0
\$\begingroup\$

I am using GPIO pins to control the switch. If OE of the 74LVC1G126 line driver is enabled, I like to switch the VBUS on by setting A to low and VBUS off by setting A to high.

enter image description here

The DMG2305UX is a P-Channel Enhancement MOSFET. Gate Threshold Voltage is -0.5 min and -0.9 max.

When Source Voltage (VCC) is 5.09V, Gate Voltage is 5.08V, I would expect there is no voltage on VBUS unless I enable the line driver and set A to low. But there is voltage around 4.43V on the VBUS when OE is disabled.

74LVC1G126 Function Table:

enter image description here

I come from a programming background and do not have much experience with electronics.

Am I designing the circuit correctly?

\$\endgroup\$
  • \$\begingroup\$ Oh yes, silly mistake. Thank you both. Unfortunately, I can only check one correct answer. \$\endgroup\$ – aobs Jun 16 '17 at 9:51
  • \$\begingroup\$ I might have clicked something wrong. The previous comment is gone. \$\endgroup\$ – aobs Jun 16 '17 at 9:52
  • \$\begingroup\$ I deleted my comment and pasted that into an answer, perhaps that is why you think it is gone. You can only accept one answer, doesn't mean the other are considered wrong. You can also +1 all answers which were helpful. \$\endgroup\$ – Bimpelrekkie Jun 16 '17 at 10:54
0
\$\begingroup\$

You have connected that MOSFET upside down.
Change drain and source pins.

\$\endgroup\$
3
\$\begingroup\$

Low side and high side switching: -

enter image description here

Note that for either FET type, the source connects to the power rail for optimum low volt-drop load switching.

But there is voltage around 4.43V on the VBUS when OE is disabled

Re-wire the p channel MOSFET as shown above. At the moment the internal bulk diode will be always forward biased and drop about 0.7 volts to produce about 4.3 volts ALWAYS on the load. This is how you have your MOSFET connected at the moment - look at the body diode: -

enter image description here

\$\endgroup\$
1
\$\begingroup\$

I see that pin 3 of the PMOS appears to be the source as it also has a wire going to the bulk (the anode side of the bulk diode).

Almost always on discrete MOSFETs bulk and source are shorted.

That would mean 3 is the source. This is wrong because the body diode will be in forward mode so Vbus will be Vcc - 0.7 V. This is not what you want.

Swap pins 1 and 3 so source = Vcc and drain = Vbus.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.