Apologies if this is really obvious, but I'm struggling to understand the datasheet of the FL5150MX IC.

I think my confusion may stem from the integrated shunt regulator.

Under the 'Recommend Operating Conditions' section it states 'Ishunt=5 mA'.

Is that the current for this IC when operated at the typical 17V?



The datasheet is telling you that the device was tested with a shunt current of 5mA. The absolute maximum shunt current appears to be 25mA.

So, you should design the circuits that supply Vs assuming that the nominal current is 5mA but also guaranteeing that the current can never exceed 25mA. The datasheet provides several examples of how to do this.


@TimWescott makes an important point.

If you draw current from the 5V VDD output of this chip then that current will also be drawn from VS and you will need to design accordingly.

The maximum quiescent current (800\$\mu\$A) is specified for VS = 12V, meaning that no significant current is passing through the zener diode. If you design for VS = 17V then the quiescent current is likely to be much larger. I suggest you design your circuits assuming that the shunt current is 5mA when VS = 17V.

  • 2
    \$\begingroup\$ To add to this (because I don't want to toss another answer into the mix): the regulator has a minimum current at which it'll do a good job, and a maximum current beyond which it'll burn up. It is up to you to design your circuit so that the maximum current is never exceeded, and the minimum current is met or exceeded whenever you want the thing to actually work. \$\endgroup\$ – TimWescott Feb 6 '19 at 19:38
  • \$\begingroup\$ Thanks Elliot, how you described it is how is how I expected. I'm glad you mentioned the IQ being at 12V I hadn't noticed that. I've studied the examples in the data sheet but they seem to be relatively wasteful resistive supplies, unless I'm misunderstanding. I'm planning to use a capacitive supply, which, whilst not as efficient as an SMPS, should still be better than the resistive design without the size and cost of an SMPS. I've seen recommendations from others to use 12V VS, and that would seem to make sense with what you've said above regarding the quiescent current. \$\endgroup\$ – J Tester Feb 7 '19 at 15:19

Current supplied to the VS pin should be at least 0.8 mA so that the attached linear regulator can supply +5V to the chip's internal logic. The zener is designed to run at an average current of 5mA. A parallel capacitor (C2) acts as a short-term reservoir. Its supply current through R1 (10k for 120V rms supply).

FL5150 data sheet, Fairchild (ONsemi)
Some parts of the 50/60 Hz cycle appear to rob current from the Zener so that it runs down to 0 mA, and charge is drawn from its parallel capacitor (C2). This can be seen from the data sheet's 'scope photos:
5150 scope (from data sheet) Even at an average of 5mA, the Zener alone will dissipate nearly 0.1W. At the maximum recommended current of 25 mA, this chip would run quite warm, and waste needless power for most applications. I would guess the 25mA limit is a heat dissipation limit for this little chip.

  • \$\begingroup\$ Wouldn't it be 5mA RMS? The 10K Ohm at 120V (figure 1) would seem to imply (120-18)/10k --> 10mA RMS or about 14.5mA peak current into C2 and Zener. R1 dissipates about 1W and the Zener about 170mW. \$\endgroup\$ – Jack Creasey Feb 6 '19 at 17:49
  • \$\begingroup\$ @JackCreasey Looks like the Zener average current changes with phase-control pot setting. While MOSfets drivers are on, the Zener seems inactive: Zener seems to only draw current while MOSfets are off. \$\endgroup\$ – glen_geek Feb 6 '19 at 18:03
  • \$\begingroup\$ Not quite so, look at D1, D2 in figure 1, there is always a Vin on R1. They do what you say in the much more complex circuit of Figure 2, 3, 4. ….but of course it costs more in components. \$\endgroup\$ – Jack Creasey Feb 6 '19 at 18:21

Pin description for Vs says Vs is clamped to 17V for the shunt regulator.


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