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???
09/01/05 20:28
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Msg Score: +2
 +2 Good Answer/Helpful
#100427 - 8 bit shift register, 8 bit parallel out
Responding to: ???'s previous message
It's a very usefull IC.

The chip has 3 separate parts:

1. A 8 bit shift register, use CLK and DATA inp to shift data in
2. A Latch, the shift register is copied into the latch with STROBE, data in the latch is not shifted when CLK toggles.
3. The latch has an output enable. OE is used to set the outputs into 3-state. For example if you use CD4021 (8 bit parallel in and shifted serial out to a microcontroller)

Your code look like:

1. set STROBE low (0), set OE to high (1)

2. take a byte,
LABEL:
set clk pin 0
set a bit on a port pin (DATA)
set clk pin 1
goto LABEL (8x for 8 bits (byte))

3. Now the bits are in the shift register
but not on the output pins Q1-Q8

set STROBE high (1), now the byte is visible on Q1-Q8
set STROBE low

After setting STROBE low, you can shift in new data.
The outputs Q1-Q8 are not affected when CLK toggles now.
goto 2.

===

When STROBE is always high, you can directly follow
the shifted data on the outputs, the shift register is
continous copied into the latch.

When you want to use two 4094 chips you can Qs from chip 1 connect to DATA of chip 2 and use one CLK signal.
So you have a 16 bit shift register.
Or 3 chips then you have a 24 bit shift register and 24 I/O pins
made with only a DATA, CLK and STROBE signal.

I tried to explain how the chip works.
Success!








List of 7 messages in thread
TopicAuthorDate
CD4094BC            01/01/70 00:00      
   a latch is not a flip flop, when the gat            01/01/70 00:00      
      May I add...            01/01/70 00:00      
   CD4094BC            01/01/70 00:00      
      Synchronous and asynchronous operation            01/01/70 00:00      
   8 bit shift register, 8 bit parallel out            01/01/70 00:00      
      Thank You            01/01/70 00:00      

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