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I want to know where are the input and the output of this circuit in LTspice. I also need the values I should use for the resistors and the transistors in order to see clear results for current limiting.

enter image description here

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    \$\begingroup\$ Perhaps just a little more explanation about the circumstances? \$\endgroup\$
    – jonk
    Commented May 20, 2021 at 10:21
  • \$\begingroup\$ This is the circuit, I just need to show that the current limiter work for a project \$\endgroup\$
    – Rida
    Commented May 20, 2021 at 10:24
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    \$\begingroup\$ Oh! In that case, read this. It covers pretty much everything you need to know. \$\endgroup\$
    – jonk
    Commented May 20, 2021 at 10:26

2 Answers 2

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I want to know where are the input and the output of this circuit in LTspice. I also need the values.

This simple limiter uses the load on the high side of a common emitter biased by R1 such that R1/load << hFE minimum so that Ic/Ib can drive it towards saturation Vce<1 where hFE will also reduce rapidly say up to 90%.

You choose Re for about 600 to 650 mV/Ie. this is the typical range for the feedback amplifier Q2 to start sinking R1’s current away from the base of Q1. The higher the hFE choice of parts , the better for limiting the rise in current due a rise in V+. If the current is large then the power dissipation in Re must be chosen to use a part rated for twice the Pd to limit the temp rise to about 60’C. Also compute max Pd in Q1 for the range of supply and voltage drops Vce x Imax to choose a heatsink if needed.

Using an NFET for Q1 can give better results, if Ron is say 1% of load, then R1 can be quite large and thus the current more constant.

There are ways the reduce the overall voltage drop and also limit current with an extra Vref and pullup R to Q2-b so that the Vsense can be reduced from 600mV @1mA for Q2 to say 100mV, but this is more difficult to calculate the right R ratios.

Also R2 is slightly redundant as normally the current limiter if working on simple loads and slow changing V+ it will not overdrive Vbe2 so it can be 0.

Ideally you want a lossless current sense but my rule of thumb is 50mV drop for Re using an op amp to limit the drive base current to Q1.

Small transistors like PN2222A are limited to about 1 ohm driven with sufficient base current, so bigger parts are needed for smaller loads.

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This is not a current limiter but a constant current sink. It will keep a downward current through R-load constant, irrespective of the (sufficiently positive) VCC and temperature to some extent. The link it jonk's comment explains its function well. Essentially you use R-sense/Vbe(Q2) to define the current and the rest of the circuit makes this robust.

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  • \$\begingroup\$ A summary of the link from the comments would make this answer better. \$\endgroup\$
    – JRE
    Commented May 20, 2021 at 10:31
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    \$\begingroup\$ Is it not actually a constant current sink? As far as the OP goes, it is effectively a current limiter, as it will be more or less saturated until the current drawn by the load reaches the limit (i.e. voltage across R-sense equals the Vbe drop of Q2. \$\endgroup\$
    – Frog
    Commented May 20, 2021 at 10:40
  • \$\begingroup\$ @Frog i changed it to sink. I think, the OP expects (like me) that current enters and exits a limiter, hence their first sentence. This circuit only sinks it away. \$\endgroup\$
    – tobalt
    Commented May 20, 2021 at 11:51

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