Switching polarity

[Jason Farmer]

Hi,

I need a bit of help. I have an application where I need to alter the
polarity automatcially without swapping the connections.

Currently I have a 16VDC 900mA power supply and a switch that brakes a
circuit. The first time I use the switch I need it to break +ive the next
time -ive ad infinitum.

Noramally I write software so I can see this shouldn't be hard, I just don't
know what I need to accomplish this!

TIA
Jason Farmer

 

[Reason]

"Jason Farmer" <jason@sapphire.co.im> wrote in message
news:cg23je$phr$1@news5.svr.pol.co.uk...

Hi,

I need a bit of help. I have an application where I need to alter the
polarity automatcially without swapping the connections.

Currently I have a 16VDC 900mA power supply and a switch that brakes a
circuit. The first time I use the switch I need it to break +ive the next
time -ive ad infinitum.

Noramally I write software so I can see this shouldn't be hard, I just
don't
know what I need to accomplish this!

TIA
Jason Farmer

Try a garage-door opener control circuit.

 

[Robert Spanton]

Reason wrote:

Try a garage-door opener control circuit.

A garage-door opener circuit would certainly do, but it seems a bit of
a waste to wreck a perfectly good garage door for that!

What I suggest is that you use a FET
(http://en.wikipedia.org/wiki/Field_effect_transistor) based H-Bridge.
An H-Bridge is simply an arrangement of 4 transistors connected as
shown in http://www.todorobot.com.ar/informacion/tutorial%20stepper/hbridge.jpg
(this shows bipolar transistors being used). The coil (big curly thing
in the middle, just in case you didn't know - marked "Bobina") is the
system that you want to apply the output to. Simply replace the coil
with it.

Replace Q1 and Q2 in the diagram with P-channel FETs and replace Q3
and Q4 with N-channel. When you get some FETs, make sure that they
will have a low enough on-resistance (Ron) at the desired control
voltages.

Connect the input (gate in FET case or base in bipolar case) of Q1 to
the input of Q3 and connect the input of Q2 to that of Q4. Call the
Q1/Q3 interface connection A and call the Q2/Q4 interface connection
B.

Now to have the LHS (as seen on diagram) of the device more positive
with respect to the RHS connect A to ground and B to the positive
rail. If you're using bipolar transistors then you'll need a resistor
in between the rail and the input. To have the RHS more positive than
the LHS simply reverse the previous values - so that A is positive and
B is grounded. If you want an "OFF" state then you need to have some
arrangement of gates that allows you to take the inputs to:

Q1 and Q2: High
Q3 and Q4: Low

Note that to have the OFF state you can't directly connect Q1 to Q3 or
Q2 to Q4 - they require some arrangement of gates too.

Also note that the diodes shown must be power diodes - signal diodes
won't do at all! I used signal diodes once, and it wasn't a pretty
sight...

I suggest that you use FETs because you won't have the junction
voltage drop problems that many simple bipolar designs have. It is
quite easy to create a bipolar H-Bridge, but it requires more
components.

Hope that helps,

Rob