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Why JK flip-flops
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ec navneet



Joined: 21 Feb 2008
Posts: 3
Post21 Feb 2008 18:55   

Why JK flip-flops
Please , if anyone can answer to my these 3 questions :-

1.Why JK flip-flops are called Jump to kill ffs?
2.What is special in referring "solid state switching devices(eg.JFET etc.)" afterall we all know that these are ofcourse in a "solid state"?
3.How the physical size of inductors or capacitors depend on the frequency at which they operate ?

Thanking You in advance ............
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avi_e-



Joined: 18 Feb 2008
Posts: 85
Helped: 5
Location: INDIA
Post22 Feb 2008 15:25   

Re: Why JK flip-flops
Hi Friend,
Here is the answer for your 2nd question.

Before the invention of transistors, early people used vaccum tubes. They were hollow in nature and the transistors which we see now are made of solid silicon or germanium. This is why it is named after. To know more about this refer to "SOLID STATE ELECTRONIC DEVICES" by BEN G. STREETMAN & SANJAY BANERJEE, Prentice Hall of India Pvt Ltd.

Thankyou., Bye Bye!!!!
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halls



Joined: 12 Nov 2007
Posts: 107
Helped: 21
Location: Basque Country
Post22 Feb 2008 23:29   

Re: Why JK flip-flops
ec navneet wrote:
1.Why JK flip-flops are called Jump to kill ffs?

Jump-Kill is to J-K, what R-S is to Reset-Set in R-S flip flops...

ec navneet wrote:
3.How the physical size of inductors or capacitors depend on the frequency at which they operate ?

The physical size affects their main parameters: capacity for the capacitor and inductance for the inductor.

The impedance of any of these elements is a function of the frequency they are working at and their parameter. For the capacitor, its impedance:

3$X_{C}(\omega)=\frac{1}{j\cdot \omega \cdot C}

And for the inductor:

3$X_{L}(\omega)=j \cdot \omega \cdot L

So as you can see, the impedance will vary depending on its parameter, thus depending on its physical geometry. In the case of a full parallel-plate capacitor, for example, the capacity is denoted by:

3$C \approx \frac{\epsilon \cdot A}{d}

Where 3$\epsilon is the permitivityof the dielectric, A is the area of the plates and d is the distance between them. As you can see, the bigger the plates, the larger the capacity. The same way, the longer the distance, the lower the capacity. So geometry affects its capacity.

For a cylindrical inductor, for example, its inductance is denoted by:

3$L=\frac{\mu_{0} \cdot \mu_{r} \cdot N^2 \cdot A}{l}

Where amongst others, N is the number of windings of the coil, A is the area of the cross-section of the coil and l is the total length of the coil... So again, contrary to the popular male belief... size matters Razz

The a further look on Wikipedia:

http://en.wikipedia.org/wiki/Capacitor
http://en.wikipedia.org/wiki/Inductor

Hope it helps.
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cole4eng



Joined: 23 Feb 2008
Posts: 2
Post23 Feb 2008 7:16   

Why JK flip-flops
Great post Halls.
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Post23 Feb 2008 7:16   

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ec navneet



Joined: 21 Feb 2008
Posts: 3
Post23 Feb 2008 18:40   

Re: Why JK flip-flops
Highly thankful for your reply....
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sundarmeenakshi



Joined: 12 Dec 2006
Posts: 322
Helped: 19
Post24 Feb 2008 4:25   

Re: Why JK flip-flops
halls wrote:
ec navneet wrote:
1.Why JK flip-flops are called Jump to kill ffs?

Jump-Kill is to J-K, what R-S is to Reset-Set in R-S flip flops...

ec navneet wrote:
3.How the physical size of inductors or capacitors depend on the frequency at which they operate ?

The physical size affects their main parameters: capacity for the capacitor and inductance for the inductor.

The impedance of any of these elements is a function of the frequency they are working at and their parameter. For the capacitor, its impedance:

3$X_{C}(\omega)=\frac{1}{j\cdot \omega \cdot C}

And for the inductor:

3$X_{L}(\omega)=j \cdot \omega \cdot L

So as you can see, the impedance will vary depending on its parameter, thus depending on its physical geometry. In the case of a full parallel-plate capacitor, for example, the capacity is denoted by:

3$C \approx \frac{\epsilon \cdot A}{d}

Where 3$\epsilon is the permitivityof the dielectric, A is the area of the plates and d is the distance between them. As you can see, the bigger the plates, the larger the capacity. The same way, the longer the distance, the lower the capacity. So geometry affects its capacity.

For a cylindrical inductor, for example, its inductance is denoted by:

3$L=\frac{\mu_{0} \cdot \mu_{r} \cdot N^2 \cdot A}{l}

Where amongst others, N is the number of windings of the coil, A is the area of the cross-section of the coil and l is the total length of the coil... So again, contrary to the popular male belief... size matters Razz

The a further look on Wikipedia:

http://en.wikipedia.org/wiki/Capacitor
http://en.wikipedia.org/wiki/Inductor

Hope it helps.

can u explain jk cllearly
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halls



Joined: 12 Nov 2007
Posts: 107
Helped: 21
Location: Basque Country
Post24 Feb 2008 13:01   

Re: Why JK flip-flops
Well it's quite easy:

JK flip flop has two inputs (J and K) and two (actually is like one) outputs (Q and negated Q).

When you activate J input, you "jump", this is, you activate the Q output. In other words: if J=1 (and thus, K=0), then Q=1 (and thus negated Q = 0). So activating J input is like setting the output to 1 (like de S input of the SR flip flop).

When you activate K input, you "kill", this is, you deactivate the Q output. In other words: if K=1 (and thus, J=0), then Q=0 (and thus negated Q = 1). So activating K input is like setting the output to 0 (like the R input of the SR flip flop).

Then there are two more possible states, because there can be up to 4 inputs:

J=0 and K=0 is used to maintain the specified value, while J=1 and K=1 is used to toggle the output (if it was 1, then it's going to be 0, and viceversa).

All these outputs only are taken on account each time a positive edge reaches the clock input. This means that if you have J=1 and K=0, Q won't be 1 until a clock signal reaches the clock input, which is then when the change of state happens.

That's what is the J=0 and K=0 input is useful for. With that input, it doesn't matter how many clock inputs we get, because the exit won't change at all. In the other hand, with J=1 and K=1 the output will continuously change with each clock edge.

I'm not sure if this is clear, but take a look at the wikipedia article: http://en.wikipedia.org/wiki/Flip-flop_(electronics)
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