EveryCircuit - 50Hz OpAmp Multivibrator

9 years ago

Nice....and for those who want easier maths I think R=1/2.2fC if R1 and R2 are the same. (Eg R=27.5Kohm for this cap to give 50Hz with any equal resistors for R1 and R2 ...say 1kohm each). also... T= 1/f as reqd.( 20ms here each wave length).

Robert_Kidd

9 years ago

Absolutely right! Keep maths simple where possible (although it's really R=1/2.19722458fC LOL).

Robert_Kidd

9 years ago

Of course I'm jesting since component tolerances will have a greater effect than the approximation to 2.2.

2ctiby

9 years ago

I had a good laugh when I saw those decimal places there. It is like telling someone how many millimeters it is from London to Paris. Mind you, conversely, it is important to use loads of decimal places when doing large calculations and then round up the last result. If people do similar calculations but round up each part as they go, then it is surprising how different their final answer can be.

Robert_Kidd

9 years ago

Again, quite correct :-)

Robert_Kidd

9 years ago

I did want to know where you buy your 27.5k resistors too !!!

Robert_Kidd

9 years ago

One good point about this circuit is that the frequency is quite stable with supply voltage changes.

Robert_Kidd

9 years ago

That's not very easy to demonstrate on EC.

2ctiby

9 years ago

Short of stringing resistors together to obtain any value such as 27.5k ohm, I think there are internet sites where you can put in values of this and that and it gives you the other. A little playing with those and you might soon find a suitable mix of components....or did someone mention a pot somewhere? Actually, a slight shift away from that 0.5 kohm makes quite a difference to your Hz result.

Robert_Kidd

9 years ago

I was only jesting. Lots of circuits get posted with odd values. I try to always use preferred values as it makes sense to. I usually work out the combination of resistor values myself to give 'odd' values when required but there are online calculators to help.

Robert_Kidd

9 years ago

Have just posted on SPST to DPDT thread.

2ctiby

9 years ago

Hurz had a real good oportunity there to avoid insults and join with everyone in friendly debate. I suppose that if a person actively wants hate etc then there will be no changing that. We who do not want that have to decide when to stop being drawn in to their den....to stop feeding their appetite and to avoid becoming like them. If he thinks that I am going to keep being his kicking post then he is wrong....he does not seem to be changing.

Robert_Kidd

9 years ago

I have often told him he's a moron. I rest my case, M'lord

Robert_Kidd

9 years ago

I will give any further comments from him with the attention they deserve. By behaving the way he does he alienates everyone except a few who gasp at his apparent knowledge.

Robert_Kidd

9 years ago

Excuse my grammar. Thinking ahead there!

hurz

9 years ago

T = 2RC x log(e) ((R1 + 2R2)/R1) this notation/expression is very missleading or even wrong. What you try to do is use of 'Log(e)(x)' as ln(x), at least i hope you mean this, else your formular is wrong. In case you mean ln(x) Its better written as f = log(e)/[2RC x log(((R1 + 2R2)/R1))] or f = 0.434/(2RC*log(1+2R2/R1)) not because Its calculating f instead of T. Its better this way to use only logarithmus to the base of 10. Hope this helps.

Robert_Kidd

9 years ago

My formula is perfectly right. Indeed I use log to the base e, otherwise know as Natural or Napier logarithms. No, it doesn't help as I happened to know exactly what I was doing.

Robert_Kidd

9 years ago

Hurz your formula containing logs to base e AND base 10 is more complicated than ever and even more likely to confuse people.

Robert_Kidd

9 years ago

Hurz, use your formula and the component values in my circuit. What frequency do you get?

hurz

9 years ago

My formular contains only log() to the base of 10. If I use logarithm to the base of e, I always write ln() or I highlite it in my description to which base the function log() is suppose to be. The notation you are using in your formular log(e)(x) will confuse many users here.

Robert_Kidd

9 years ago

You have e in your first equation but calculate its log to base 10 to give 0.434, thereby removing it for the final equation. That's fine. The point made by 2ctiby is very valid too. By having R1 & R2 of equal value the formula is simpler still with no logs at all. P

Robert_Kidd

9 years ago

If people don't understand natural logs they can ask. That, after all, is how we learn.

hurz

9 years ago

I get 49.6Hz don't know why you get something different. Check your calculation again

Robert_Kidd

9 years ago

No, I got that too. Made an error first time around.

hurz

9 years ago

Nice to see you agree to something. But why not updating the description? Use loge() or ln() or mention your construct log(e)(x) is equal to log with base to e? Is it because you do not know how to update? Or is it a way to keep some Droid knowledge and look more clever then others? Tell us, the community is listening!

Robert_Kidd

9 years ago

I do agree with something when it is correct. No one has asked for the formula to be explained. You don't give them enough credit. I needn't have given any formula but added it in case people wanted to alter the design. Why would I set out with the intention of baffling them? We were having a pleasant conversation up to this point.

hurz

9 years ago

Log(e)(x)

Robert_Kidd

9 years ago

Meh!

hurz

9 years ago

Log(e)(x) were x=4711 gives {4711, 2046, 8.458} three possible solutions. I don't like knowledge hiding as you do here! If you think you have something for the community to be valueable, do not hide information.

Robert_Kidd

9 years ago

You really are an infantile idiot. You set yourself up as 'the people's champion' accusing people of trying to 'look clever' when all you do is try to big yourself up and seek praise whilst trying to make yourself look good by downing those around you. Grow up, stop your childish tantrums. You can't always have it your way and throwing your toys out of the pram doesn't work.

hurz

9 years ago

[BLOCKED]

Robert_Kidd

9 years ago

You think there's any community looking in on this conversation? I doubt there's much interest out there for arguments created by you. Anyone out there? Any views welcomed.

hurz

9 years ago

[BLOCKED]

Robert_Kidd

9 years ago

As you are wrong about my circuit what you say is irrelevant. Just like you.

Robert_Kidd

9 years ago

Despite you crying over my circuits the descriptions will not change. Get a life you sad old fart.

Robert_Kidd

9 years ago

You are too thick to realise that you must turn off mains powered equipment before sticking your fingers inside. How can anyone help a pratt that thinks a circuit unsafe because they want to be able to play with it while it's live? Dickhead!

Robert_Kidd

9 years ago

If you ever manage to make any remark worth commenting on I will respond. I doubt that will happen.

hurz

9 years ago

You see, you start all over the same discussion we already had. And you are agressive as the last time. And the conculsion was last time "Robert, you are a fool", you remember?

Robert_Kidd

9 years ago

Meh!

Robert_Kidd

9 years ago

Thanks for highlighting my circuits for all to enjoy. Appreciated.

ViVBling

9 years ago

Oh, I see what you did there…

Robert_Kidd

9 years ago

Hi there :-)

kiani

6 years ago

[BLOCKED]