EEVblog #1285 - How to do Design By Inspection

[EEVblog]

What is "Design by Inspection"
Dave answered what he thought was an obvious forum question, but to beginners it's not so obvious. Rather than build something up and test, let's do design by inspection.

[johnlsenchak]

This video is like when you use to do  "fundamental  Friday"  which I always  liked.  You need to do more  videos like this !

[beenosam]

It's always seeing videos about fundamentals things, especially as a new engineer.

[nctnico]

I found 'design by inspection' an interesting term and Googled it. It doesn't seem to be a very common term. I did find a thread on a construction engineering forum ( https://www.eng-tips.com/viewthread.cfm?qid=350761 ) and on there they defined 'design by inspection' more like design re-use. For example if you need to calculate beams for a floor construction you can calculate one beam and say the other beams see the same load so if they have the same dimension they are OK. If the floor is loaded un-evenly or has a non-rectangular shape then calculate the worst case beam and from there it is safe to assume the other beams (with the same size) will be able to carry the load.

If you translate that to electronics: if you have tested & verified a sub-circuit (for example a switching DC-DC converter) it is relatively safe to use it in other designs as well for as long as it stays within the test & verification limits. I don't think copying a circuit from a datasheet or application note counts as 'design by inspection'. You'll still need to build & verify the circuit.

[Alti]

Design by inspection aside, I have some remarks:

"HV regulator versus capacitor with zener solution" and "orders of magnitude less current" and "use vastly less power".

First remark is that for end user it is mostly irrelevant how much mains current a device for domestic purposes draws. This is an interesting problem with minimal practical value.

Second remark is that I think that you forgot to start from underlining a peculiar requirement that narrows down all claimed advantages of your concept (EEVBlog #1284), namely that you are looking for a solution that draws 10uA 99.9% of the time and 15mA 0.1% of the time, 15VDC, IIRC.

Three orders of magnitude 

As for the proposed circuit itself - capacitive dropper comes with inrush current limiter and the HV regulator does not. Why? At startup there is 324V accross input diode. 

[mcovington]

What happens when the alarm sounds and starts drawing 15 mA or so?  The low-Iq linear regulator would then be dropping something like 400 V at 15 mA, which means it would dissipate 6 watts.  Can it do that?  It doesn't have to be efficient then -- these alarms don't spend much time sounding -- but it had better work!

As for the previous poster's question whether this whole thing is relevant, it certaintly is when a country is going to install millions of them, and when there might be hundreds in a single large building.  Factor-of-1000 inefficiencies are not to be sneezed at.

[EEVblog]

Second remark is that I think that you forgot to start from underlining a peculiar requirement that narrows down all claimed advantages of your concept (EEVBlog #1284), namely that you are looking for a solution that draws 10uA 99.9% of the time and 15mA 0.1% of the time, 15VDC, IIRC.
Three orders of magnitude 

If your smoke alarm is going off for 86 seconds every day I think you'd better lay off smoking the wacky weed.

[Alti]

Second remark is that I think that you forgot to start from underlining a peculiar requirement that narrows down all claimed advantages of your concept (EEVBlog #1284), namely that you are looking for a solution that draws 10uA 99.9% of the time and 15mA 0.1% of the time, 15VDC, IIRC.
Three orders of magnitude 

If your smoke alarm is going off for 86 seconds every day I think you'd better lay off smoking the wacky weed.

Second remark is that I think that you forgot to start from underlining a peculiar requirement that narrows down all claimed advantages of your concept (EEVBlog #1284), namely that you are looking for a solution that draws 10uA 99.9% of the time and 15mA 0.1% of the time, 15VDC, IIRC.
Three orders of magnitude 

I do not know how long the smoke alarm should stay beeping when tripped, this definitely does not happen periodically and definitely not daily. The percentage values given indicate the time relationship in between two varying modes, the core assumption of your concept that you (I presume) forgot to point out.

Anyway, thank you for your remark.

[nctnico]

What happens when the alarm sounds and starts drawing 15 mA or so?  The low-Iq linear regulator would then be dropping something like 400 V at 15 mA, which means it would dissipate 6 watts.  Can it do that?  It doesn't have to be efficient then -- these alarms don't spend much time sounding -- but it had better work!

That is a good catch. You can't really use a purely linear regulator for these purposes. But a linear regulator would be a good solution as a post-regulator behind a capacitive dropper.

As for the previous poster's question whether this whole thing is relevant, it certaintly is when a country is going to install millions of them, and when there might be hundreds in a single large building.  Factor-of-1000 inefficiencies are not to be sneezed at.

But you can say the same about standby energy consumption of mobile phone chargers, TVs, settop boxes, stereos, etc, etc. There is very little equipment nowadays which has a hard power switch.

[EEVblog]

I do not know how long the smoke alarm should stay beeping when tripped, this definitely does not happen periodically and definitely not daily. The percentage values given indicate the time relationship in between two varying modes, the core assumption of your concept that you (I presume) forgot to point out.

I did not forget:

[nctnico]

Dave, I hate to say but you have dug yourself into a bit of a hole here. Take a deep breath and count to 10. The wacky weed remark is totally uncalled for. The basic NCP786A circuit just isn't suitable to power the beeper. Given the thermal performance of the package 'Engineer this right' comes down to using a different solution alltogether. Perhaps this isn't the best example of 'design by inspection' after all.

[mcovington]

In fact, regarding the 15-mA question (what to do when the alarm is sounding), I see that the NCP786A can deliver 10 mA and has no strict limit on power dissipation; rather, a thermal shutdown temperature of 145 C.  So with a generous heat sink, it's close to being what we want, and there is surely a similar chip that is a bit bigger and can definitely do it.

[Alti]

(..)First remark is that for end user it is mostly irrelevant how much mains current a device for domestic purposes draws. This is an interesting problem with minimal practical value.

(..)
But you can say the same about standby energy consumption of mobile phone chargers, TVs, settop boxes, stereos, etc, etc. There is very little equipment nowadays which has a hard power switch.

Ok, I'll put some bolds here.
Minimizing a mains current draw of a domestic appliance (like smoke detector) has minimal practical value because it is the active power we are charged for and the network is loaded with mostly inductive loads. Of course it is nice to have a gear that draws several uA and has built-in power factor corrector but from practical point of view, even if it has low PF - this is neither a problem nor a cost that can be worth saving.

As for active power (current*voltage*PF), that is a different story.

[nctnico]

In fact, regarding the 15-mA question (what to do when the alarm is sounding), I see that the NCP786A can deliver 10 mA and has no strict limit on power dissipation; rather, a thermal shutdown temperature of 145 C.  So with a generous heat sink, it's close to being what we want, and there is surely a similar chip that is a bit bigger and can definitely do it.

Let's assume there is a slightly beefier chip. I don't know the requirements for smoke alarms but I'd design them for (at least) 60 degrees ambient. Next step: 145 degrees is the thermal shutdown temperature of the device. You don't want to operate it near that temperature due to component variation. It is likely to go off a couple of times in it's life. So I would try and stay below 120 degrees for reliability. Worst case mains is about 265V which translates into 375V DC. That results in 375*15mA=5.4 Watt. With 60 degrees of temperature margin you'd need a thermal solution which is better than 11 K/Watt.  BTW the thermal limit of the NCP786A is given in the datasheet: the datasheet says that junction to ambient has a thermal resistance of 82 K/Watt while the chip is mounted on a 25mm x 25mm copper plane. There are SMD heatsinks of around 11K/Watt (like this one: http://www.assmann-wsw.com/wo/en/products/thermal-management/smd-and-copper-heat-sinks/detail/1059624/ which costs around .70 US$ in 5k quantities) but the question is whether you can get all the heat from thermal pad of the chip through the board into the heatsink. All in all this is not a route I'd like to take.

[EEVblog]

In fact, regarding the 15-mA question (what to do when the alarm is sounding), I see that the NCP786A can deliver 10 mA and has no strict limit on power dissipation; rather, a thermal shutdown temperature of 145 C.  So with a generous heat sink, it's close to being what we want, and there is surely a similar chip that is a bit bigger and can definitely do it.

And add in a resistor dropper.

[EEVblog]

Dave, I hate to say but you have dug yourself into a bit of a hole here.

Nope.

[mcovington]

In fact, regarding the 15-mA question (what to do when the alarm is sounding), I see that the NCP786A can deliver 10 mA and has no strict limit on power dissipation; rather, a thermal shutdown temperature of 145 C.  So with a generous heat sink, it's close to being what we want, and there is surely a similar chip that is a bit bigger and can definitely do it.

And add in a resistor dropper.

Bingo!   Good idea.

[dcac]

You can also allow for some "excessive" input ripple by choosing a smaller rectifying cap. So in the 15mA stage perhaps let it ripple 50V-100V, or even more, to cut down the total power losses.

But all in all there probably will be some 3-4 watts total loss and that tend to generate some heat in a small device.

[nctnico]

In fact, regarding the 15-mA question (what to do when the alarm is sounding), I see that the NCP786A can deliver 10 mA and has no strict limit on power dissipation; rather, a thermal shutdown temperature of 145 C.  So with a generous heat sink, it's close to being what we want, and there is surely a similar chip that is a bit bigger and can definitely do it.

And add in a resistor dropper.

Bingo!   Good idea.

No. Run some numbers on it first. The minimum voltage you need to drop in the regulator is 375 (Vmax) -265 (Vmin) =110V * 15mA is still 1.65W (which is over twice as much as you can realistically dissipate without a heatsink). And this is without having any margin for drop-out and ripple on the rectifier capacitor.

[kkessler]

In the transformerless power supply DaveCad'ed in this video, isn't the zener diode forward biased with each half cycle? I'd think that would be bad, and maybe the zener should be after the normal diode.