i have function generator and that is of range upto 3 mhz. Is there any possible way to get above 6mhz signal using my 3mhz function generator .Thank you in advance

 

Output from IC2-a then you will get a double frequency, and no need the IC1-b, R2, C2, you also need to try the values of R1, C1.

 

i have function generator and that is of range upto 3 mhz. Is there any possible way to get above 6mhz signal using my 3mhz function generator .Thank you in advance

Phase locked loop.

 

i have function generator and that is of range up to 3 mhz. Is there any possible way to get above 6mhz signal using my 3mhz function generator .Thank you in advance

Yes, but it isn't easy. For a digital (pulse) output, the circuit in #2 will work at 6 MHz but the output won't be a symmetrical square wave unless you adjust the R and C values for a specific frequency. A phase locked loop is a more complex circuit, but it can be made to track the input over a range of frequencies. For an analog output like a sine wave you can use a balanced mixer circuit to double the frequency.

ak

 

that will work for pulses, but for other waveforms, it wont work. triangle, sine, and sawtooth waves will not go through those circuits. an rf converter might work hetrodyning the output of the function generator up to the frequency you need.

 

What brand and model of function generator do you have?

 

Yes, but it isn't easy. For a digital (pulse) output, the circuit in #2 will work at 6 MHz but the output won't be a symmetrical square wave unless you adjust the R and C values for a specific frequency. A phase locked loop is a more complex circuit, but it can be made to track the input over a range of frequencies. For an analog output like a sine wave you can use a balanced mixer circuit to double the frequency.

ak

Another method that I didn't try, maybe it can solve the problem that you described, that is double and double frequency and using D flip flop to divide by 2 and then it will be a 50%/50% duty cycle as F= (3Mhz *2 * 2)/2 = 6Mhz.

 

I think that works only if the pulse stretchers are set to 50% of the input pulse width. Otherwise the divided output will not be symmetrical.

ak

 

I think that works only if the pulse stretchers are set to 50% of the input pulse width. Otherwise the divided output will not be symmetrical.

ak

 

I think that works only if the pulse stretchers are set to 50% of the input pulse width. Otherwise the divided output will not be symmetrical.

ak

Just double it twice and feed it through a divide 2 flip-flop.

 

Yes, an asymmetrical pulse train, lets say a square wave with a 10% duty cycle, will, after going through one flip flop, produce a 50% square wave at 1/2 the original frequency. But that works only with 1 degree of asymmetry. If it isn't tuned for a specific frequency, a boxcar-type frequency doubler (example: R-C and XOR gate) will produce a wave that has a narrow pulse at each edge (both positive and negative) of the incoming wave. If the incoming wave is 10% duty cycle, then you get two pulses close together, a long gap, then two more pulses close together. The overall frequency (number of positive edges per second) is doubled, but the resulting wave has three distinct frequency components. If you run this through a FF you do not get a 50% wave, you get the original 10% wave. Running the original wave through two doublers before the FF makes things worse, not better.

And, as I tried to say in my first post, even if this worked it wouldn't work across a wide freq range, or with any other wave shapes.

ak

 

Yes, an asymmetrical pulse train, lets say a square wave with a 10% duty cycle, will, after going through one flip flop, produce a 50% square wave at 1/2 the original frequency. But that works only with 1 degree of asymmetry. If it isn't tuned for a specific frequency, a boxcar-type frequency doubler (example: R-C and XOR gate) will produce a wave that has a narrow pulse at each edge (both positive and negative) of the incoming wave. If the incoming wave is 10% duty cycle, then you get two pulses close together, a long gap, then two more pulses close together.

The state of the flip-flop outputs only change on either the positive or negative edge on the input.

If for instance its negative edge clocked - it just doesn't care where the positive edge is in relation.

 

The state of the flip-flop outputs only change on either the positive or negative edge on the input.

If for instance its negative edge clocked - it just doesn't care where the positive edge is in relation.

Unless you have two pulses close, then a long delay, then two close together again.

 

Unless you have two pulses close, then a long delay, then two close together again.

I don't think any of the frequency doubling solutions suggested so far can produce that situation.

With a *REGULAR* pulse train....

A negative edge is a negative edge - unless you mean a pulse narrower than the flip-flop propagation delay.

 

I don't think any of the frequency doubling solutions suggested so far can produce that situation.

With a *REGULAR* pulse train....

A negative edge is a negative edge - unless you mean a pulse narrower than the flip-flop propagation delay.

I was just trying to clarify what I think AK was trying to say above...

If the incoming wave is 10% duty cycle, then you get two pulses close together, a long gap, then two more pulses close together.

I assume he means something like this...

 

I was just trying to clarify what I think AK was trying to say above...


I assume he means something like this...

View attachment 80395

Not insurmountable - add one each extra x2 and divide 2 stages.

And don't use a doubler with a peculiar output.

 

OK, lets try this again. Attached is a PDF file with two sets of waveforms. The PDF export does not pickup the gridlines, so you'll have to take my word on the duty cycle percentages.

In the upper group, the frequency of a 50% duty cycle waveform (A) is doubled by running it through a bidirectional pulse former. This turns both positive and negative edges into positive pulses. The pulse former is set to 40% duty cycle to show the effect of asymmetry. That wave is doubled again (C), this time with exactly 50% duty cycle. Notice that while the average frequency of C is 4x that of A, it has two distinct frequency components. Passing this through a positive-edge-triggered T flipflop yields (D). The average frequency of D is 2x that of A, but it has the duty cycle distortion of B.

The lower group highlights the phase distortion by using smaller duty cycles. Again, B is perfectly recovered from D, but the phase distortion inherent in B is uncorrected. This always will be the case, and it doesn't matter what the duty cycle of the original waveform is. Without a servo to modulate the charging current of the delay capacitor, this type of frequency doubler can not preserve the duty cycle of the input in the 2f output, and cascading the doublers only makes it worse.

Note that for both groups, adding a 2nd divider after D will get you the original waveform no matter what phase distortions are introduced by B and C.

ak

 

Thanks for you reply.Today my professor said we dont need any funciton generator we can also use the received radio signal to work with circuits (like how we use function generator ).is that making any sense to you peoples