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General Vintage Technology Discussions For general discussions about vintage radio and other vintage electronics etc. |
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9th Apr 2019, 7:16 pm | #61 |
Dekatron
Join Date: Sep 2007
Location: Oxfordshire, UK.
Posts: 4,311
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Re: Maths help....Again...exp?
My Windows10 calculator looks like the attached pic. What you have to do is to open the menu A and select Scientific from the drop-down. Then click on the arrow B until you get the screen in my pic. Then enter the number you want, say 0.03334, and click the e^x button, C, which will get you exp(0.03334). If you want exp(-0.03334) then you can use the plus/minus key, D, when you enter the number.
Cheers, GJ
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http://www.ampregen.com Last edited by GrimJosef; 9th Apr 2019 at 7:24 pm. |
9th Apr 2019, 7:43 pm | #62 |
Dekatron
Join Date: Apr 2011
Location: Cornwall, UK.
Posts: 13,454
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Re: Maths help....Again...exp?
Brill, many thanks, had the scientific already but didn't know there were more keys, thought it was just exp not ex as well, anyways just done that exercise for 0.03334 and it returns the correct answer, eureka!
Next, a step by step instruction for the procedure to do the calc using the original formula, can you help with that? Original formula...Vr=(Vp-0.7)(1-exp(-T/CR)) I can do Vp-0.7 no problem, I can also do T/CR no problem, the answer should be 0.47 volts or thereabout using that formula with the following inputs...Vp=15 volts...T=0.01667 secs...R=5,000 ohms...C=100uF Lawrence. |
9th Apr 2019, 9:01 pm | #63 |
Rest in Peace
Join Date: Jun 2006
Location: Chard, South Somerset, UK.
Posts: 7,457
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Re: Maths help....Again...exp?
I use MS Excel for all sorts of formulae that arise in electronics. This attachment might help . . . .
Al. |
9th Apr 2019, 11:10 pm | #64 |
Dekatron
Join Date: Apr 2011
Location: Cornwall, UK.
Posts: 13,454
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Re: Maths help....Again...exp?
I think I've figured it out now using the Windows 10 scientific calc now that I've found the ex function and I find it faster using the formula derived by GJ from my original alternative method, is that coincidence or is there a faster way using the original formula?
Lawrence. |