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I am trying to use the following MOSFET (N-Channel) to turn an IC on and off. http://www.diodes.com/datasheets/ZXMS6004FF.pdf

In my test circuit I connected 5VDC to the drain of the MOSFET and then connected the source to the V+ supply pin on the IC. The Gnd pin of the IC remains tied to ground. For some reason when I apply positive voltage to the gate of the MOSFET it turns on but I only measure about 2.5VDC at the supply pin of the IC and this is not enough for the IC. Any ideas what I am doing wrong here?

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As Oli says - you need the gate to be more positive that the source by a certain amount to turn the device on. (The level varies with current - for this IC 2 volts is usually enough - see datasheet). It's a very nice part, just not suited to how you are using it.

If your circuit will allow it you can use this part as a "low side driver" which is what the datasheet says it is used for.

Connect source to ground.
Connect drain to load -ve.
Connect Load positive to V+.
Drive gate high to turn on.

This circuit has the advantage of allowing the load to be operated from up to 36 Volts while activating it with eg a 3 Volt supply powered device.
It has the disadvantage that the load is at V+ supply potential when turned off (rather than at ground potential.)

enter image description here

Shown above with a lamp as a load but this can be whatever you are powering. The diode is needed only if the load has an inductive component (to provide a path for the "flyback" reactive energy when the FET is switched off. )

As Oli also notes - IF you can drive the gate to several volts above V+ then your circuit will work.

As Oli also also notes, a P Channel FET will work for you (source to V+, drain to load, load ngative to ground,) with gate high (= V+) to turn off and low (= ground) to turn on. Maximum V+ is supply voltage of driver if you do not use an extra driver stage (1 extra transistor usually).

enter image description here


This is probably the best choice overall:

Using one extra transistor allows you to use a low voltage control signal to drive a load at up to near the FET rated Vmax.

enter image description here


This very nice device may meet your need well - depending on current and voltage requirements. Only 3.6V max Vin :-(. It's an intelligent high side driver with low side logic level control $1.22/1 at Digikey in stock.

enter image description here

The 8 pin dip version of this IC, an ST TDE1898, also a logic level driven high side driver costs $%3.10/1 at Digikey but allows 18-35V supplies. There will be others with weird supply voltage ranges - BUT a P channel FET and single transistor as above probably do what you'd need.


Level shifting:

You MAY be able to switch a 5V high side P channel MOSFET with a 3.3V mcu but the design would either be marginal or tricky. If you swing the drive signal 0/3.3V and have a 5V high side supply the FET sees 5V/1.7V relative to +5V. A MOSFET with a Vth of >= 2V would notionally work. Better Vth > 2.5V or > 3V. As Vth gets higher the on margin decreases. Data sheet max and min values need to be considered. Doable but tricky.

In 2 transistor circuit above use a "logic transistor" (internal R1) to eliminate one resistor. Extra is then one, e.g., 0402 :-) Resistor and one eg SOT23 transistor pkg. // Using a zener in output from MCU can reduce Vmax to safe levels and allow a high side 5V P FET to be driven. "Mickey Mouse" :-).


Use of a resistor divider from output of MCU to high reduces the minimum voltage from high side gate to V+ but also reduces the maximum drive. this may be acceptable.

Example only:
8k2 V+ to P Channel gate
10k P channel gate to mcu pin.
33k mcu pin to ground.

mcu pin is pulled high when OC to 33/(33 + 10 + 8.3) x 5 = 3.2V.
When mcu is at 3.2V, gate is at 3.2 + 1.8 x (10/(10+8.2)) = 4.2V.
When mcu pin is at ground gate is at (10)/(10 + 8.2) x 5 = 2.75V
So relative to V+ gate swings from 0.8V to 2.25V.
This would be OK for some FETS but max and min gate values need to be OK.
Very tricky to get right.

2 transistor circuit is much preferred.
Low side N Channel drive is better still if acceptable.

Both the ICs referred to do the whole task in one IC with no extra components. In both cases the voltage used are limited (<= 3.6BV in one case and 18-35V in the other) but there are certainly ICs that handle a wider range of voltages. www.digikey.com and www.findchips.com are both good places to look.

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  • \$\begingroup\$ Russell, I am now looking at using ZVP2120ASTZ P-Channel MOSFET. The gate is being supplied by my MCU that operates at 3.3V. I need the MOSFET to switch 5VDC on and off. Are you saying that I cannot do this without adding an additional transistor? The datasheet says the gate-source threshold voltage is -1.5 to -3.5V. I really need to avoid adding components to this circuit. I could configure the port pin of my MCU to use an open collector output if that would be of use. \$\endgroup\$ – PICyourBrain Sep 6 '11 at 15:16
  • \$\begingroup\$ Although using the open collector output wouldn't help because I just checked and my output port on this pin is not 5V tolerant. \$\endgroup\$ – PICyourBrain Sep 6 '11 at 15:21
  • \$\begingroup\$ hmm maybe I can find a small SSR instead of a MOSFET \$\endgroup\$ – PICyourBrain Sep 6 '11 at 15:25
  • \$\begingroup\$ You MAY be able to switch a 5V high side P channel MOSFET with a 3.3V mcu but the design would either be marginal or tricky. \$\endgroup\$ – Russell McMahon Sep 6 '11 at 15:34
  • \$\begingroup\$ The eg ADP194 is effectively an SSR. What voltage and current do you need. Is my low side drive suggestion not suitable? (1 x MOSFET, NO other parts). \$\endgroup\$ – Russell McMahon Sep 6 '11 at 15:45
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This won't work as when the supply pin rises, the voltage from gate to source that is turning the MOSFET on drops (as the gate voltage stays constant but the source voltage rises), so it will start to turn the MOSFET off again and settle somewhere between Vdd and GND, depending on the Vth/Ron, how much current the IC is sinking and what voltage the gate is at. If you can set the gate to something >Vth above Vdd (e.g. Vdd + 2V) then it would work (e.g. a pullup to a higher supply).
A better way is a P-channel MOSFET, source to Vdd, drain to IC supply pin. To turn on you pull gate to ground.

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  • \$\begingroup\$ The datasheet for this component says that it's common application is switching power to microcontrollers. Is this normally done by cutting the GND then? \$\endgroup\$ – PICyourBrain Sep 6 '11 at 14:08
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    \$\begingroup\$ It says "general purpose switch driven from microcontrollers", which is a bit different. \$\endgroup\$ – Oli Glaser Sep 6 '11 at 14:13
  • \$\begingroup\$ so it does. should have read that more carefully \$\endgroup\$ – PICyourBrain Sep 6 '11 at 14:42
  • \$\begingroup\$ A great example of this is in the UM232R datasheet. It shows you how to connect a P-Channel MOSFET to USB Power to drive a MCU, and it uses the USB Standby state to drive the gate. ftdichip.com/Support/Documents/DataSheets/Modules/DS_UM232R.pdf \$\endgroup\$ – Erik Sep 6 '11 at 15:39

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