why does the 3525 IC produce half frequncy?

Thread Starter

imbaine13

Joined Oct 6, 2013
67
Hello everyone,
I know that the oscillator the 3532 IC feeds into a flip-flop, but can't seem to understand why when the oscillator is oscillating at 100Hz, the flip-flop out puts give 50Hz. May I get a simple explanation for this?

Secondly, could someone help me with the --- dead - peak(+) - dead - peak(-) times of 50Hz and 60Hz modified sine waves which will give 230V RMS, when the peak voltage is 325V?

Appreciate your time.
 

crutschow

Joined Mar 14, 2008
27,890
Hello everyone,
I know that the oscillator the 3532 IC feeds into a flip-flop, but can't seem to understand why when the oscillator is oscillating at 100Hz, the flip-flop out puts give 50Hz. May I get a simple explanation for this?

......................
A flip-flop triggers on one edge of the input waveform not both. Thus it changes stage on one (say rising) edge of input and then changes stage again on the next rising edge, giving an output frequency of 1/2 the input. This characteristic is used in typical counters to count down the input using multiple FFs connected in sequence.
 

alfacliff

Joined Dec 13, 2013
2,458
the flip flop following the oscilator is to produce a 50 50 duty cycle in the output. the oscilator might not have a 50 50 duty cycle.
 

crutschow

Joined Mar 14, 2008
27,890
the flip flop following the oscilator is to produce a 50 50 duty cycle in the output. the oscilator might not have a 50 50 duty cycle.
The only way the FF can generate a 50-50 duty-cycle is by halving the frequency. So you need an oscillator at twice the frequency that you desire at the FF output.
 

Alec_t

Joined Sep 17, 2013
12,248
A 50:50 duty cycle is important if, for example, you are using the output for push-pull drive of a transformer. Ratios other than that can lead to core saturation.
 

Thread Starter

imbaine13

Joined Oct 6, 2013
67
Vrms=Vpeak/(2^½)
Vrms=325/1.414=229.8; round it up to 230 V.
Thanks, I know that, I just wanted to know the time spent in each of the states. In otherwords, for how long will the outout be dead (0v), then for how long it will be high(+325v), an then how long dead again and then low (-325) ... etc, for a period of 0.02seconds (50Hz AC). A modified sine wave.
The intervals are such that the peak voltage is 325v and should give 230v RMS.
Thanks for your input though.
 

Thread Starter

imbaine13

Joined Oct 6, 2013
67
A flip-flop triggers on one edge of the input waveform not both. Thus it changes stage on one (say rising) edge of input and then changes stage again on the next rising edge, giving an output frequency of 1/2 the input. This characteristic is used in typical counters to count down the input using multiple FFs connected in sequence.
Thanks for that. I with that logic, I would think that only one of the outputs gives half the frequency. If both outputs are used to drive say the power stage of an inverter, wouldn't the frequency get doubled taking us back to the original frequency? I'm a little confused?:confused:
 

shteii01

Joined Feb 19, 2010
4,644
Thanks, I know that, I just wanted to know the time spent in each of the states. In otherwords, for how long will the outout be dead (0v), then for how long it will be high(+325v), an then how long dead again and then low (-325) ... etc, for a period of 0.02seconds (50Hz AC). A modified sine wave.
The intervals are such that the peak voltage is 325v and should give 230v RMS.
Thanks for your input though.
Voltage=325 volts*sin(2*pi*50 Hz*time)
So.
Set Voltage=0 volts:
0=325*sin(2*pi*50*t)
You have one unknown, t, solve for it.
And since it is sine, then you will have 0 volts at 0 seconds mark.

Same for the others.
Set Voltage=-325 volts:
-325=325*sin(2*pi*50*t)
you can actually make a leap here, -325 on the left must be equal to 325*(-1) on the right, so sin(2*pi*50*t) must be equal -1, you got one unknown, t, solve for it.

Set Voltage=325 volts:
325=325, sine must be a 1, so sin(2*pi*50*t)=1, one unknown, t, solve for t.


Then you can do the same for 60 Hz.
 

crutschow

Joined Mar 14, 2008
27,890
Thanks for that. I with that logic, I would think that only one of the outputs gives half the frequency. If both outputs are used to drive say the power stage of an inverter, wouldn't the frequency get doubled taking us back to the original frequency? I'm a little confused?:confused:
Both outputs of the FF have the same frequency, just inverted (180° out-of-phase).
 
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