6
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Here's a simplified diagram of a circuit I have:

schematic

simulate this circuit – Schematic created using CircuitLab

The battery is 24V. The relay is rated at 8A. My load draws 2A maximum. The capacitors are thousands of µF worth of filter capacitors.

In hindsight it's pretty obvious why I burned two relays with this. So I tried to figure a way to prevent the current spike when the capacitors are empty and the relay is switched on. I finally added a small low-value power resistor in series between the battery and the relay so the max current does not exceede the relay specs. It works fine and generating heat while the capacitors are charging is fine but generating heat all the time is inefficient.

I need the battery to last as long as possible.

Any suggestion?

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  • 1
    \$\begingroup\$ Back in the day, we just threw a choke and a resistor at it. (About 100 g of mass.) The bigger the better. Of course, today, that isn't cheap or small. The traditional ideas such as thermistors and relays and just plain old resistors are all discussed in this just released patent. It's interesting to read and I thought others might enjoy it, even if it doesn't necessarily apply here. \$\endgroup\$
    – jonk
    Commented Jul 25, 2021 at 4:20
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    \$\begingroup\$ the problem is the arcing during bouncing. you just have to delay the surge until after the bounce, eg with a series FET after the relay with a slow gate RC. \$\endgroup\$
    – tobalt
    Commented Jul 25, 2021 at 5:14
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    \$\begingroup\$ As tobalt implied, a bouncing input voltage to a an inrush limiter based on heating a thermistor eventually comes unstuck and you find that there will be situations where you will still have a large inrush current. Even MOSFET based inrush limiters (if not designed correctly to account for the many and various power situations) will also come unstuck. \$\endgroup\$
    – Andy aka
    Commented Jul 25, 2021 at 11:56
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    \$\begingroup\$ @JohnDoty it's a simplified diagram. The capacitors are actually part of an array of buck converters. \$\endgroup\$
    – Jerther
    Commented Jul 25, 2021 at 12:33
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    \$\begingroup\$ OK, next question: why a relay rather than a MOSFET? \$\endgroup\$
    – John Doty
    Commented Jul 25, 2021 at 13:04

5 Answers 5

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The solution to this is a thermistor. Alternatively, have another relay to bridge your resistor after an amount of time has passed.

Of course, you could also put the capacitors on the other side of the relay.

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  • \$\begingroup\$ Isn't a thermistor also inefficient? Moving the capacitors would be tricky but I like the idea. Thanks \$\endgroup\$
    – Jerther
    Commented Jul 25, 2021 at 3:34
  • \$\begingroup\$ The thermistor only has high resistance initially, that's the point! \$\endgroup\$
    – mmmm
    Commented Jul 25, 2021 at 3:46
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    \$\begingroup\$ @Jerther They make thermistors specifically for this exact purpose, called ICLs ("Inrush Current Limiters"). \$\endgroup\$
    – Hearth
    Commented Jul 25, 2021 at 3:48
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Since you asked me to expand on my earlier comment: ...

In the olden days, we used heavy iron a lot for handling inrush currents -- big inductors. A resistor was often added in parallel to the heavy inductor and the pair was used in series between the power source and the capacitor bank. Something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The peak current in the resistor, \$R_1\$, will happen when the capacitance is discharged. So \$I_{R_{_\text{PEAK}}}=\frac{V_1}{R_1}\$. And that means we can select \$R_1\$ based upon what we want the peak current in the resistor to be.

Let's compute that one right now. I'm going to pick \$I_{R_{_\text{PEAK}}}=4\:\text{A}\$. This seems like a safe margin vs your relay's current limit.

Note: Now, pause. There will also be an increasing inductor current. And perhaps this limit isn't good enough. But in practice it will be. The reason is that while the inductor's current is climbing, so also is the limit-resistor's current declining. Done right, these will just almost perfectly cancel each other out so that if we design each for the same current limit, then the sum of their currents will also be similarly limited.

So let's work out that \$R_1=\frac{24\:\text{V}}{4\:\text{A}}=6\:\Omega\$ and choose either \$R_1=5.6\:\Omega\$ or \$R_1=6.8\:\Omega\$. I'm going to lean towards the safer side, so I'll select \$R_1=6.8\:\Omega\$. This means my new current limit will be about \$3.5\:\text{A}\$ when working out the inductance.

Which brings us to the inductor. To deal with the question about the peak current in the inductor, I'll re-draw the above in a slightly different, but equivalent, form:

schematic

simulate this circuit

In the above, I've "noticed" that \$R_1\$ and \$R_{_\text{LOAD}}\$ make up a voltage divider with an equivalent source voltage of \$V_{_\text{TH}}=V_1\cdot\frac{R_{_\text{LOAD}}}{R_1+R_{_\text{LOAD}}}\$ and \$R_{_\text{TH}}=R_1\cdot\frac{R_{_\text{LOAD}}}{R_1+R_{_\text{LOAD}}}\$.

What's really nice about the new arrangement is that it is easier to see it is an RLC circuit. The general idea here is that the peak inductor current happens in the first quarter of a period of time that is determined by a combination of these three items.

Roughly, you can compute \$L_1\approx \frac{16}{\pi^2}\cdot R_{_\text{TH}}^{\,^2}\cdot C_1\$. And since \$R_{_\text{TH}}=\frac{V_1}{I_{_\text{LOAD}}+I_{R_{_\text{PEAK}}}}\$, this is works out to: \$L_1\approx \frac{16}{\pi^2}\cdot C_1\cdot \left[\frac{V_1}{I_{_\text{LOAD}}+I_{R_{_\text{PEAK}}}}\right]^{\,^2}\$. Assuming \$V_1=24\:\text{V}\$, \$I_{_\text{LOAD}}=2\:\text{A}\$, and \$I_{R_{_\text{PEAK}}}=3.5\:\text{A}\$ (the new value), then \$L_1\approx 1.62\cdot 10\:\text{mF}\cdot \left[\frac{24\:\text{V}}{2\:\text{A}+3.5\:\text{A}}\right]^2=309\:\text{mH}\$. I'd pick \$L_1=270\:\text{mH}\$ as a very close, available value. (Of course, you might just as well pick \$L_1=330\:\text{mH}\$, too.)

I've no clue what's going to happen when I pop this into LTspice. (I'm going to cheat just a little bit to keep the schematic simple, as I'll have LTspice "turn on" the power supply on its own by checking off a box.) So let's see:

enter image description here

(Click on the above picture to see more detail.)

Here you can see the behavior resulting from the above approach. It's almost exactly as expected. You can see that the sum (the red curve) that is being delivered through the relay only slightly peaks over \$4\:\text{A}\$. So we have kept to our limit. But this also tells you that there will be a little more current than you target in the design, due to the fact that both the resistor and the inductor are supplying current during the early quarter-cycle (damped) phase. (Had I decided to round up and use \$L_1=330\:\text{mH}\$, then the peak current through the relay would have been about \$3.8\:\text{A}\$.)

And yes, the output voltage did ring a bit. It reached a slightly higher voltage than \$24\:\text{V}\$. Back in the day, that was fine. We were using diodes the size of your fist (selenium rectifiers with lots and lots of fins) and vacuum tubes. A little extra voltage never hurt anyone. ;) Today, you have to think about that more.

If you can accept a higher peak voltage and a higher peak inductor current, you can reduce its magnitude. Say, by half or so. If the peak voltage is critical and you cannot accept a high value, then you need to increase its magnitude. Matching up the current limit of the inductor with the current limit of the resistor usually gives the better response, though, and is a nice balance for the design.

Anyway, that's a way of doing it. The way it once was done.

Oh, and a last thought. The current-limit resistor's peak heating will all take place in the first tenth of a second. It might absorb several Joules during that turn-on period. Some resistors are better than others at handling these stresses. You can look for "surge resistors" or else wire-wound (which can handle this kind of thing well.) And definitely read the datasheets, where possible, to make sure. If the datasheet rates the resistor for a surge or a certain number of Joules in a short period, then that's likely a good thing. Finally, do some testing, as well.

On the Inductor

Given my hobbyist state of ignorance, inductor design by manufacturers can seem like rocket-science. I believe many important practical details go into designing a commercially competitive device. I'm merely a hobbyist, so I can only stand back and appreciate from some distance and with my sincere respect what a manufacturer applies in designing products.

But there are some basics, too. In the above case, we can work out the energy being stored in the inductor once equilibrium is reached (a second later, at the latest.) The inductor current is DC -- it's not varying much. The energy in the choke inductor is \$E_{_\text{L}}=\frac12\,I_{_\text{L}}^{\,2}\,L_1=\frac12\,I_{_\text{LOAD}}^{\,2}\,L_1\$. In this case, that's about \$540\:\text{mJ}\$.

Webers is the Joules per Amp, so in this case we can work that out as \$\Phi_1 = \frac{540\:\text{mJ}}{2\:\text{A}}=270\:\text{mWb}\$. If you know the \$B_{_\text{MAX}}\$ of the core material and the number of turns, \$N\$, wound on the core, you can work out the the cross-section area as \$A\gt \frac{\Phi_1}{N\cdot B_{_\text{MAX}}}\$. If we are using a good quality iron core with \$B_{_\text{MAX}}=1.1\:\text{T}\$ and if \$N=1000\$, for example, then: \$A\gt \frac{270\:\text{mWb}}{1000\,\cdot\, 1.1\:\text{T}}\$. This suggests that the cross-section area must be \$A\approx 2.5\:\text{cm}^2\$. The 1000 windings will take up some magnetic path length to achieve, so the resulting inductor will have some significant mass.

I may be wrong about the quantitative details. Inductor design is more a matter of dimensional analysis to me as a hobbyist and it's possible I've gotten a factor wrong. But that's how it looks to me. I'll take whatever criticism experts lodge, with appreciation.

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  • \$\begingroup\$ it solves the problem and it was good in the old days but unless the choke is needed for ripple rejection, a simple mosfet will probably be cheaper, smaller, more versatile. The same can be said about Mosfet v relay though. \$\endgroup\$
    – tobalt
    Commented Jul 26, 2021 at 4:21
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    \$\begingroup\$ @tobalt Yeah. I pretty much think everyone is aware of all that. The reason I posted this up is because the OP specifically asked me to elaborate on a comment I made. No other reason. If you go read my comment, made earlier above, you'll also see that I added a reference to a modern (2021) patent that relates to this, as well, using modern parts and techniques. \$\endgroup\$
    – jonk
    Commented Jul 26, 2021 at 6:23
  • \$\begingroup\$ I didnt see that comment. thank you for your detailed elaboration. \$\endgroup\$
    – tobalt
    Commented Jul 26, 2021 at 7:22
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    \$\begingroup\$ @Jerther It wasn't done at the time. The inductor current (and associated magnetic energy) just died through the parallel resistor. In the above case, the inductor is only holding half a Joule when you flip the switch open and operating at about 2 A, as you know. So this just means about 14 V, worst case across the parallel resistor. The whole thing is over within a few tens of milliseconds and no dangerous or damaging voltages involved. And the heating in the resistor is much less than what happens at turn-on. \$\endgroup\$
    – jonk
    Commented Jul 26, 2021 at 16:56
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    \$\begingroup\$ Note that the magnetic calculations match pretty closely to something the scale of a microwave oven transformer -- indeed, one could probably use one this way, opening the core a bit for airgap. Hm, maybe not; one I have here measured 57Ω DCR on the secondary. So you'd have to rewind it. I'm guessing DCR < 1Ω would be desirable here. It is NOT a compact, affordable inductor, in any case! \$\endgroup\$ Commented Jun 23, 2022 at 11:33
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An NTC, Inrush Current Limiters (ICL) rated for the energy being switched is one option 1/CV^2 = E.[J]. Some ICL OEM’s will specify this.

Spec’s: 24V battery nominal ( assume 29 max on charger ).
Cap size: say 5mF.
Max current : 2A load

E=1/2CV^2= 5mF/2 24^2 = 1.44J nom. , 1.7J on charger

Suggested solution : CL-21 to CL-150

I Amp min-max range must include load.
E range > 1.7J

Initial R @ 25’C needs to meet Relay limits e.g. 4 Ohms min.

Final R depends on %Imax but also drops voltage and heats up ICL continuously e.g.120’C preferably shunted by a secondary time delay relay after charging to >80% .

Conclusion: use 8A Relay 5 Ohm ICL and 2nd relay delay to shunt voltage drop.

Cap ESR determines surge currents.

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  • \$\begingroup\$ I read NTCs tend to fail short, and have a somewhat short lifespan. Since you're talking about a 2nd relay to shunt the voltage drop, is there a significant advantage for the NTC over, let's say, a simple 4 ohms power resistor in this situation? I guess it'd be more efficient but is it really worth it for T = RC = 5*(2000 * 10^-6) = 10ms? \$\endgroup\$
    – Jerther
    Commented Jul 25, 2021 at 14:32
  • 1
    \$\begingroup\$ You chose the time delay. Yes it significant as the ?Ax4 Ohm drop means the caps don’t charge up to desired >80% and the 2nd shunt relay will get the surge next. The ICT saves the 1st relay or FET and the 2nd can bypass both to save FET or relay losses and NTC ICT extended high temp. So you get soft start, low R and low loss with 2 relays. \$\endgroup\$ Commented Jul 25, 2021 at 16:40
  • \$\begingroup\$ I understand. There's one thing that occured to me though. Let's say the capacitors have just been charged and the 2nd relay just kicked in so the NTC is still hot, and then the battery is immediately disconnected and the capacitors discharge faster than the NTC cools down (and I believe they actually do). If the battery returns immediately after that, the NTC resistance is going to be low so there's going to be a surge. This could happen for instance when connecting the battery. Am I right? \$\endgroup\$
    – Jerther
    Commented Jul 26, 2021 at 0:32
  • 1
    \$\begingroup\$ Yes that’s why PC PSU’s disable startup for at least 1/2s after shutdown or disconnect with other timers so that intermittent connection with switch on doesn’t surge those using ICL’s for main power \$\endgroup\$ Commented Jul 26, 2021 at 1:30
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I'll show you a naive MOSFET based current limiter, but it may be more of a liability than a solution, depending on how big those capacitors are.

schematic

simulate this circuit – Schematic created using CircuitLab

This works by using a bipolar transistor Q1 to sense current in R1. When the voltage across R1 rises to 0.7V, Q1 begins conducting, and "pinches off" the source-to-gate potential difference of MOSFET M1 (thus switching it off).

In practice, the current is clamped rather than cut-off, because the system finds an equilibrium, regulating current at some maximum. That maximum will be the current required to develop 0.7V across R1, which in my example here is:

$$ I = \frac{V_{R1}}{R_1} = \frac{0.7V}{90m\Omega } = 7.8A $$

Here are the plots of output current and voltage (taken at the drain of M1). The relay is closed at time \$t = 100ms\$:

enter image description here enter image description here

Current is clamped at a constant 8A or so, until the capacitors are fully charged, at which point the only energy consumer remaining is the load R3, and current drops to 2A. With constant current through them, the capacitors must naturally charge with constant \$\frac{dV}{dt}\$, which is why the voltage across them rises linearly.

What are the caveats? Glad you asked. Here's the plot of power dissipated in M1:

enter image description here

Shocking, isn't it? Power rises instantly to nearly 200W, but quickly diminishes to nothing after about 40ms. That might be OK, if you choose the right MOSFET, but it's a bit of a kick in the teeth for the poor thing. This gives you some idea of the abuse that your relays have had to tolerate.

There's really not much you can do with a linear system to mitigate this. As I said in my comment to your question, whatever energy you manage to store in the capacitors, the same amount (not a share of it, this is additional energy) will also be dissipated in the path of current used to charge them. In fact, you can see this relationship in the above power graph. The area under it is the total energy dissipated in M1. Some will have been lost to R1 as well, but it's small in comparison, and I'll ignore it. That area is roughly triangular:

$$ E_{M1} = {1 \over 2} \times 33ms \times 180W = 3.0J $$

The amount of energy stored in capacitor C1 is:

$$ E_{C1} = {1 \over 2} CV^2 = {1 \over 2} \times 10mF \times (24V)^2 = 2.9J $$

I hope I haven't discouraged you too much. I think Jonk's answer promises better results, because the energy is not "lost" in the charging path, rather it is stored in the big inductor's magnetic field. What happens to that energy in the long term is a different issue, but it's possibly less "violent" than a purely resistive solution such as the one I show here.

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  • \$\begingroup\$ This is a good approach -- just mind a complete solution requires SOA calculations to ensure the transistor can handle at least one pulse. Also mind that it isn't tolerant of overloading, or peaky loads (for a peak factor more than 4, as shown -- which is not too bad, probably that's adequate for, say, a motor drive of this size), or repetitive action (no thermal limiting). Protective features quickly get tedious to implement with discrete components, so if you need them, consider a load, hot-swap or wired-OR controller IC. \$\endgroup\$ Commented Jun 23, 2022 at 11:50
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I have used a simple current limiter circuit such as the following, which limits charging current to about 8 amps:

Current Limiter NPN PNP


Upon further investigation, I found that this would not be useful when the capacitor is being charged via a full wave bridge and having a load. The circuit tries to limit the charging pulses and in so doing dissipates a destructive amount of power in the series transistor Q1. LT Spice shows the problem.

LT Spice simulation

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    \$\begingroup\$ PStechPaul - Hi, (a) Please review the tour, the help center and, when you want more details, the Stack Exchange (SE) FAQ. (b) You commented on your own answer to explain a problem. That isn't the correct action on SE. Part of the SE approach is that Q and A can always be (and should be, when necessary) edited by their authors to improve them. You also added a new "answer" just to add an image that you wanted to add in a comment. Again, that's incorrect on SE. I have moved both your comment & "non-answer" to be edits to your original answer. \$\endgroup\$
    – SamGibson
    Commented Jun 23, 2022 at 9:25
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    \$\begingroup\$ [continued] (c) Supplementary information (for a Q or A) must never be added as a new answer here on SE. In future, to add more images to your A you should edit it. (d) Comments don't display images, but image links are possible. In the rare cases when you want to upload an image for use in a comment, the technique explained here should be used. Note that the draft answer (or question) must not be submitted. It is used only to upload the image and is then discarded. || If you need more help with this, please ask on our Electrical Engineering Meta. Thanks. \$\endgroup\$
    – SamGibson
    Commented Jun 23, 2022 at 9:26

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