Different Octopus Circuit - Page 1
Forum section: electronics / beginners

author: Bryan
author_rank: Frequent Contributor
timestamp: May 03, 2014, 09:05:59 am
content: Hello:Was looking at this variation of the typical Octopus schematic and wondering what your thoughts are on the regulator and bridge rectifier. From what I can tell the regulator and pot are configured to regulate current, but what is the point of having a DC output, or maybe it's a half wave?http://www.next.gr/meter-counter/meters/Simple-Curve-Tracer-Circuit-l14760.html

author: Rory
author_rank: Frequent Contributor
timestamp: May 03, 2014, 04:21:01 pm
content: Follow the current paths through the diode bridge/regulator. It doesn't rectify but limits current flow in both directions.Google search brings up page 300 of Encyclopedia of Electronic circuits vol. 5, which gives this schematic plus a block diagram of the circuit.This image cropped but credit in Graf's book is Popular Electronics.

author: Rory
author_rank: Frequent Contributor
timestamp: May 03, 2014, 05:37:59 pm
content: Wondering if this is Charles Hansen's Semiconductor Tester, published in Popular Electronics, May 1999 p. 31.Can't find it online. I spent about an hour searching the web and not a single pdf copy of that issue. Lots of references, a link to a walled off archive, but what ever happened to the collections of scanned magazines? Who owns the copyright to the various incarnations of P-E? ZD sold to Gernsback, Gernsback went out of business. This stuff should be easy to find at archive.org but it isn't.Sometimes the Internet sucks as an archival source of information.

author: smpowell
author_rank: Contributor
timestamp: May 08, 2014, 04:24:39 am
content: >From what I can tell the regulator and pot are configured to regulate current, but what is the point of having a DC output, or maybe it's a half wave?The output (as seen by the part being tested) isn’t DC or half wave rectified, it’s distorted AC. The four diodes form a bridge rectifier so the LM317 only sees current going one way, but the current goes both ways outside of the rectifier bridge.I doubt if many people ever built this version for the following reasons:1. It puts a peak voltage of about 18 volts across the part being tested. This is way too high for most parts.2. When the current limiter is switched on (but not current limiting), it will be roughly equivalent to a pair of diodes with a voltage drop of about 3 volts in either direction, This will distort the results.3. R5 would have to take all the current flow, so make sure it has a high enough wattage rating.I used to have a web page about Octopus testers, if I were making a new one, I’d look at using an old MP3 player (or CD player or old laptop) instead of a transformer. That way it could use any frequency in the audio range, rather than being limited to line frequency.Web Archive of my old Octopus pagehttp://web.archive.org/web/20101216092412/http://octopus.freeyellow.com/octopus.html

author: Bryan
author_rank: Frequent Contributor
timestamp: May 09, 2014, 06:23:52 am
content: Quote from: smpowell on May 08, 2014, 04:24:39 amI put it together and yes you are right, puts out a lot of voltage to the component. I measured 35v peak to peak using a 12v transformer. Adjusting R5 pot limited current "Y axis". Your "My first Tester v1.0" looks interesting. Couple questions.Why separate resistors for R1 and R2?Any recommended suggestions for resistor values for a 12v transformer. I for the life of me can not figure out the voltage dividing math for ACCheers

author: smpowell
author_rank: Contributor
timestamp: May 10, 2014, 03:13:01 am
content: It’s been about 15 year since I built it, but I think:If you take a close look at the bottom most photograph the web page, you will see that R1 – R4 are each bundles of parallel resistors. I would have picked the individual resistors out of whatever I had on hand to get the voltage drops & minimum power dissipation that I needed.There is no deep meaning as to why labeled R1 & R2 as separate resistors. About 70% of the power dissipated in the voltage divider is in R1 & R2, so I probably needed a lot of ¼ watt resistors.I wanted the current flowing through the voltage divider R1-R4 (93ma) to be at least 10X the maximum current through the part (7.8ma). This was so that the voltage divider would act more or less like an ideal voltage source and not have the voltage vary depending on the current flowing though the part being tested.The 10X is probably over conservative; the classic tester uses a lot less.>Any recommended suggestions for resistor values for a 12v transformer.If you want to leave R6 at 1000 ohms, then the maximum current through it is 12ma. So I’d look for at least 5 times that through the voltage divider (60ma minimum).1.Since V=IR, 12 volts= (0.06 amps)X(200 ohms). In other words, the sum of R1 – R4 should be 200 ohms to get 60ma through the divider.It's late and my mind is too fuzzy at the moment to finish the math.....maybe tomorrow.

author: Bryan
author_rank: Frequent Contributor
timestamp: May 10, 2014, 08:35:38 pm
content: Thanks for the reply. Gives me a starting point.

author: smpowell
author_rank: Contributor
timestamp: May 11, 2014, 01:07:04 am
content: >I for the life of me can not figure out the voltage dividing math for ACPeak voltage = 1.41 X VAC, other than that you use the same formulas as DC.2.If you wanted to use the same voltages that I used ( 3 volts peak & 0.3 volts peak), then:3 volts peak/ 1.41 = 2.13 VAC0.3 volts peak/ 1.41 = 0.21 VACSo you need a voltage drop of 12.00 – 2.13 = 9.87 VAC across R1 & R2.Since R=V/I, R1 + R2 = 9.87 VAC/ 0.06 amps = 164.5 ohmsFor a current of 60ma, that means that R1 + R2 = 165 ohmsSince P = VI, the power dissipated in R1 & R2 is9.87 VAC X 0.06 amps = 0.59 wattsYou can use the same principle to calculate R3 & R3.When you double check your work (assuming you stay with 60ma) , R1 + R2 + R3 + R4 should be about 200 ohms.

author: Bryan
author_rank: Frequent Contributor
timestamp: May 11, 2014, 10:31:26 am
content: Thanks for the information. I got caught up in thishttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/vdivac.htmlI note your circuit has the option of inputting from a signal generator. Saw this circuit which thought was interesting.http://www.learningelectronics.net/circuits/curve-tracer-adaptor_17.html

author: smpowell
author_rank: Contributor
timestamp: May 11, 2014, 06:20:49 pm
content: >Thanks for the information. I got caught up in thishttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/vdivac.htmlWith a resistor divider, you don’t have to deal with complex impedances. When you test a capacitor or inductor, you are adding a complex impedance but because the current going through the divider is much larger than the test current, the affect is too small to be concerned about.>I note your circuit has the option of inputting from a signal generator.You could use any sine wave source as long as the output impedance is low enough.If I were updating my web page, I’d look at using an obsolete MP3 player (or CD player or Laptop) as an alternative to a transformer. Being able to use various frequencies (like 100 Hz, 1000 Hz, & 10,000 Hz) would be useful.Probably try it with my “99 cent ESR tester” (http://web.archive.org/web/20101125091850/http://octopus.freeyellow.com/99.html) also, although 100 kHz square waves are a better choice for testing ESR.