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I'm designing an aftermarket widget that needs to accept an input voltage range of 8 to 70V DC and power a NE555 timer driving a very high impedance load. The supply voltages will appear as <100ms pulses between 48 and 70V. If the input is 8V then the pulse will be continuous.

Really want to avoid switching regs as any BOM component buzzing away at over 9 kHz makes me, to my understanding, subject to EMC requirements.

I.e. The pulse for voltages > 8V will be up to 100ms, and then drop to ~20% duty cycle @60-600 Hz. For the fringe 8V case, there is an initial 50V burst followed by sustained 8V. My widget creates a faster PWM drive to efficiently buck power to the main load as an inline-plug-in-thing, hence why I want to sip a bit of power off the main rail intended for the main load (which is huge so wont miss a couple of watts). The main circuit works great for OEMs running 48V pure PWM systems, but I found another OEM that does this 50/8V DC thing and I want my widget to support their stuff too, if possible. There are some systems running at 70V - my widget doesn't directly support those systems, but I want the widget to at least survive being plugged into one of them.

From the datasheet the most the 555 will draw in switching is about 15mA. So I made a simple series active zener simulation. Life would be simple if the voltage would always be >40V (add some series zeners to drop a lot of the voltage off before input to this circuit), but there is this pesky 8V fringe case I need to support.

The only way I can think to do it is with a series bjt (ignore 2222 in schematic - that will not survive), as I need as much of that 8V as possible to drive the gate of a MOSFET at the output of the 555. This is why R_inrush needs to be so small too, plus thats just enough in combination with the 18V zenner to keep the voltage across the BJT at 50V (maximum for cheap power 15W SMD BJTs I'm looking at) if some curious individual attempts to plug in a 70V system.

Is there a way I can use a switch, to say, divert current through some series zeners if the voltage > 20, otherwise pass straight to this ground referenced regulator? Open to any other suggestions. Ignore efficiency. It is totally irrelevant compared to the switched load.

enter image description here enter image description here

enter image description here

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  • \$\begingroup\$ Please give more details about the pulsed supply typically how the pulse width alters with supply peak voltage and, at what frequency is oscillates. \$\endgroup\$
    – Andy aka
    Sep 6, 2023 at 11:31
  • \$\begingroup\$ The pulse for voltages > 8V will be up to 100ms, and then drop to ~20% duty cycle @60-600 Hz. For the fringe 8V case, there is an initial 50V burst followed by sustained 8V. My widget creates a faster pwm drive to buck power \$\endgroup\$
    – learnvst
    Sep 6, 2023 at 11:53
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    \$\begingroup\$ Since the output current is low, have you considered an LMC555 or 7555 -- the CMOS version? There are much newer, better timers out there, too. \$\endgroup\$ Sep 6, 2023 at 13:18
  • \$\begingroup\$ @TimWilliams . . sure. CMOS is one option. I had in mind, but I was wondering if I was missing a trick. Like a way to divert the current based on voltage \$\endgroup\$
    – learnvst
    Sep 6, 2023 at 13:35
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    \$\begingroup\$ "Really want to avoid switching regs as any BOM component buzzing away at over 9 kHz makes me, to my understanding, subject to EMC requirements." Err... putting anything containing electronics on the market usually makes you subject to EMC requirements in most countries. Where did you get this 9kHz limit from? Also: harmonics. \$\endgroup\$
    – Lundin
    Sep 7, 2023 at 11:18

5 Answers 5

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I would recommend a depletion MOS limiter:

enter image description here
From: https://electronics.stackexchange.com/a/237572/311631

For a zener of say 9V, and a suitably rated depletion NMOS, output voltage will range from 9V (at IDSS) to maybe 11-12V (VZ minus whatever VGS(off) is), which is fine for gate drive purposes, or can be regulated tighter with an LDO proper. Zener current can be quite small; with a 50µA-rated zener family, nominal voltage is had for R1 in the 100k range.

Note that filtering and additional protective components apply. Probably you want a bulk capacitor on the output. Clamp diodes to prevent output reversal and excessive VGS are likely. A polarity diode at the input is likely, and perhaps a TVS or MOV to handle transient overvoltage.

Efficiency can be further improved by replacing R1 with a constant current source (which can in turn be constructed with a depletion MOS). This probably doesn't matter here.

For higher currents, a switching regulator is desirable; a wide input range isn't necessarily available though, in which case both circuits can be combined, the D-NMOS serving to limit to the switcher's max input, in which condition input current will be much lower than final/output load current is, so its dissipation is greatly reduced for high input voltages.

This is also a good approach in applications with large surge voltages, like automotive load dump.

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  • \$\begingroup\$ Interesting reading. Will investigate depletion mode NMOS. TVS on input or output side? \$\endgroup\$
    – learnvst
    Sep 6, 2023 at 14:17
  • \$\begingroup\$ Good solution. +1 Was just looking at using a DN2450, which is less than 10Ω. \$\endgroup\$ Sep 6, 2023 at 14:35
  • \$\begingroup\$ Simulations look good. Voodoo component \$\endgroup\$
    – learnvst
    Sep 6, 2023 at 18:26
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This is more an "other suggestions" thing. Today, there are a plethora of small monolithic buck converters that would provide a regulated output voltage over that range, from the TI LM5009 or LM5164 (100V) to parts such as the 80V Shanghai Siproin SSP9480, with little added complexity, and no heat dissipation issues.

enter image description here

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    \$\begingroup\$ This is the modern way to do things, but there is a gotcha I didn't mention. This would be a device with a switching frequency of over 9 kHz on the BOM, which would then - to my knowledge - require my device to go through all of the EMC requirements to get it CE stamped. \$\endgroup\$
    – learnvst
    Sep 6, 2023 at 13:28
  • \$\begingroup\$ .... but then that's a whole other can of worms. I will definitely investigate the components referenced \$\endgroup\$
    – learnvst
    Sep 6, 2023 at 13:30
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    \$\begingroup\$ You can buy the monolithic version of your diode+transistor from outfits such as Diodes Inc - it's almost ideal (15mA characterized and up to 100V or so) but they have chosen a base resistor that doesn't guarantee low dropout. Maybe you could use a CMOS 555 for a few pennies more and get down to < 1mA. \$\endgroup\$ Sep 6, 2023 at 13:35
  • \$\begingroup\$ @learnvst CE is self-certified, so in theory you don't need to do tests. While I wouldn't do that for your own layout, there are pre-made modules you can buy, and quality ones come with EMI information. IFF the module is the only > 9khZ part in your device, skipping tests doesn't sound that bad. \$\endgroup\$
    – jaskij
    Sep 7, 2023 at 18:26
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The simple buck converter in the following SE answers may be useful.
I have used it to produce 12V out from 12-200V in.
It acts as a low dropout linear pass element below regulation voltage and startssmoothly as Vin exceeds Vreg.

It's switching frequency can be arbitrarily low with due care.
As shown in the examples it's switching is semi-chaotic.
This can be changed by adding some formal poitive feedback.
Various people have said that it cannot work. It does :-)

High side FET used here
Can be bipolar if desired.

enter image description here

Bipolar version here

enter image description here


My 2001 design challenge related to the above circuits here

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  • \$\begingroup\$ Dear god, 66nC gate charge, through a 100k resistor!? \$\endgroup\$ Sep 6, 2023 at 14:25
  • \$\begingroup\$ @TimWilliams Final details differed :-) . AND implementation is mine. The requirement was for an efficient enough psu with Vin from 12v-200v, low cost, well behaved below 12V in, ... . At 200V in you'd get "somewhat more drive" from the 100k than you'd usually get. I don't now recall (20+ years on) what the final value of RBUK4 was, but it needs to be high enough to allow "not too much" division with RBUK3. Value of RBUK3 is constrained by dissipation at 200V. (About 1 Watt here). || This was in an exercise bike. Initially children would set load to zero and pedal flat out and destroy the FET \$\endgroup\$
    – Russell McMahon
    Sep 7, 2023 at 1:28
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    \$\begingroup\$ ... I changed the software to limit the maximum voltage by adding load when Vin rose too high. This did not affect serious users. This would affect the R4:R3 allowed ratio. The final version added an output current limiter. At 200V Vin efficiency was about 50% (compared to 6% for a linear regulator) BUT at usual voltages it was very much higher. Overall it worked very well indeed. That was the highest voltage PCHannel FET at sensible prices available in Taiwan at the time. || A gate turn off clamp could have been added - but the extra cost and complexity was not needed in practice. \$\endgroup\$
    – Russell McMahon
    Sep 7, 2023 at 1:32
  • \$\begingroup\$ @TimWilliams I found [this] (electronics.stackexchange.com/questions/173819/…) post open on my PC - maybe from a GSR search. In 2015 I said " ... The final version may have added (as far I recall after 14 years) a high side emitter follower to speed up the FET switching. ..." --> So it looks like I agreed with you even then :-). || Longish December 2004 discussion of the design here \$\endgroup\$
    – Russell McMahon
    Sep 7, 2023 at 5:12
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If your circuit is a 555 driving a MOSFET, I'd replace the NE555 with one of the CMOS variants, for example LMC555 which has a supply current of less than 500µA, or TS555 which has even lower supply current. It accepts up to 15V supply and has enough output current capability to drive your PWM MOSFET (10-100mA). In addition the low input currents mean you can use high value resistors to save power.

This will greatly reduce supply current. You should check how much by including the MOSFET drive current, resistor dividers and other parts.

Then you can use DN2540, or... once the current draw of the circuit has been reduced to less than 1mA, a simple BJT. It's on the left. I used a current source to bias the Zener diode to get lower dissipation at high input voltage. To get lowest voltage drop with BJTs, I'd use a simple Zener shunt regulator (on the right) powered by a constant current source.

enter image description here

Due to the CMOS push pull output stage, these 555 variants can drive your MOSFET with the full supply voltage. The plain NE555 has a dual NPN output stage, which can drop about 2V from the supply, so it will only drive your MOSFET with about 6V at your minimum supply voltage of 8V.

This means even with the (low) voltage drop due to the current source in the shunt regulator proposed above, MOSFET drive voltage will be higher on low supply voltage compared to NE555 plus DN2540.

In order to avoid driving the FET with a low voltage that would make it dissipate too much power, I'd recommend driving the 555 Reset input with a power supply supervisor "reset chip". Here's an alternative using only passives: the two Zener diodes ensure Reset pin will only receive voltage when proper supply is present. However Reset threshold is not very accurate on 555, so that's the cheap solution.

enter image description here

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  • \$\begingroup\$ Thanks for the amazing input. The question on my mind now is, is there any advantage to the current-source-shunt over the DN2450 based solution if both are used with a CMOS 555? Pure cost vs space? \$\endgroup\$
    – learnvst
    Sep 7, 2023 at 17:29
  • \$\begingroup\$ DN2540 has lower voltage drop. Current source is a bunch of cheap parts that are always in stock ;) \$\endgroup\$
    – bobflux
    Sep 7, 2023 at 17:37
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The simple solution is just to buy a linear regulator that can tolerate the maximum input voltage, while having low enough dropout to work at the minimum voltage.

For example TPS7A43 uses a two-stage linear regulator structure to achieve the combination of 85V input voltage range and dropout voltage of 800 mV.

The regulator will still need to dissipate the wasted energy (about 1W) as heat, so make sure to lay out a large copper area on PCB for that. If you can minimize current draw by switching to a CMOS version of NE555, that will reduce the heating and energy waste.

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