decimals,division..ohms,kilohms...amps,milliamps.. I am so confused I need help!!!!!

[doityoureself]

I have a big problem with the answers to my equations being wrong because of the

decimal point. I have a good understanding of conversions of all the units to decimals, that

is not the problem. But I often come up with 200ohm instead of 20ohm for example, or it

could be .5A instead of .05A or vice versa. I am reading "Electronics..A Self Teaching

Guide" 2nd edition by Harry Kybett. I just do not see a rhyme or reason to much of the

way the author lays out the questions sometimes he will have 50mA÷100 and then

he will use .05A÷100, or 20V ÷ 12mA or 20V ÷ .012A???? I have tried converting

everything to decimals but it just does not always work out, sometimes his answers are

right keeping it in mA and sometimes they are not unless I convert them to decimals. I

feel like there is a rule about what to do in cases such as these that makes sense of all

of this, I feel like an idiot, I always did great in math but its been so long(30+yrs). I am in

the 4th chapter on transistors. In this chapter, it teaches steps on how to turn on and off

transisters as a one, two, or three transistor switch circuits, and how to calculate the right

resistors, Please any help will be greatly appreciated.

 

[davenn]

You should always convert to decimal

show some of his examples where you got different answers to him and see if we can see where the problem lies

cheers
Dave

 

[doityoureself]

examples of confusing equations

Thanks for viewing my problem, I really hope someone can help me and I hope I have made my problems clear enough.

As an example I am using two very similar equations, both 10v circuits with one npn transistor in each, with gain values given, a collector resistor value that is given, and a base resistor value that is NOT given. On the first equation I get the SAME ANSWER as the author, and on the second equation I get a DIFFERENT ANSWER than the author, my problem in the second equation is the decimal position in my answer is different than the author. .

Rc = 1k-ohm, gain(B)=100, 10v circuit...........find Rb

10v ÷ 1k-ohm = 10

10 ÷ 100(B) = .1

10v ÷ .1 = 100k-ohm

Rb = 100k-ohm (same as author)

Now this answer agrees with the authors, I did not convert anything to decimals or worry about designating the current units(mA, A, ect.) Simple enough I thought.

But here is the next equation where again, I did not convert anything to decimals or worry about designating the current unit symbols(mA, A, ect.). My same exact order of calculations bring different results. Same exact circuit with npn transistor and two resistors.

Rc = 22k-ohm, gain(B) = 75, 10v circuit........ find Rb

10v ÷ 22k-ohm = .4545

.4545 ÷ 75(B) = 0.006

10v ÷ 0.006 = 1666.66

Rb = 1666.66 (MY ANSWER) Rb = 1.65M-ohm(AUTHORS ANSWER)

I know this is going to be difficult to follow, I tried my best to make it understandable, I will be very grateful if someone could take the time to try to find where I am going wrong, thanks to all the contributors to this wonderful site.

 

[duke37]

You know that the trouble is that you do not bother with multiples or submultiples. If you work in fundamental units - A, V, Ohms iin your calculations then things will turn out right. With more experience, you can work in multiples, shifting the decimal point usually three places.
In the case of your second example
10/22000 = .004545A
.004545/75 = .000006A
10/.000006 = 1666666Ohm = 1.66MOhm
Your answer of 1666.66 should be 1666.66k = 1.666M and ocurred because you divided by 0.006 instead of 0.000006A or 0.006mA
However, there is an easier way. To get the required current through the transistor, with a gain (B) you need a base resistor which is B times the collector resistor
i.e. 22000*75 = 1650000 = 1.65M

Using the multiples or submultiples can lead to fewer errors since there are fewer digits to play with. You cannot just not bother with the odd factor of 1000 (1k) or 1000000 (1M)

 

[doityoureself]

Thank you very much for taking the time to clear up the confusion I was having and to explain things so very clearly 'duke37'. The explanation on using multiplication to get the resistor size instead of going through the extra steps is also enlightening.

 

[duke37]

Glad that I could be of help. Early IBM computers could not deal with anything other than simple upper case characters and used a system of numbers with E followed by the number of 0s to be multiplied. Thus 1000 became 1E3 and 1/1000 became 1E-3. This is the system I often use and my calculator accepts it. To multiply, multiply the numbers before the E and add the numbers after the E. e.g 2E3 * 3E3 = 6E6 = 6M,
To divide, divide the fist two numbers and subtract the numbers after the E. Thus 1E6 / 5E3 = 0.2E3 = 200
Hope that this does not cause more confusion. The more you use any system, the easier it becomes. Best wishes.

 

[duke37]

I have had an e-mail giving another post but it has not appeared here. Your equation for resonant frequency includes the number 3.3*10E-3. This should be written 3.3E-3 which is 3,3mH. the 3.3 is indeed multiplied by 10 to the power of -3 (divided by 1000) but adding the 10 only increses confusion.
If you wish to make part of the equation subject to the square root, it should be surrounded by brackets, then when the equation is evaluated, the bracketed section is evaluated fist.
Now, perhaps you could help me. How do you get the root sign, multiply sign and divide sign to print with only a standard keyboard?

 

[doityoureself]

yes, there was another post by me but I accidently deleted the whole thing while trying to edit it to say that I realized what I needed to do to and no longer needed assistance, I figured it out when I realized the author was adding the 10 instead of doing just what you explained, anyway, to get the root sign or any other is kinda a pain but I just find it on another page and copy and paste.

 

[nbw]

I have had an e-mail giving another post but it has not appeared here. Your equation for resonant frequency includes the number 3.3*10E-3. This should be written 3.3E-3 which is 3,3mH. the 3.3 is indeed multiplied by 10 to the power of -3 (divided by 1000) but adding the 10 only increses confusion.
If you wish to make part of the equation subject to the square root, it should be surrounded by brackets, then when the equation is evaluated, the bracketed section is evaluated fist.
Now, perhaps you could help me. How do you get the root sign, multiply sign and divide sign to print with only a standard keyboard?

I don't know if using Insert-SYmbol in Microsoft Word, the copy-paste would work?