# Can you help me understand these numbers ?

Discussion in 'General Electronics Chat' started by Hextejas, Dec 21, 2017.

1. ### Hextejas Thread Starter Member

Sep 29, 2017
187
11
I have stepped back from the inverter and am now working on this.
I got it working ok,no smoke but I am trying to relate the answers that I get from the equation, f= 1/RC*4.4, with what I measure and see.
I used 2 different combinations of R and C with the following results.
1) R=2977 ohms, C=100 uf. When I measure the OSC freq it measures 1.32hz and I can see the light blinking at a bit faster than once a second. The Q and Q* lights are blinking at roughly 1/2 the speed of OSC. The result of the equation yielded .007634287. How does that equate to what i measure and observe ?
2) R=47k, C=100uf. Audioguru recommended these values to get close to 50hz. The ORC measures at 97hz and the lights are on solid. I guess to be expected if 97hz. The result of the equation yielded .47846. How does this equate to 50hz ?

Thanks
George KG5TKY

2. ### wayneh Expert

Sep 9, 2010
15,000
5,453
Watch your units and order of operation. The formula likely expects R and C in ohms and farads respectively although you should check that. Be sure you’re applying the 4.4 factor properly. And the realize the combined errors of components values and measurements can give measured values wide of the calculated values.

And yeah, you can’t see 100Hz.

3. ### AlbertHall AAC Fanatic!

Jun 4, 2014
6,679
1,575
47k and 100nF gets you close to 50Hz.
47k x 100n x 4.4 = 20.68mS

4. ### MrChips Moderator

Oct 2, 2009
17,349
5,344
I did not check your numbers.

Yes, the Q and /Q outputs will be exactly one-half the frequency of the oscillator. The chip is designed to do exactly that.

As @wayneh implies, an LED flashing at 100Hz would look as if it is solidly ON, even though it is flashing.

5. ### Hextejas Thread Starter Member

Sep 29, 2017
187
11
I think I did the order properly they are all *. The number of 000s is something else again.

6. ### Hextejas Thread Starter Member

Sep 29, 2017
187
11
It was 100uf rather than 100nf and How do you get 50hz out of 20.68 ms?

Thanks

7. ### Audioguru Expert

Dec 20, 2007
10,606
1,182
My schematic shows 47k and 100nF. 100nF is 0.1uF, not 100uF.
Then the oscillator runs at 96.7Hz and the two outputs are 48.4Hz.
1/20.68ms =48.4Hz.

8. ### AlbertHall AAC Fanatic!

Jun 4, 2014
6,679
1,575
The period of 50Hz is 20mS: 1/0.02 = 50

9. ### Dodgydave AAC Fanatic!

Jun 22, 2012
7,606
1,261
I would say if you want 50Hz at the Q/Q- outputs, use a 45K resistor with a 1K preset, and 100nF cap.

Output F= 1/ 4.4 x RC

Last edited: Dec 22, 2017
10. ### Hextejas Thread Starter Member

Sep 29, 2017
187
11
Thank you all for your replies. Since I am getting close to getting this to work, my next questions are:
I don't have a scope yet but isn't the output of OSC a square waveform sorta like AC ?
And can't I turn that into a proper AC waveform ? ( whatever proper means ).
I will want it to be a lot less than 110 volt.
Another question. Whenever I see an inverter schematic, they use Q and Q* to feed the xformer.
Why not use OSC ?

11. ### AlbertHall AAC Fanatic!

Jun 4, 2014
6,679
1,575
The OSC output is not necessarily 50:50 mark space ratio and would result in a net DC component in the transformer primary which they do not like. The Q and Q* will be 50:50 and so is good for driving a transformer.

12. ### MrChips Moderator

Oct 2, 2009
17,349
5,344
OSC output is 0-5V.
Using Q and Q* on the transformer would constitute a bridged output. Hence you get twice the voltage on the transformer.

13. ### Audioguru Expert

Dec 20, 2007
10,606
1,182
A squarewave inverter is an extremely simple circuit. Its output will drive heating appliances or incandescent light bulbs that are also heaters. 110VAC electricity is a sinewave with a peak voltage of 110V x 1.414= 155.5V which many modern electronic products need. But you do not want a squarewave inverter to produce 155.5V peaks because then it will burn out heaters and incandescent light bulbs that use a 110V sinewave.

A good modern inverter produces a sinewave. A cheaper modern inverter produces a modified sinewave that is actually a modified squarewave like this:

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