Mosfets - Diode as pull down resistor? - Page 1
Forum section: electronics / beginners

author: nukie
author_rank: Frequent Contributor
timestamp: September 17, 2014, 09:08:10 am
content: Hi all,I am working on a ultra low consumption battery management system for a battery pack. I have this idea (perhaps a very dumb one) to use an ultra-low vf schottky diode as a pull down resistor for a N-channel mosfet.With a 100K pull down resistor, when the Mosfet is turned ON it consume about 30uA@3.8V.With the diode cathode connected to the gate and anode to source/ground the ON current consumption drops to <1uA.The gate seems to discharge quickly which is good.I am not switching the Mosfet fast, it's used to disconnect the battery from the load when battery is below certain voltage.Any thoughts?

author: Simon
author_rank: Unknown
timestamp: September 17, 2014, 09:18:05 am
content: I'f I'm not mistaken your diode is not conducting anyway as if it was you would have a short when you signal the gate, maybe it's high reverse resistance is enough to keep the gate discharged but if that is the case then use a 1-10M resistor.

author: nukie
author_rank: Frequent Contributor
timestamp: September 17, 2014, 10:39:24 am
content: You are right the diode is not only blocking the gate signal voltage from reaching the source but also prevents discharge of the gate. The gate must be discharging slowly through the diode reverse voltage. With a 1M pull down resistor, the current consumption drop to ~3uA. I wasn't sure if a high resistance pull down is good enough for completely discharging the gate in the first place. I don't want a half floating gate draining the power.

author: Simon
author_rank: Unknown
timestamp: September 17, 2014, 10:47:06 am
content: The reason you need a pull down is because static can charge the gate and turn it on or it can charge up through the body perhaps.The resistor just makes sure the gate capacitance is discharged and prevents any build up of charge. try a 10M and measure the drain to source with your multimeter and you will have your answer. I've never personally used 10M but if you want such low power consumption.the only other thing you can do is actively hold the gate low with another mosfet which basically means work out a half bridge low power mosfet driver.

author: bobcat
author_rank: Regular Contributor
timestamp: September 17, 2014, 11:39:32 am
content: What are you using to drive the MOSFET? Can you post a schematic?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 17, 2014, 12:03:04 pm
content: Junction leakage is all over the place,orders of magnitudefor each of the degrees of freedom: manufacturing variation, with respect to temperature, with respect to voltage, and even with other ambient conditions as applicable (light on glass body diodes, RF fields being rectified, etc.).Unless you're driving it from a switch or relay contact, the same will be true of the driver as well. This is literallyworse than "suicide biasing".Tim

author: ConKbot
author_rank: Super Contributor
timestamp: September 18, 2014, 04:42:50 am
content: Use a P-channel Jfet as a pull down, Jfet source gets tied to your existing gate signal, Jfet drain gets connected to ground/source though a current limiting resistor. The Jfet gate gets tied directly to the ground/source connection.When your gate signal is high, the Jfet gate is biased negative, and the Jfet is clamped off other than a couple uA of leakage. This leakage pulls down the big mosfet, slow at first, but as the voltage declines, the resistance of the Jfet decreases also, increasing how fast it pulls the voltage down.When you turn back on, you have to fight the Jfet at first, (hence the series resistor) but once you get enough voltage on the gate to pinch off the Jfet, its back to a nice low-current state.Now, what Jfet to use? I cant give solid advice on that, most datasheets I find give currents at a Vds of 15V, so at 4-5v leakage at cut-off could be vastly different, or maybe not.The MMBFJ176 shows a 10nA drain current with 15Vds, and the cutoff between 1.0 and 4.0 volts, which unfortunately is a pretty wide range, given that this depends on the VGS(off) to set the point at which the pull down changes from low to high current.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 18, 2014, 09:50:32 am
content: Quote from: ConKbot on September 18, 2014, 04:42:50 amThis doesn't work because source and drain are symmetrical. There's no substrate to reference the threshold voltage against, it's always relative to the lesser of the two terminal voltages.You can make a lambda diode with two JFETs connected much like an SCR (or the equivalent with one BJT and one JFET), which will exhibit negative resistance. But you're still at the mercy of JFET variation and voltage scale (the peak and valley voltages are relative to pinchoff or whatever).Another negative resistance trick is to use a pair of CMOS inverters chained (or a single buffer), with a large resistor from output back to input. The input state remains latched, until sufficient current is drawn from the input to push the voltage below the logic threshold, which causes the buffer output to change, and it will again hold the new state. But, this isn't useful for pulling down against an active-high-only signal.Which is why it would be nice if the OP would elaborate on the circuit, because lots of things have active outputs that don't need pull-up/down resistors at all. Especially for battery management, where CMOS reigns supreme.Tim

author: Clear as mud
author_rank: Regular Contributor
timestamp: September 19, 2014, 02:13:42 am
content: I've used a 10M resistor for a pull-down on a MOSFET gate, when I wanted to slow down the turn-off time to a few hundredths of a second. I did it to provide a simple method of switch debouncing without requiring extra components. I haven't actually looked at the output on an oscilloscope, but the circuit seems to operate just fine.