Current limiting circuit understanding

Question

I have found this circuit which is supposed to be a current-limiting circuit:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

'R1 is used as a current's resistor. It monitors the current flowing through Q1. The voltage drop across R1 increases as the current through Q1 increases. If the voltage at the top of R1 reaches 0.65V Q2 begins to turn on. Q2 diverts some of the current from the base of Q1 and sends it to ground. This reduces Q1's collector current'.

However, I am having trouble understanding this:

When Q2 is on the voltage drop of R1 will become less and Q2 must be off despite we say it is on and this will create a closed loop of paradoxes.

Where am I wrong?

limits the current through *what*, i.e. where do you attach your load? — Marcus Müller, Jul 20 '20 at 07:50
and again, this can't work at all: VCC is a voltage source, so there's no "diverting" current from Q1; the source can source infinite current. — Marcus Müller, Jul 20 '20 at 07:51
true but current gets divided while Q2 is on and R3 is only part of 1 branch. — Helena Wells, Jul 20 '20 at 07:53
when Q2 is on current flows both through Q2-R2 and R1-Q1-R3. — Helena Wells, Jul 20 '20 at 07:54
yes, so what? the fact that Q2 starts conducting doesn't at all influence the current through Q1's base – again, VCC is a voltage source. The voltage at the base of Q1 is fixed at VCC. — Marcus Müller, Jul 20 '20 at 07:56
Yes you are probably right . I just took it from this website:http://www.physics.unlv.edu/~bill/PHYS483/current_lim.pdf — Helena Wells, Jul 20 '20 at 08:01
Here's a red flag: there are two base-emitter junctions between VCC and ground, with no resistors between. Q2 emitter is 0V, so Q2 base is 0.6V, and Q1 emitter is 0.6V, so Q1 base is 1.2V -- and that connects directly to VCC. — MarkU, Jul 20 '20 at 08:06
@HelenaWells: Your circuit is **not** the same as the one in the linked PDF. Your circuit **can't** work, though the one in the PDF does work. — JRE, Jul 20 '20 at 08:08
R2 needs to connect to Q1 base and Q2 collector (R3 is the load resistor). The other side of R2 is a lower voltage control signal (the linked paper has +12V for the load and +5V for the R2 control signal). — MarkU, Jul 20 '20 at 08:10
@HelenaWells Also see [here](https://electronics.stackexchange.com/a/481317/38098) for some added thoughts. — jonk, Jul 21 '20 at 04:44

Transistor · Answer 1 · 2020-07-20T13:12:12.910

4

You have transcribed the circuit from physics.unlv.edu incorrectly.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. (a) What you have drawn. (b) The corrected version. (c) The original circuit.

Your R2 should be providing the bias for Q1 and Q2's collector should be connected directly to the base of Q1. This way when the voltage across R1 reaches 0.65 V or so Q2 will start to steal the bias from Q1.

In its modified form of Figure 1c the enable signal provides the bias. The 5 V supply is adequate because Q6's emitter will be at 0.65 V and its base at about 0.65 V above that - so 1.3 V above ground.

From the comments:

OP wants to know about the edge case transition at about 0.65 V and why it is continuous and stable, which this answer does not really address.

schematic

^{simulate this circuit}

Figure 2. The equivalent circuit.

I don't think there's much point in discussing the stability of this circuit. It's a pair of PN junctions shorting out V_CC to ground. It won't last long.

edited Jul 20 '20 at 13:12

answered Jul 20 '20 at 08:14

Transistor

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Good error spot +1 – Andy aka Jul 20 '20 at 08:16
Though you noticed an important correction, this doesn't seem to answer the question at hand. OP wants to know about the edge case transition at about 0.65 V and why it is continuous and stable, which this answer does not really address. – Groger Jul 20 '20 at 12:46
@Groger: The OP's circuit is a pair of series-connected PN junctions shorting out the power supply. It won't be stable as there will be continuous temperature rise until Q1 or Q2 burns out. Do you see some other solution? – Transistor Jul 20 '20 at 13:00
I agree that there is a problem in OP's circuit, which you addressed well. This should be handled as a question comment or edit, perhaps. This answer does not address the actual question from OP, "When Q2 is on the voltage drop of R1 will become less and Q2 must be off despite we say it is on and this will create a closed loop of paradoxes. Where am I wrong?". – Groger Jul 20 '20 at 13:09
See the update. – Transistor Jul 20 '20 at 13:13

score 0 · Answer 2 · answered Jul 20 '20 at 12:57

The difficulty you have in resolving this paradox stems from the treatment of transistors as binary, digital elements. In reality, transistors are analog, and have gradual, continuous behavior. In the case of switches, like the realm of digital, this middle region is ignored, and only saturation and cutoff behaviors are considered, in which case we have our typical amplifiers.

A more general treatment of transistors would be with this equation

\$i_C=I_Se^{\frac{v_{BE}}{V_T}}(1+\frac{V_{CE}}{V_A})\$

In this case, a transistor can always be considered to have a collector current flowing (except with all grounded pins), rather than a drastic transition. For general purposes, the electrical engineering community accepts \$0.65 V\$ or \$0.7 V\$ as the "turn on" voltage, but in reality, current flows long before that.

In the case of your circuit, \$Q_2\$ will always be taking a little bit of current from \$R_1\$, but for voltages less than the turn on voltage, this is miniscule enough that it can be ignored. As the voltage across \$R_1\$ increases,this current diverting effect becomes more dramatic, limiting the current through \$R_1\$.

Current limiting circuit understanding

2 Answers2