the current 1 mA.

schematic

simulate this circuit

Potentiometer current source: Let's now simulate this arrangement by a potentiometer P.

schematic

simulate this circuit

When sweeping its transfer ratio K (wiper position) from zero to 1, its complementary resistance halves crossfade, and the current does not change. Ignore the yellow horizontal line here and below; I put it only to "cheat" the CircuitLab autoscaling.

STEP 1.5

Real circuit: In the simplest transistor current source, the transistor Q acts as the dynamic resistor Rce above but it does much worse (because of the Early effect). As can be seen, the result is the same...

Voltage-supplied transistor: ... when the voltage across the base-emitter junction is kept constant...

schematic

simulate this circuit

STEP 1.6

Current-supplied transistor: ... or when the current through the base-emitter junction is kept constant.

schematic

simulate this circuit

STEP 1.7

In both cases, when the load changes from zero to 10 kΩ, the transistor changes its "resistance" from about 10 kΩ to zero.

With negative feedback

The current source above "blindly" maintains the current without "caring" about its value, i.e. there is no negative feedback. Its advantage is that the voltage across the load can even reach the supply voltage. A more precise way to maintain current is through negative feedback.

Conceptual circuit: To implement it, we insert another constant resistor Re (1 kΩ), and keep the voltage across it constant (1 V) by adjusting Rce. The rest part of the circuit is the same as above; only Vcc is increased by 1 V to compensate for the voltage drop across Re.

RL = 5 kΩ, Rce = 5 kΩ: Initially, the total resistance is 11 kΩ and the current is 1 mA.

schematic

simulate this circuit

RL = 6 kΩ, Rce = 4 kΩ: If the load increases its resistance with 1 kΩ, we see that VRe decreases and begin decreasing Rce eventually with 1 kΩ. As a result, the total resistance is again 11 kΩ, and the current, as before, 1 mA.

schematic

simulate this circuit

RL = 4 kΩ, Rce = 6 kΩ: Conversely, if the load decreases its resistance with 1 kΩ, VRe increases and we begin increasing Rce eventually with 1 kΩ. The total resistance is again 11 kΩ, and the current, as before, 1 mA.

schematic

simulate this circuit

Real circuit: This idea can be implemented by an emitter follower Q with constant input voltage Vref.

schematic

simulate this circuit

The result is excellent; the Early effect is compensated by the negative feedback.

STEP 2.4

Grounded Vref: Vref above is floating which is inconvenient. We can replace it with a grounded voltage source with a complementary voltage Vcc - Vref.

schematic

simulate this circuit

The result is the same.

STEP 2.5

OP's version

Finally, let's make sure that this circuit should not be current driven (through a base resistor).

schematic

simulate this circuit

The result is unsatisfactory (as in Schematic 1.7).

STEP 2.6

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