Understanding circuit for inrush current limiting

[AngraMelo]

Hey guys,
Im building a couple of linear PS and they are fairly beefy, both around 300W.
The capacitor bank is also pretty  hefty so ill need a inrush current limiting.

On the Texas Instruments website they have the attached circuit but I do not understand it.

The position of that zener holding the 12V point but at the same time tied to ground doesnt make sense to me. Also the way contacts 1 and 3 of the relay are tied up together seems very weird.
Im not criticizing the circuit I just dont understand it.

So why is diode 103 there? Wouldnt it be simpler to have the anode of the tied to ground (with a current limiting resistor)?
Can someone help me with this, please?

https://e2e.ti.com/blogs_/b/powerhouse/archive/2015/03/31/powertips-how-to-limit-inrush-current-in-an-ac-dc-power-supply

[AngraMelo]

So this is my case:
I have a transformer that is rated at 25A and has 21V on the secondary. It has 30,000uF capacitor bank.
So the inrush current would be Vrms minus the diodo drops and that divided by my series resistance right?
Then we get 30V - 2V divided by 82R/10W resulting inrush a current of around 340mA. Now, I used that value because thats what I have for 10W resistors. Will that be enough? If I use the circuit above and after a couple seconds trigger the relay and bypass the resistors wont I have a huge current dump?
Or should I use a lower value like a 10R/10W (that I also have laying around) and use a 555 to wait one sec and then trigger the relay and ignore the power on the resistor given that it is just for 1 second?
That would give me still only 2.8A of inrush current and Im not convinced that bypassing the 10R resistor after one second wont give me a huge current spike.

[Benta]

As it is not apparent what "GND2" and "12V2" is, impossible to say. More info is needed to answer your question.

[AngraMelo]

As it is not apparent what "GND2" and "12V2" is, impossible to say. More info is needed to answer your question.

I cant, that is all it is provided on an article title"Power Tips: How to limit inrush current in an AC/DC power supply"

The link is here

https://e2e.ti.com/blogs_/b/powerhouse/archive/2015/03/31/powertips-how-to-limit-inrush-current-in-an-ac-dc-power-supply

[AngraMelo]

But that is fine, I think the main idea of the circuit is to bypass the resistor after some time has passed and some current has raised the voltage of the caps.
So what am I searching for here? Is there a specific current that I should let pass or a minimum voltage for the caps to reach?
What is the science here?

[rstofer]

D103 seems to be a kickback diode to dissipate the collapsing field of the relay when it is de-energized.

Inrush current of a transformer can be determined by measuring its DC resistance.  You shouldn't be surprised if it is 10 times nominal full load.

I wouldn't spend much time worrying about the rating of the inrush resistor.  It will be switched out of the circuit soon after turn-on.  The resistor value isn't magic either.  Too low and the inrush will still be high, too high and the downstream circuitry won't get fully past inrush.  I would have a tendency to build it the way it's shown unless I can prove it won't work - and I can't.

That strange relay connection doesn't do anything, the designer figured he had to do 'something' with the unused contact and that's what he chose.

There's no charging resistor shown for C109 so the relay may pick up very fast depending on the output impedance of the 12V2 source.  I'm not sure where that source comes from and how long it might take to have enough voltage to pick up the relay.  If 12V2 is derived from the downstream PS then it will take some time (maybe a few cycles) before it is up to voltage.  That's pretty much what you want!

[AngraMelo]

D103 seems to be a kickback diode to dissipate the collapsing field of the relay when it is de-energized.

Inrush current of a transformer can be determined by measuring its DC resistance.  You shouldn't be surprised if it is 10 times nominal full load.

I wouldn't spend much time worrying about the rating of the inrush resistor.  It will be switched out of the circuit soon after turn-on.  The resistor value isn't magic either.  Too low and the inrush will still be high, too high and the downstream circuitry won't get fully past inrush.  I would have a tendency to build it the way it's shown unless I can prove it won't work - and I can't.

That strange relay connection doesn't do anything, the designer figured he had to do 'something' with the unused contact and that's what he chose.

There's no charging resistor shown for C109 so the relay may pick up very fast depending on the output impedance of the 12V2 source.  I'm not sure where that source comes from and how long it might take to have enough voltage to pick up the relay.  If 12V2 is derived from the downstream PS then it will take some time (maybe a few cycles) before it is up to voltage.  That's pretty much what you want!

Im sorry if I was not clear, that is not my circuit. that is just the example shown by texas of how to limit inrush current. My situation is completely different. I described it on reply#1

[rstofer]

No, it wasn't clear.  You linked the TI paper and embedded the schematic in your first post.  Your specific questions in the OP were all related to the TI diagram.

Then, in your Reply #1, you talk about the resistor in terms of the secondary voltage as though that is the voltage seen by the resistor.  Are you just trying to throttle the capacitor charging current?  Or the transformer inrush?

[rstofer]

As to charging the capacitors (if that's what you are limiting), Tau = R*C and it takes 5*Tau to get close to the final voltage.
Tau = 82 * (30,000 *10^-6) = 2.46 seconds
5*Tau = 12.3 seconds.

[floobydust]

That TI circuit is wrong, two 75k ohm resistors R104, R105 in series with a 12V 30mA relay coil???
Needs over a kV to pull in!

[AngraMelo]

No, it wasn't clear.  You linked the TI paper and embedded the schematic in your first post.  Your specific questions in the OP were all related to the TI diagram.

Then, in your Reply #1, you talk about the resistor in terms of the secondary voltage as though that is the voltage seen by the resistor.  Are you just trying to throttle the capacitor charging current?  Or the transformer inrush?

You are absolutely right, my post was a mess. Sorry about that. At first it seems to me that the 30mF capacitor bank is a much bigger problem than the transformer inrush current, I could be totally wrong so please point that out if it is the case.
I dont see a lot of inrush current limiting resistors in simple project schematics that use a transformer. Are they always needed?
if we consider that a load wont be connected to the supply as it turns on then it doesnt really matter the power rating of the transformer, right? The transformer in that case would pull the same current be it a 1W transformer or a 1000W transformer, correct?

Now, if we consider a capacitor bank that size as a load then I think it might be usefull to limit the transformer inrush current, but Im kinda talking out of my ass now. I think my question is this.
Given that Im building a linear PS and the capacitor bank is rather large, what is the considerations about inrush current?
I did some research and it seems that limiting it with a power resistor and then bypassing the resistor is the way to go on PSs with the power rating Im working on. Other sources suggest using a NTC but it gets reaaaally specific and I dont have a large selection to choose from where I live.

[AngraMelo]

As to charging the capacitors (if that's what you are limiting), Tau = R*C and it takes 5*Tau to get close to the final voltage.
Tau = 82 * (30,000 *10^-6) = 2.46 seconds
5*Tau = 12.3 seconds.

I did not know about that! that is awesome! thank you!
So do I need to wait for 5*Tau or can I have a minimum charge to trigger a bypass relay?
I guess it is like you said on reply#5, that the resistor value isnt magic. So what could be a common sense approach to that?
If Im expecting them to charge up to around 30V, should I let the relay go around 25V?

[rstofer]

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html


The charging current falls off pretty fast.  Again, Tau = R*C or 2.4 seconds

I = V/R*(e^-t/Tau) so it will be reduced to 37% in 2.4 seconds.  The maximum current occurs at t=0 and is equal to V/R.

You know the maximum inrush, V/R, now all you need to decide is how long should you wait.  I would think 1 Tau is about right.

Assuming you are just limiting the capacitor current.

[AngraMelo]

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html


The charging current falls off pretty fast.  Again, Tau = R*C or 2.4 seconds

I = V/R*(e^-t/Tau) so it will be reduced to 37% in 2.4 seconds.  The maximum current occurs at t=0 and is equal to V/R.

You know the maximum inrush, V/R, now all you need to decide is how long should you wait.  I would think 1 Tau is about right.

Assuming you are just limiting the capacitor current.

One last question,
If I use the resistor and relay circuit on the primary side of the transformer will I achieve inrush current limiting for both capacitor bank and transformer?

[rstofer]

One last question,
If I use the resistor and relay circuit on the primary side of the transformer will I achieve inrush current limiting for both capacitor bank and transformer?

Yes...  It's hard to say, without seeing your circuit, just how effective the current limiting will be.  Your transformer is rated about 500 VA and at 120V that is about 4A full load.  Starting inrush MIGHT be as high as 10x or 40A.  But that would only last a couple of cycles.  Still, the entire line voltage and the inrush current are working on the resistor.  Think in terms of P = E²/R = 1440 Watts.  Not for long but it just seems like the wrong way to do things.

You could model your design in LTSpice.  You'll have to search the web to find a transformer model, there isn't one in the default setup.

Another approach:

https://www.ametherm.com/inrush-current/ptc-thermistors-for-inrush-current-limiting

[T3sl4co1l]

What kind of transformer do you have?

Toroidial: high inrush current, low leakage.  A double-whammy.
"Shell" (conventional, windings in layers on top of each other on an E core usually): modest inrush, modest leakage (still fairly low).  Very common.  No one really cared, back in the day; why do you?*
Bank wound (bobbin is divided, windings are side by side): Low inrush, high leakage.  Very easy to start even a high capacitance load... but, the voltage droop under load is higher (poorer regulation), and then you need more headroom for a given output voltage at max current.

*Which is a good question, why?  If just for curiosity in avoiding that turn-on "clunk" or dimming of the lights...  ?  (I mean, don't forget that doing it as an exercise, so you know how to do it when you finally need to, has value!)  Is it to meet a standard?  Is it even more practical than that, e.g. you've had inrush knocking out your household appliances or generator or something, and want to make something that'll play nicely in that limited environment?

Anyway, regarding startup resistors: the power rating needs to be adequate to handle the starting energy.  That's contradictory (power is not energy), so bear with me.  Generally, more powerful resistors are also bigger resistors.  For example, a 100W resistor is usually good for 10-100J (5ms pulse).  The dissipated energy is more-or-less equal to the stored energy in the capacitors.

So, 30mF at 30V is 13.5J, so you should expect to dissipate about as much in the resistor during startup.

Probably a 10 to 50W resistor would be adequate, depending.  (I'd say, check the datasheet, but often, they don't say -- and then you need to read a lot of datasheets to get a feel for what resistors of a given style/type may be capable of.  Yeah, it's tedious; well, in short, that's one of the things I get paid for...)

Now, how to control the resistor?  Preferably, it should be in circuit for:
1. Only a brief moment, during startup;
2. Until the output is fully charged (say to 90% of nominal); and
3. Removed from circuit if condition 2 is not met in the expected time frame.

Why 3?  Consider what happens if 2 never occurs, or the bypass relay never closes.  That poor resistor is left supplying the entire circuit in operation.  It quickly burns out (hopefully without making a smoking or sparking mess -- vitreous/enameled resistors are a good choice here), leaving you with a broken supply.  Or, it burns out somehow, and the relay (which would be timed in this case) closes anyway, dumping inrush current to begin with, but also slowly damaging its contacts due to the surge current.

This is easily arranged with a few more transistors, resistors and capacitors.

You don't have to go the extra mile in this case -- I wouldn't be afraid to use the first attached circuit, with suitable changes of course (the resistors as all have noted).  That implements #1 and #2.  This final #3 step is only what you want for industrial reliability.

Tim

[AngraMelo]

Hey Tim,
Great considerations! I was thinking about using a 5W because Im using a 1.2sec delay to bypass the resistor and in that time the resistor does not show any signs of getting even remotely warm. But given your message Ill stay on the safe side and use a 10W.
Now, I dont think I have enough skills to make a circuit to save the PS in case the relay goes to shit.
Would you point me in the right direction so I could start researching that?

thank you very much

[soldar]

On the Texas Instruments website they have the attached circuit but I do not understand it.

It is quite simple. Initially R103 limits the inrush current. When the PSU is outputting 12V then relay K100 is activated and shorts the resistor.

D102 and D103 are there to absorb the flyback current when the 12V is switched off. Not entirely necessary IMHO.

IMHO that circuit is good but overengineered and can be simplified quite a bit. In fact, most computer PSUs just use a NTC to limit inrush and, depending on your needs, that may be enough for you too.

[AngraMelo]

On the Texas Instruments website they have the attached circuit but I do not understand it.

It is quite simple. Initially R103 limits the inrush current. When the PSU is outputting 12V then relay K100 is activated and shorts the resistor.

D102 and D103 are there to absorb the flyback current when the 12V is switched off. Not entirely necessary IMHO.

IMHO that circuit is good but overengineered and can be simplified quite a bit. In fact, most computer PSUs just use a NTC to limit inrush and, depending on your needs, that may be enough for you too.

Soldar,
what is that zener doing there? if you hold both ends of the coil at 12V potential how can there be current flow?
Also, 2 75k resistors in parallel gives the coil a current of 0.3mA, you would need 1.2kV to drive that relay!
I still have no idea of what is going on.

[soldar]

Yeah, I overlooked the two 75K resistors which make no sense and should be shorted.

The diode is there to dissipate the energy stored in the relay coil when it is turned off. It is a very common and standard protection. The Zenner serves the same purpose and can be omitted and the circuit would still work correctly.

[T3sl4co1l]

The zener and diode aren't necessary for a 12V supply, but they are helpful for a switched source.  It's an appnote with other apparent errors, don't read into it too deeply.

Tim

[soldar]

Yeah, it seems like they extracted that from somewhere and did not adapt it correctly. I would ignore it as there are simpler solutions.

For 300 Watt probably a NTC would suffice. Most computer PSU are of higher wattage and just use an NTC.

If you want to go the relay way you can just power the relay from the filter caps.  Suppose the input is 230Vac. Once rectified you will have in the order of 300Vdc.  You could use a relay for 200~300 Vdc, maybe with an appropriate resistor in series. The relay will be open until the caps voltage reaches a certain point and will then close.

A relay solution is more intellectually "better" but a relay can give you problems much more probably than a NTC.

In any case you need a good filter to block noise and transients in both directions.

If cost and space are not at a premium you can look into fitting a filter with some serious inductance which would, not only seriously filter the regular current but also the inrush. I have occasionally used the primary or secondary of a transformer for this purpose.

[AngraMelo]

thank you guys for helping.
What I would like to know is, if I use a NTC or the resistor/relay on the primary side of the transformer will I effectively be limiting the inrush current for the transformer as well as the capacitor bank?
It is way easier to find a NTC (and relay) to the primary side currents then the secondary.

[soldar]

It is way easier to find a NTC (and relay) to the primary side currents then the secondary.

I am confused by your post. What do you mean?

[spec]

thank you guys for helping.
What I would like to know is, if I use a NTC or the resistor/relay on the primary side of the transformer will I effectively be limiting the inrush current for the transformer as well as the capacitor bank?
It is way easier to find a NTC (and relay) to the primary side currents then the secondary.

Hi AngraMelo,

Yes, in most cases the NTC thermistor is placed on the mains side of the PSU transformer and it limits the power to all circuit elements after the NTC thermistor.

Bear in mind that the NTC thermistor will run very hot and should be mounted off the PCB and with a length of self-lead so that the NTC thermistor heat does not crystallize the solder joint or discolor the PCB. You can get stand-off ceramic beads for this purpose. Sometimes high melting point solder is used for soldering thermistors and other power devices to a PCB.

Also bear in mind that the NTC thermistor, even when at a high operating temperature, will add resistance to the primary winding which will contribute to the overall resistance/impedance of the secondary winding.

There are a whole range of NTC thermistors available and there will be one to suit your application just fine. But unless you are using a torodial transformer, in-rush current control may not be required.   

https://www.digikey.co.uk/products/en/sensors-transducers/temperature-sensors-ntc-thermistors/508?FV=fff803f7,fff80044