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In the below schematic, I am trying to use the shunt pin (normally connected to ground when a barrel plug is not inserted), to allow the switching regulator, U3 to turn off (pulling the EN pin to ground turns the regulator off).

When the barrel plug is inserted, the gate of the N channel FET, Q3 will go high through the resistor, R13, causing the FET to pull the EN pin to ground.

I tried this schematic with a low gain NPN transistor: http://www.mouser.com/ds/2/308/DTC143E-D-278514.pdf (hFE of 15-30) and I noticed the regulator didn't turn fully off when the barrel was inserted, it was outputting about 0.8 Volts or so. In normal operation the regulator is set to output 5 Volts.

With the N channel FET in the schematic below, would I still have the same issues as I did with the NPN BJT?

enter image description here

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  • \$\begingroup\$ Full relevant circuit, test equipment5, load testpoint(s) please. \$\endgroup\$
    – Russell McMahon ♦
    Jul 10, 2017 at 9:19

1 Answer 1

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ACT4088 datasheet here

"Output" is from "SW" pin 1.
Voltage appearing on SW when the IC is off depends on load to ground, which you have not stated, and on internal impedance or leakage from Vin to SW when the IC is off.

SW leakage is given on datasheet page 3 in table as 1 uA typical, 10 uA max. If you were using a 12V supply then 1 uA "looks like" a resistor of
R = V/I = 12V / 1 uA = 12 megohm resistance.

With infinite SW load V_SW could be 12V.
With a 10 mOhm meter - about 5V.

To get ~= 0.8V then Rload on sw could be about
12 Meg x 0.8V/(12-0.8V) ~~~= 860 kOhm.
With 10 uA leakage it could be about 100k.
In fig 2 page 8 the example cct has a feedback divider of about 60 k total and a reverse biased Schottky diode with Ireverse of Murphy_knows kOhms.

ie Without seeing the rest of your circuit and the test conditions it's not possible to be certain, but it sounds like you are seeing about what you'd expect to.

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EN pin switching.

According to the datasheet, the IC is turned off when the enable pin is grounded and on when EN is floating OR when it is above 1.24V when sourcing 2 uA. That implies that the effective on resistance of the switch needs to be less than
R = V/I = 1.24/2 uA =~ 600 K.

It seems nearly certain that that is not your problem.
But, any NMOS FET ever made is liable to have an Rdson of << 600k if you choose to use one. Even though it probably won't help - see above.

If you really care about the 0.8V and want it lower AND if the cause is as above, then adding eg a 10 K load should reduce it to about 0.1V, at the loss of Vout/10k when operating.
To get below that you may wish to add a clamp transistor on Vout also driven by the jack. Add a small series resistor to allow it to survive switchover mis-timings :-).

No guarantees :-).

R8 + R7 form a load of about 60k so the 10 uA max leakage should give
V = I.R = 0.6V

This is similar to the result I obtained above for maximum leakage.
Reverse leakage on D3 would reduce this voltage but is probably very small.

The difference between using an MOSFET and a wire shunt is puzzling. The effects of instead using a say 1 Ohm or 0.1 Ohm resistor have not been advised.

enter image description here

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  • \$\begingroup\$ If I pull EN to ground directly using an SPST switch, the SW output is zero volts, so I was trying to figure out if the NPN transistor I was using is the issue, and hence wanting to try an N channel FET instead. The gain of that NPN I was using was pretty low. \$\endgroup\$
    – klcjr89
    Jul 10, 2017 at 0:14
  • \$\begingroup\$ @aviatorken89 It's always nice to be given all the available useful info before spending time answering :-). FETs with Rdson of < 0.1 Ohm are extremely commonplace. Rdson < 10 milliOhm is common. By using a resistor to ground you can find out what will work and use an appropriate FET BUT what you are using SHOULD work. Transostor datasheet here. R13_9k53 + Rinternal_4.7k ~= 14.2K. Ib ~= 0.8 mA. Vsat = VCe (page 3) is 0.25V max at Ib = 1 mA so hardly more at 0.8 mA and that's at Ic = 10 mA. SO your V_EN low is << the 1.24V needed SO.. \$\endgroup\$
    – Russell McMahon ♦
    Jul 10, 2017 at 7:35
  • \$\begingroup\$ @aviatorken89 ... it is probably the leakage. A FET may well help - try say 1 Ohm or 10 ohms to ground on EN to see result. If they work then FET will too. | Is your transistor polarity correct? What is the voltage on EN with transistor on? \$\endgroup\$
    – Russell McMahon ♦
    Jul 10, 2017 at 7:36
  • \$\begingroup\$ Voltage on EN pin is about 4mV with the NPN bjt transistor active and barell plug inserted in the jack. Voltage on EN pulled directly to ground through the SPST switch was of course 0 Volts. Also, the current draw from the battery is about 11.3mA with the transistor active and the barrel plug inserted (not plugged into a charger or anything, just testing the shunt pin). Current draw measures 0 when pulled to ground directly with the switch. \$\endgroup\$
    – klcjr89
    Jul 10, 2017 at 8:40
  • \$\begingroup\$ @aviatorken89 I summarise your input as: (1) With 0.000 v on the EN pin, Vout on unspecified location with unspecified load and unspecified test instrument connected between unspecified Vout and unspecified reference point is 0.000V. | (2) With 0.004V on EN Vout_unspecified location etc as above is 0.080 V. || A few more specifieds and the whole actual circuit load meter measure-between ... MAY help BUT at present what you are reporting sounds to be VERY VERY VERY unlikely. ie 4 mV ris on EN causes significant change on Vout-unspecified .... . Yes? \$\endgroup\$
    – Russell McMahon ♦
    Jul 10, 2017 at 9:19

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