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Subject:
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RE: mosfet H-bridge
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lugnet.robotics.handyboard
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Date:
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Mon, 20 Jan 1997 17:56:20 GMT
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Original-From:
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Chuck McManis <cmcmanis@freegate.net!NoMoreSpam!>
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1278 times
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Sure, I've built several. The only "trick" to MOSFET bridges is that if you use
all N-channel FETs you need a charge pump on the upper FETS.
(look out for typos, this was done 'off the cuff' so to speak.)
Chuck's Simple MOSFET Tutorial
-------------------------------------------------
Power MOSFETs (also known as Enhancement-mode MOSFETs) make
excellent high current swtiches because their voltage drop is proportional
to the current they are carrying times their 'on' resistance (which is typicall
less than .28 ohms). This also gives them their high current capactity because
the power they dissipate is simply the current squared times the resistance
(P = I^2R)
MOSFETs are "voltage controlled" devices (whereas regular transistors are
current controlled), and when a MOSFET it completely turned on (its resistance
is as low as it will ever get) it is said to be in "full enhancement mode". This
is exactly analagous to being in saturation mode for a regular transistor.
The pins on a MOSFET are labelled "Gate", "Source", and "Drain". One way
to remember which pin is which on a TO-220 package (thats the square one
with a heat sink tab) is to remember the phrase "God Damn Snakes" or Gate,
Drain, Source. Its crude, but it works for me. Remember this only works when
looking at the "front" of the package (the heat sink is on the back)
The voltage difference between two pins on a MOSFET is represented as
Vxy where x is one of g, s, d and y is one of g, s, d. So Vgs is the voltage
measured between the Gate and the Source pins. Similarly the value Rds
is the resistance between the "drain" and the "source" pins. A lower Rds
means the MOSFET can carry more current. (see reason above)
To turn an N-channel MOSFET "on" you need a Vgs that is some number
of volts higher than Vds. So if you connect it like so :
V+
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Light Bulb
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|----+ (Drain)
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Gate ---|<-+ N-channel MOSFET
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|--+-+ (Source)
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Ground
You can compute the voltage between the drain and source as simply Rds * I
where I is the current the light bulb uses (say 1Amp) then for an Rds(on) of .28
ohms Vds is .28volts, and if the "on" voltage is 10v (typical unless it is a
'logic level FET') then the voltage at the Gate must be 10.28 volts in order for
the FET to be turned on completely. NOTE: It doesn't matter if V+ in this situation
is 12Volts ore 2Volts, Vgs must still be 10.28 volts if the FET is to be turned
on, and carrying 1amp. Generally you are safe putting up to 20v across Vgs which
would turn the FET fully on in all situations.
Now if you reverse the situation and put the light bulb on the "bottom" of the FET
you get this situation:
V+
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|----+ (Drain)
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Gate ---|<-+ N-channel MOSFET
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|--+-+ (Source)
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Light Bulb
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Ground
And now the situation gets a bit tougher (in the N-channel case). In this situation
all off the voltage drop will be across the light bulb and so the voltage at the
Source pin is now (V+ - .28v). But the requirement is still to have Vgs be
10 volts higher than Vd so you need a gate voltage of (10.28v + V+) to turn
on the FET completely. That is 10.28 volts *more* than your power supply
for your light bulb (or robot motor).
This used to be a really tough engineering problem with multiple power supplies
etc, but now there is a company called Maxim that makes single chip charge
"pumps". A charge pump uses capacitors to store up charge to create an
output voltage that is higher than the input voltage. (sometimes much higher)
Because of the way FETs are built, there is very little current flow into the
gate (but a high capacitive effect) and so you don't need a lot of current from
your high voltage supply. These two items combine so that building an H-bridge
is pretty easy.
Maxim sells a part numbered the MAX621. This part has four logic level inputs
and four 'gate driver' outputs. You power it with V+ from your motor supply and
the gate driver outputs, when on, go to V+ + 11v (very nice, turns our FETs
completely on, every time)
To build an H-bridge get four N-channel MOSFETs (pretty much any TO-220
FET will have enough current to drive most motors, there are somewhat
harder to find FETs that can handle 30 - 40amps) and get a MAX621 (if you
ask nicely Maxim will send you one as a sample for free, see their web
site http://www.maxim-ic.com) You can also read their data sheets online.
Now since a MOSFET has a natural snubber diode built into it, you don't
need to bother about an external snubber network unless you are switching
them on and off very quickly (ie > 1Mhz).
Now connect two of your MOSFETs with the source pins connected to your
motor supply, and their drain pins to either side of your motor. Connect two
more MOSFETs with their sources connected to either side of the motor and
their drain pins to ground. Now depending on your motor voltage (is it > 11 volts?)
connect two of the gate driver outputs of the MAX621 to to the gate pins of
your two "top" MOSFETS (the ones connected to V+), now if your motor
power supply is > 11v, using an open collector driver IC or simply a couple of
transistors, to connect the motor voltage to the gates of the 'lower' MOSFETS.
The transistor version (using 2907 PNP transistors) would look like:
V+
|
\
/ 47K
\
|----- to MOSFET gate
| /
1K |/ (emitter)
Computer I/O -\/\/\---| PN2907 PNP transistor
|\ (collector)
| \
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Ground
This circuit works as follows, when the Computer I/O pin is at a logic level of '1'
its output will be about 5volts, which will create a current flowing into base of
the PNP transistor which will turn it *OFF*. That will allow the voltage V+
to flow through the 47K resistor, into the Gate of the MOSFET and turn
the MOSFET *ON*. When the computer I/O pin is at a logic '0' it gets connected
internally to 'ground'. This will allow current to flow out of the base of the PNP
transistor turning it *ON*. When the transistor turns on, it will connect its
emitter (the top pin) to its collector, which is conencted to ground, this will
in turn cause the Gate of the MOSFET to be connected to ground which will
turn it *OFF*.
Now the reason you can't just connect the computer pin to the MOSFET directly
is that the voltage wouldn't be high enough to turn on the MOSFET unless it was
a 'logic level' mosfet (harder to find).
Now if the motor voltage is less than 8.5 volts, you can skip the transistor circuit
and connect all four gate driver pins of the MAX621 to the MOSFETs, that is
because the lower gates will see a voltage of 11v + Vmotor and an 8.5 volt
motor voltage is 19.5 volts which is less than the 20 volt maximum (remember
that since the upper MOSFETs connect to the motor, rather than ground they
only see 11v additional voltage provided by the MAX chip). For 6 volt motors
this is a good way to wire it up.
To turn your H-bridge on, turn on one of the upper MOSFETS and one of the
lower MOSFETS. The trick is to make sure that the pair you turn on are connected
to opposite sides of the Motor. If you turn on two on the same side you will
toast the MOSFETs.
You can use some simple logic (and gates) to produce a 'direction' and an 'enable'
bit, the truth table is as follows:
Direction Enable High-Left High-Right Low-Left Low-Right
X 0 0 0 0 0
1 1 1 0 0 1
0 1 0 1 1 0
It is a straight forward job to build this logic circuit.
Now hook up your motors and your off and running!
Last minute caveat, MOSFETs are static sensitive, so be careful when you
handle them. You can damage their current carrying ability without completely
destroying them. This is easily detected though, if one of your FETs gets a lot
hotter than the rest, it has probably been zapped by static.
Enjoy,
--Chuck
----------
From: Scott Sherman[SMTP:sherman@plains.nodak.edu]
Sent: Sunday, January 19, 1997 11:52 PM
To: handyboard@media.mit.edu
Subject: mosfet H-bridge
Hello,
does anyone know how to build a mosfet hbrige and, if so, could they pass
that on to me? It is my understanding that the mosfet hbridge does not
heat up near as much as the transistor hbridge nor does it have any loss
of voltage or current like the transistor hbridge does.
thanks,
scott
sherman@plains.nodak.edu
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