Current Limiting for a Linear Power Supply - Page 1
Forum section: electronics / projects

author: rsduhamel
author_rank: Contributor
timestamp: June 19, 2018, 12:18:35 am
content: I received a question about my video series on building a linear power supply. I gave it a quick answer, but the asker wants more detail. I thought I'd put it out to the EEVBlog community.Here's the video series playlist:https://www.youtube.com/playlist?list=PLL_nf1OmixTTKhw2oVBtCZUYoMecz-YW5.The current limiting circuit is a 0.5-ohm resistor in series with the load. This resistor is connected from the base to the emitter of an NPN transistor. That resistor's collector is connected to the base of the pass transistor. When the current approaches 1 amp the current limiting transistor will start to pull current away from the pass transistor, limiting the power supply current. Here's the circuit:Keep in mind that this video series is not about building the best power supply, but is about how a few components can work together to make a useful circuit. It's for beginners who just learned about how the individual components work.Here's the conversation so far:Tamojit Sahawhat if I want a high output current limit, around 8A then how to improve the circuit?RSD AcademyPretty simple. Just a lower value for the current sense resistor. 0.5 volts / 8 amps gives you 0.0625 ohms. Don't forget the power rating. 8 squared times 0.0625 gives you 4 watts, so 5 watts should do it.﻿Tamojit SahaThanks for your fast response. Such low-value resistors are hard to get.Anyway, how to calculate the short circuit power dissipation in this case?﻿Tamojit SahaAlso in this circuit, you are assuming the transistor will turn on at 0.7v and by that, you are calculating the value of R_sense with the known current limit. So, 0.7v/100mA = 7ohms. If the typical turn-on voltage is 0.5V then you 0.5v/100mA gives you 5ohms. So every time you need to adjust the value of R_sense with the turn-on voltage of the transistor. This turn-on voltage is not same for all transistors and therefore precise current limit is not possible.So how can I achieve precise current limit with reference to your designed circuit?﻿

author: JS
author_rank: Frequent Contributor
timestamp: June 19, 2018, 12:27:23 am
content: While your circuit is good for reliable and fast current limit not so much for precission and is most lilely you want a completely different approach for that.When current gets higher you will probably use multiple transistors in parallel, sensing then from a single transistor base and it's resistor will cut the current for all of them, so with 4 transistors doing 2A each and using 0.25Ω resistors could do for 8A, given you 0.5V are a good estimation.For precission is much harder to get fast and reliable, usually they need to be slow and have some overshoot. I'm trying to design something to do something like that but circuits aren't easy. Simulator is giving quite a fight.JS

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 19, 2018, 02:35:09 pm
content: 1. For more current, you should probably use more 2N3055s (or better yet, a modern transistor) in parallel. There are also darlingtons in that current range, which saves the trouble of the 2N4401, but meh.2. For accurate current limits -- no. It's simply not the right circuit for that.For that, you need to change the circuit by adding a current sense path, and an error amplifier. The outputs of the two error amplifiers need to be reconciled (i.e., wired-OR), and the result applied to the follower.When op-amps are wired-OR with diodes, one saturates to the opposite rail while the other does operation. This causesintegrator wind-up, i.e., it takes a dozen microseconds for the saturated amp's output to rise back up to operating level. In that time, the output over/undershoots. It's a high-level version of diode recovery: instead of current overshooting due to semiconductor physics, it's current overshooting due to delays in the control system. The overall effect is much the same, though!You need to use a wired-OR configuration that prevents windup, like this:https://www.seventransistorlabs.com/Images/Limiter2.pnga discrete op-amp wrapped around the one error amp, so that the error amp is always in control, but its inputs are manipulated to perform one duty or the other. Note that this circuit doesn't have amazing gain on the limiter loop (too much gain there, and you have the same problem as always: oscillation), which also gives it a somewhat soft threshold, and less precision within that limit. Still, it's much better than a single Vbe.It's also worth noting, your circuit is drawn as almost DC-only. That is, no consideration has been made for AC characteristics, aside from the rectifier. This may be a matter of representation -- that it's not a practical circuit, and "details are left to the student", so to speak.But if it's not being represented in that way, then it is disingenuous, and those details should be added.To that end:- There should be some compensation across the amplifier (OUT to -IN)- The output should not be overdriven (add series, maybe 1kohm, from OUT to 2N4401-B)- The current limiting transistor should, itself, have a current limiting resistor in series with its base, maybe 1k; this prevents damage under sudden load conditionsTim

author: schmitt trigger
author_rank: Super Contributor
timestamp: June 19, 2018, 04:07:46 pm
content: I always applaud the fact that someone takes the time to learn how to use 1970s electronics. This gives you a great insight on how components actually operate, and the pitfalls associated with non-ideal components.But to expect a tight limits from such simplistic circuit? No, no way.Another pitfall is the transistor's power dissipation under short circuit conditions. You may want to research "foldback current limiting".

author: bson
author_rank: Unknown
timestamp: June 25, 2018, 04:47:22 am
content: I think the questions try to do things with your instructional circuit that it's not really suited for, such as a precision current cut-off, CV/CC, or adjustable limits. I think it would be a perfectly fine answer that there are better ways to accomplish those things.

author: David Hess
author_rank: Super Contributor
timestamp: June 25, 2018, 10:49:56 pm
content: Quote from: T3sl4co1l on June 19, 2018, 02:35:09 pmOperational amplifiers which expose their transconductance output for external compensation can be clamped through that point to prevent windup.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 25, 2018, 10:57:13 pm
content: Quote from: David Hess on June 25, 2018, 10:49:56 pmPreciselyit's too bad there are so few of them out there. Dominant pole everything...Tim

author: David Hess
author_rank: Super Contributor
timestamp: June 26, 2018, 12:54:43 am
content: Quote from: T3sl4co1l on June 25, 2018, 10:57:13 pmI have been compiling a list of suitable parts but unfortunately only a few support external compensation and only some of those bring the transconductance output out. Even worse, many potentially suitable parts are from companies like Analog Devices and Texas Instruments who do not supply equivalent schematics so there is no way to know if they can be clamped or not from their incomplete datasheets.And then there is a practical aspect; do I really want to publish a design which requires an operational amplifier type of limited availability? It *is* feasible to design in discrete clamping as in your example; it is just annoying. On the other hand, it is not like the LM301A and LM308 derivatives are going away.

author: m3vuv
author_rank: Super Contributor
timestamp: September 12, 2019, 03:49:34 pm
content: since when has a resistor had a collector?