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By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. The load has a variable impedance. My goal is to make an alarm of sorts when that impedance exceeds a threshold, and the way I want to do that is to detect when the is in current limiting mode.

When it’s passing less than 1 mA of current, I would expect it to be acting like a very low-value resistor, meaning that the voltage drop across it should be low. When the 1 mA of current is reached, it should begin to increase its resistance, thus increasing the voltage drop across it. I’d like to measure the voltage drop and light an LED really an optoisolator when it exceeds a threshold.

At least, that’s my thinking at the present.

If anyone else has an idea of how to light an LED when an LM is actually limiting the current through it, that would work datasheeet. In the schematic, I1 represents the LM I don’t see a way in circuitlab to represent an actual LM, so the simulation likely won’t operate correctly.

R1 represents the ground impedance – the goal is to detect when it’s value exceeds 10k. From there, it’s just a matter of tuning either the set resistor of the LM or the voltage threshold to datashete to the setpoint.

Note as well that the actual circuit ground is datsaheet the top side of R1 – Vout is the ground potential for the purpose of this circuit. The actual datssheet is on the far side of R1. Any real current source will have a voltage limit beyond which it will not deliver a constant current.

In the case of the LM, it is rated with a voltage drop of 2. The typical performance is shown in the datasheet in this graph:.

National Semiconductor

So for low currents, it might be delivering more-or-less constant current at 0. This will vary over temperature and from unit to unit, that’s what they mean by ‘typical’.


So if you monitor the voltage across the LM perhaps with a comparator you can have a good idea of whether it is working or not. If the voltage is more than 2. If it’s less than that it may or may not be working there is no minimum voltage below which is it is guaranteed to not work.

So if the LM is connected low side and sinking 1mA, and you want to know when the forward voltage drops below 1V you can use a comparator with a 1V reference. The reference could be derived from a regulated supply voltage such as 5V with a voltage divider.

The comparator could be something like half an LM At 25C, the LM has a minimal drop-out voltage of about 0. Othewise, the load is getting the maximum voltage that the source can provide, and is not running at full current. The simplest way to detect that would be by biasing a PNP transistor with a scaled-down version of the drop out voltage.

This isn’t superbly accurate, but might work well enough. The divider produces about 0. You can increase the resistance of the divider if you can use a larger load resistor R3 or a higher-gain transistor. By clicking “Post Your Answer”, you acknowledge that you have read our updated terms of serviceprivacy policy and cookie policyand that your continued use of the website is subject to these policies.

Home Questions Tags Users Unanswered. How to tell when an LM is limiting the current? I have an LM based constant current source configured to supply a max of 1 mA. Usually when a current supplied less than the current required, there is a voltage drop.

Nov 13 ’15 at I’d expect it to be minimal, however, since the should be trying to pass as much current as it possibly can. If it is a “constant” current source, why is the current “variable”? Tyler Think of it rather as a current limiter.

If the load impedance is high, then the current will be much lower than the limit. If the load impedance is low, then the current will hit the limit. The LM in this case does not have an infinite voltage with which to overcome the load impedance. The typical performance is shown in the datasheet in this graph: Spehro Pefhany k 4 All of that is true. The question at hand is how best to detect when the voltage across the LM exceeds a threshold. It depends on GREATLY on how the LM is connected high side or low side, for example and how much your current is and how much error you can tolerate any monitor circuit will lower the output impedance.


One option would be a TL or LMV, but that’s just one of an almost infinite number of possibilities. It’s not possible to give a good answer without much more information such as a complete schematic and accuracy requirements.

The detection threshold would be something like 1V. The circuit in this case is a Ground Continuity Monitor. A small current from an AC supply is intentionally “leaked” to ground.

If that current doesn’t flow, then that’s an error. In the old circuit, the output drives an optoisolator’s LED, but that circuit isn’t accurate or sensitive enough. The trip point is intended to be no higher than ohms per volt – 10k. The intent now is to connect the output to ground and measure the voltage drop across it instead.

Please edit your question and delete the comment. It belongs in the question. Kuba Ober 1, 10 Can R3 be replaced with an opto-isolator diode and series resistor? I need to use an opto between this circuit and the logic systems for safety.

Nope, it won’t work.

National Semiconductor – datasheet pdf

You need a different design, and I question the need for a current source. A couple of resistors and a comparator with built-in reference would do the job in a much simpler fashion. The current limiter is intended to limit the amount of ground leakage. It’s also intended to at least attempt to desensitize the circuit to the tolerance variation of the k resistor and the supply voltage. Datashet that this circuit must act appropriately during faults – and then the LM is a short.

So what you do is connect R2 directly between L and PE terminals. R2 completely determines the behavior of the system, when looked at from outside, and sets the leakage current.

Furthermore, I think it’s rather crazy that you allow such high ground impedance 10k. Sign up or log in Sign up using Google.

Sign up using Facebook. Sign up using Email and Password. Post as a guest Name. Email Required, but never shown. Post Your Answer Discard By clicking “Post Your Answer”, you acknowledge that you have read our updated terms of serviceprivacy policy and daatsheet policyand that your continued use of the website is subject to these policies.