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Author Topic: Flip Dots, Shift Register, and HBridge  (Read 891 times)
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Hi,

I have an array of flip dots which I am building a driver for.  Essentially a flip dot is an array of electro mechanical circles (each circle is black on one side and white on the other).  Changing the polarity of the current through each dot flips the dot one direction or the other. 

I wanted to first tackle addressing only 8 dots.  I have one 74HC595 Shift register thinking that I can just connect all the "+" pins of each dot to output 1-4 and the "-" pins to out put 5-8 on the shift register.  Outputing 00001111 would theoretically flip all the dots on one side and hence the same color.  This didn't work as planned.  The pins that get the "0"s seem to be floating ground and don't flip the dots.

My new thinking is that I need to attach an HBridge to the Shift register but I can't get my head around how that would work.  How would I make the output pins in the shift register show the Hbridge output (either + current or - current).

Thanks for any help.  I am new to this and feel like I am very close!

HB
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IIRC the 595 isn't particularly powerful, can it drive the flip dot solenoids?

Do you have a data sheet for the FDs?

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Rob
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My new thinking is that I need to attach an HBridge to the Shift register
Yes good idea.

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How would I make the output pins in the shift register show the Hbridge output
A H-bridge has two inputs, current flows when they are at  different logic levels, so you could just connect them like you tried before.
However if you wanted to save on shift register pins you could invert the output of each shift register bit with a 74LS04 (6 bits inverted in one IC), and feed the signal and it's inverse into the H-bridge. In order to save current you could have one of the shift register bits connected to a master enable line for all the H-bridges.
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Here is the data sheet:

http://www.flipdots.com/EN/electromagnetic_displays/products/page-5/dot_stripes.html

RE: Grumpy_Mike:

If I arranged it this way then I would need an HBridge for each dot that I am flipping?  It seems overkill if I start arraying these strips.
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If I arranged it this way then I would need an HBridge for each dot that I am flipping?
Yes.
The data sheet says:-
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coil current 350mA minimum - burst time 1 ms coil voltage 7.5V

So each Dot requires 350mA, you are not going to get that without a driver of some sorts. You could feed your shift registers into a ULN2008 to pull down and a high side driver like the Allegro 2981 to source the current.
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nominal driving parameters:
coil current 350mA minimum - burst time 1 ms coil voltage 7.5V
implies minimum coil voltage og 6.4V, dot mechanical turning time is within 70ms max, average 50ms.
coil resistance 18.2Ω ±10% (in 22°C ambient temperature)

(18.2+10%) * 0.35A = 7.007V, plus voltage drops across transistor(s), will need more like 9V supply, unless using relays to control current direction.
Using a DPDT relay to control current flow that was discussed recently, altho I suspect a bank of L293s controlling 2 dots each might be more efficient.
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