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New to using 4013 flip flop ICs

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chadj2

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I had a quick question about using the 4013 flip flop IC. I am new to using flip flop ICs and have had trouble finding a good tutorial on the net about the detailed operation of this IC. If anyone knows where a decent tutorial that describes the detailed operation of this IC or shows various circuits that can be built with such an IC please provide the link if you can. All I am trying to do with the 4013 right now is to use it to simply invert my input signal. For example, when the signal input goes high I would like a low output from the 4013. From what I was able to read so far I am supposed to tie the set and reset inputs to ground, but I am not sure what to do with the data and clock inputs. Thanks for the help in advance and sorry for the noob question. Here is the datasheet link **broken link removed** .

Chad
 
If all you want is inversion, you should use a 7404 TTL logic chip.

The 7404 is a chip contains 6 NOT gates, NOT gates simply invert an input. So if the input was 1 the output is 0. Or vise versa.
Here is a digram of the chip.

There is a really cool logic simulator here if you need one.
http://math.hws.edu/TMCM/java/xLogicCircuits/index.html

If you just want to use the flip-flop, here's a site that has circuits, and explains the chip.

**broken link removed**
 

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I wish I would have posted here before I ordered the parts from Mouser. I thought I could use the 4013 to invert the signal. There were also 2 characterisitcs I liked about the 4013 also. The first thing is that it can deal with higher voltage than regular logic chips. The second thing was that it could put out decent current (almost 10mA). I guess I will try to improvise with what I have. Thanks
 
wmmullaney said:
If all you want is inversion, you should use a 7404 TTL logic chip.

Just use a NPN transistor and a couple of resistors.
 
Don't use old fashioned high input current 7404 TTL inverters that need a 5V regulated supply.
Instead use 4069 Cmos inverters that have no input current and work from an unregulated supply from 3V to 18V like the 4013 Cmos flip-flop.

The output current from Cmos ICs depends on the supply voltage and how much voltage loss you can tolerate. With a 10V supply and a 5V output the output current is typically 12mA.
 
Yes the 4069 is a simple and effective way of doing what you want, if you want your outputs to go (0) low when you have a high (1) input

Use a 10k pulldown resistor from your input to ground, this will keep your output high when there is nothing connected to it, and when you send a high to your input the output will go low, and as soon as you disconnect that high from the input, the 10K pulldown resistor will sent the output high again.

if you want to have it the other way around (if you want to sent a low (0) to the input in stead of a high, you can connect a 10K resistor to +V, and this will keep your output low, until you connect a low to your input)
 
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That sounds like a better idea then. I guess I will place another part order. Thanks for all the help guys.
 
Here is a schematic of one using a 2N3904 transistor. If the input isn't floating then you do not need to use R2. I did not know what voltage you are using so I calculated values for Vcc of 5V and 12V. When the input is high the output is low, when the input is low the output is high.
 

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k7elp60 said:
Here is a schematic of one using a 2N3904 transistor.
I think the base current is too low for the transistor to saturate well.
When it has a 10mA load then its max saturation voltage is 0.2V when its base current is a whopping 1mA. Compromise with a base current of about 0.5mA. Your base current is only 68uA because you are throwing away too much current to ground.
 

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