If you add LC filters with large inductors you can very easily get issues with large voltage overshoots at switch on. You can mitigate this by carefully placed additional resistance to reduce the Q of the LC filter but this will often reduce its filtering effect as well. An alternative is to use zener diodes to clamp the rail but beware that there will be a large difference between the zener clamping voltage and its "zero" current voltage.

This is the power supply from one of my designs - L2 and L3 are Bourns SRR0735A-681M and C90 and C91 are 100uF 10V polymer electrolytics with ESR about 0.016 ohms. (Kemet A758EK107M1AAAE016).

The zeners are 5.1V BZX84.

MK

Hi Michael,

Despite what I said in my last comment about leaving the -5V rail with a RC filter, and as I'm going to re-do the PCB, I'm feeling inclined to build up two versions of the +5V and -5V rails.  I've been looking at your circuit and I think it would be interesting to compare the two.  I did have some questions though to help my understanding.

You did say you liked "real filters with big inductors and caps"!  Do such large sizes have any other implications besides the overshoots (and ignoring cost/board space implications)?  The cut-off frequency of the filter is 610Hz and the resonant frequency is also 610Hz - was the values selected to do that?  I assume that any noise at that frequency will be magnified but higher frequency noise is going to be attenuated so where the noise is high-frequency, the low resonance frequency isn't much of an issue?

The 3.3V regulator is fed from the filtered +5V output so I assume its output has the same low ripple/noise as the +5V output?

You mentioned "...but beware that there will be a large difference between the zener clamping voltage and its "zero" current voltage".  I don't really understand what you mean and I haven't been able to figure it out by reading about zeners in more detail.  Could you explain a bit more please.

You'd shown this on an earlier post and gave the values of the output caps on the regulators as 470uF.  I'd not even considered such high capacitance values on the output of the regulators - were these chosen for the requirements of your supply or something you again "like to have"? 

The TPS60403 is an unregulated charge pump inverter and you are feeding it from your 5V regulator (pre filter.)  In my circuit I actual run the +12V through a charge pump inverter and then a -5V regulator.  This is to generate a -5V rail for opamps - a 'typical load' in the TPS datasheet.  Although the output is unregulated, given its purpose I don't see a problem with doing the same with my circuit and saving a number of components (the -5V regulator and its supporting capacitors) - I presume the output LC filter handles the noise and ripple well?

Not to worry  

I'd still like to understand about the zener zero current.

I also asked about the +3.3V being fed from the filtered output of the +5V but realise I'd misread the schematic!  I assume the quality of the 3.3V output was sufficient for your need here, but would there have been an issue with feeding that from the +5V filter?  I would have thought doing that would provide a quality output on V3.3.  This is what I was thinking as equivalent for my need:

12Vout --|-> LC Filter |-> LM7805 -------------------> +5V output

               |                   |--------------------------------> +12V output

               |-> Chg Amp Inv -> LM7905 -> LC Filter -> -5V output

It depends on the application - in mine the 3.3V is for logic (mainly processor) which is not fussy re. noise but will use lots of current, the +/- 5V supplies are for analogue and are susceptible to noise.

Sharing the filter is OK but only if the total current is not too high, on my design the 3.3V current was way higher the the 5V current so the 200mA rated inductor would have struggled.

MK

Ok, makes sense - I can work that through my design.

I'd still like to understand about the zener zero current.

This is from a Vishay datasheet for the BZX84 series of Zener diodes.

You can see that, for the true zeners, the clamping is quite soft. It gets better above 5V, where the avalanche-breakdown effect dominates.

So, the 3.3V zener might clamp to 3.3V at the test current of 5mA, but it's over 4V with even just 50mA flowing through it.

Thanks Jon, that's useful.  So a rail with 40mA on it would be clamped to 4V by a 3.3V Zener; with a very small amount of current it would clamp it to 2V.  For a 5.6V zener, at close to zero current it would be clamping the rail to 5V.

"An alternative is to use zener diodes to clamp the rail but beware that there will be a large difference between the zener clamping voltage and its "zero" current voltage."

This is the sentence I am/was struggling to get my head around in relation to using a zener to clamp a rail against a high inrush voltage.  Let's say I have a 3.3V rail with an inductor.  At turn on, inrush voltage briefly rises to 6V so I use a 3.3V zener after the inductor to (in theory) clamp the rail to 3.3V.  The chart shows that actually a 3.3V BZX84 zener would clamp from 2V @ 'zero' mA to 4.2V @ 50mA and above, give or take. 

From that chart and your explanation, I think the implication is that if I have a load that draws say, 700uA at 3.3V then the zener will clamp the rail to 2V and thus my load isn't going to work. I'd need to, say, add a resistance in parallel to the load to ensure that 5mA, preferably, was being drawn in order to ensure operation of my actual load?  In other words, before putting a zener into the circuit, ensure I understand its breakdown characteristics vs the load and design accordingly?

Thanks Jon, that's useful.  So a rail with 40mA on it would be clamped to 4V by a 3.3V Zener; with a very small amount of current it would clamp it to 2V.  For a 5.6V zener, at close to zero current it would be clamping the rail to 5V.

"An alternative is to use zener diodes to clamp the rail but beware that there will be a large difference between the zener clamping voltage and its "zero" current voltage."

This is the sentence I am/was struggling to get my head around in relation to using a zener to clamp a rail against a high inrush voltage.  Let's say I have a 3.3V rail with an inductor.  At turn on, inrush voltage briefly rises to 6V so I use a 3.3V zener after the inductor to (in theory) clamp the rail to 3.3V.  The chart shows that actually a 3.3V BZX84 zener would clamp from 2V @ 'zero' mA to 4.2V @ 50mA and above, give or take. 

From that chart and your explanation, I think the implication is that if I have a load that draws say, 700uA at 3.3V then the zener will clamp the rail to 2V and thus my load isn't going to work. I'd need to, say, add a resistance in parallel to the load to ensure that 5mA, preferably, was being drawn in order to ensure operation of my actual load?  In other words, before putting a zener into the circuit, ensure I understand its breakdown characteristics vs the load and design accordingly?

I think you're taking Michael too literally. I presume he meant the low current that the voltage is characterised at, in this case 5mA, rather than actually zero. If you were using the diode as a voltage reference, that's the current you'd run through the part to give a reasonably accurate voltage. But if you're trying to use the device to clamp a transient, where the instantaneous current that might pass through the device is potentially quite high, then the voltage that it limits to is going to be a good bit higher.

He was just warning you to not expect it to be a rock-hard clamp at the voltage written on the device.

As you've understood, the zener voltage is going to have to be higher than the rail voltage or the diode will be conducting significantly all the time and, with a small zener like the BZX84, that would then lead you to problems with the device dissipation, even if it didn't drag the rail down.

In a situation like this [transient suppression on a power rail], you might also want to look at TVS devices. They are zeners too, but they're engineered to be more robust.

Thanks Jon, got it now.