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Thread: [MINI TUTORIAL} Philips BUK100-50GL MosFET issue
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picon face BY : Russell McMahon email (remove spam text)



Adrian,

Expanding on Steve Baldwin's answer

The FET you have chosen is an N Channel FET.

terminals are, left to right, pins down, viewed from label side
g    gate (control terminal)
d    drain (similar to collector)
s    source similar to emitter

This can be thought of as SIMILAR TO an NPN transistor with a few very
important differences.

To turn it on the gate (input) must be greater than the source terminal.
You have chosen a "logic level" FET which means you only  need about 5 volts
to turn it on.

When the FET is "on" the drain and the source form a low(ish ) resistance
connection.
Unlike a transistor this is a true resistive connection so current passing
in either direction will see a true ohmic connection - NOT a diode junction.

However, and this is the bad news, when the FET is OFF, in one direction the
FET appears as a high resistance (drain more positive than source) but in
the other direction there is a diode so when drain goes more negative than
source the diode conducts. This diode is an inherent part of the FET design.

In your design there are two choices.

1.    FET as "low side" switch breaking the ground path to the load.
   Load from Vcc to FET drain
   FET source grounded.
   Input voltage to gate (high = on)

2.    FET as high side switch.
   Load from FET source to ground
       (which it appears is what you want)
   FET drain to Vcc (via resistor)
Now the awkward part
   FET gate to Vcc+5volts to turn FET on
   FET gate at Vcc to turn FET off.

Getting the gate above Vcc can be a nuisance in 5 volt circuits.
If no high voltage supply is available AND the FET is to be on for a limited
period this can be done with a simple diode pump.
I won't use ASCII art.
Draw this

   PIC output to "bottom" of a capacitor.
   Diode Anode to Vcc.
   Diode cathode to "top" of capacitor

When PIC output is low top of cap charges to Vcc.- diode drop
Take PIC output high to Vcc.
Top of cap is driven to Vcc + (Vcc - diode drop)

Connecting this point to the FET gate in case 2 will turn the FET on.

This is the basic circuit outline and there will be more needed in practice.

In your case use of a P Channel FET (rarer, dearer, fewer available in logic
level drive) will allow you to use the mirror image of circuit 1 directly
driven by the PIC.

**** EASIEST ****

Using an pnp transistor amay also attractive (emitter to Vcc, collector to
load, load from collector to ground.
Drive base low (via resistor) to turn on. There are many transistors which
will do this. To drive directly off PIC beta needs to be at least 2A/20mA  =
100 at 2A. Ensure beta is at least this high - preferably a bit higher. As
noted elsewhere, Zetex make some very nice small (TO92 variant) package
transistors with high current and high beta. I use ZTX749 in a similar
circuit - this from memory is only be a 1A part with a beta of 100 at 1A - a
bit low for you.

Here we are

ZTX788B
PNP
Farnell 707-790
3A max cont (8A peak)
15V max !!! (take care with inductive spikes)
Beta = 300 at 2A !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
est UK#0.65 ish ??? in 1's
0.45v Vsat at 2A = 900 mW dissipation
E-line (improved TO92 pkg)

Worth a look:

The ZTX705 (Farnell 432-933) is a 1A darlington with a beta of 3000 at 1A.







     Russell McMahon
_____________________________

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What can one man* do?
Help the hungry at no cost to yourself!
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(* - or woman, child or internet enabled intelligent entity :-))


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