| Electronics > Beginners |
| Support in learning the art of electronics |
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| MrSlack:
1N5294J is obsolete and commands a high price. It's just a JFET and a resistor in a diode package. You can read AoE and make your own in about 10 mins flat with a resistor and a stock JFET. The book also describes how to make a current mirror instead of a limiting diode. YMMV but I'd ignore the shopping list and buy a chapter at a time with suitable substitutes. |
| BobsURuncle:
--- Quote from: MrSlack on April 17, 2016, 08:18:01 pm --- YMMV but I'd ignore the shopping list and buy a chapter at a time with suitable substitutes. --- End quote --- To each his own, but personally I'd risk buying a few wrong or less than cost efficient parts for which I may or may not find some other use rather than pay huge shipping and handling charges. I don't know what you mean by a chapter at a time in this context as the book is divided into Labs. There are 25 labs and a half dozen or so suppliers. Assuming an average of 2.5 suppliers per lab and $6.00 shipping charge each that adds up to $375. Even if you buy supplies for one of the 6 "Parts" of the book at a time with an average of 3 suppliers each, that is over $100. And you are going to be waiting months for some of the parts which makes it impossible to go through the book in a reasonably orderly and timely fashion. Actually this discussion ought to be in the "Learning The Art of Electronics - Parts BOM" post since this thread was supposed to be about helping people who get stuck on certain technical topics in LTAOE. So I desist. |
| MrSlack:
You missed the $1000+ of test gear as well... This is still pretty cheap for an education. University here in the UK costs ~$12,745 a year. |
| rstofer:
--- Quote from: orolo on April 14, 2016, 09:14:12 pm ---In your drawing you will measure essentially the 10 volts of the battery, and your 2 Volts measurement range will overflow. Instead of using the multimeter in parallel with the big resistor, try using it in series. What will happen? --- End quote --- At their heart, voltmeters are measuring current through their internal resistance (plus scaling resistance). In the case of an analog meter, the current in the meter winding deflects the needle. So, putting the meter in series with the 10V battery and the 1GOhm resistor will cause 9.9 nA to flow (10V / (1GOhm + 10MOhm) ). The meter itself has a full scale of 200 nA (2V / 10MOhm) so the meter should read 9.9/200 of full scale or about 0.05V. Does that seem about right? |
| orolo:
--- Quote from: rstofer on April 18, 2016, 02:33:20 pm ---So, putting the meter in series with the 10V battery and the 1GOhm resistor will cause 9.9 nA to flow (10V / (1GOhm + 10MOhm) ). The meter itself has a full scale of 200 nA (2V / 10MOhm) so the meter should read 9.9/200 of full scale or about 0.05V. Does that seem about right? --- End quote --- :-+ Very good reasoning, save for a small error: 9.9/200 of a scale of 2 volts, is about 2*0.05 = 0.1V. |
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