#1886

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  • Опубликовано: 27 сен 2024
  • Episode 1886 chip of the day
    constant current diodes
    Be a Patron: / imsaiguy

Комментарии • 53

  • @Manf-ft6zk
    @Manf-ft6zk 3 месяца назад +13

    I heard of that before, that they repack the leftover Easter bunnies as Santa clauses. Now I see it also happens with Junction FETs.

  • @MattBaker1965
    @MattBaker1965 3 месяца назад +5

    This is news to me. Great work.

  • @MrSlehofer
    @MrSlehofer 3 месяца назад +4

    Diotec in europe still makes these, such as the CL40M45. Lovely little parts, like zener diodes they exploit the reverse bias properties of diodes, but instead of a specific reverse breakdown they have a specific somewhat constant reverse leakage current. Forward characteristic is basically identical to a normal diode.

  • @KarlAdamsAudio
    @KarlAdamsAudio 3 месяца назад +2

    One of my favourite tricks with a popular condenser microphone circuit that uses an emitter follower to drive the output transformer was to figure out the current in the emitter resistor, then substitute a current regulator diode of appropriate value.

    • @VEC7ORlt
      @VEC7ORlt 3 месяца назад +1

      What does that achieve?

    • @KarlAdamsAudio
      @KarlAdamsAudio 3 месяца назад

      The emitter follower will have lower distortion, especially at larger voltage swings, and will also get marginally closer to unity gain. In practical terms, not much difference at all, but the theory behind it is sound, and it's an easy substitution to do.

    • @VEC7ORlt
      @VEC7ORlt 3 месяца назад +2

      @@KarlAdamsAudio oh, active load, yeah, that makes sense!
      I'd use 2 transistor ccs - seems a bit easier to tune.

    • @KarlAdamsAudio
      @KarlAdamsAudio 3 месяца назад +2

      Yes, the constant current diode only makes sense as an easy retrofit to an existing PCB, replace one two-terminal device with another - not something you'd do in a clean-sheet design.

  • @stamasd8500
    @stamasd8500 3 месяца назад

    This is exactly why I keep a bunch of J310 on hand, they make nice constant current sources which you can tailor to your needs with one resistor. Specifically the J310 is good for that because it has a high Idss and that gives you a wide range of adjustment. In fact, the Idss of these parts is so high (30-60mA in the datasheet) that it confuses those cheap component testers, doesn't read it as a JFET but rather either as a resistor, or as two diodes back to back. :) Try it, it's fun.

  • @ats89117
    @ats89117 3 месяца назад +2

    Using multiple JFETs in series with all the bases tied together and a source resistor can provide a much better current source. A J111 feeding a J112 feeding a J113 string gives really good results with sufficient burden voltage.

    • @KarlAdamsAudio
      @KarlAdamsAudio 3 месяца назад

      I've used this idea too - it can perform really well, although as I recall there is also a bandwidth tradeoff in adding extra JFETs. Impressively constant current vs voltage, though.

  • @PTigris
    @PTigris 3 месяца назад +1

    The LM134, LM234 and LM334 are very good current sources too.

  • @SeanBZA
    @SeanBZA 3 месяца назад +2

    Current is very variable, depending on doping, so thus the huge number of binnings as to current.

  • @markp5726
    @markp5726 3 месяца назад +1

    Recently, I was looking at what jfets digikey has. I was surprised to see the power ratings - 63A at 1.2KV, for example. I guess the process parameters for SiC favor making JFETs rather than bipolar or MOS.

  • @andymouse
    @andymouse 3 месяца назад +3

    This is the sort of stuff you can only pay for in Gold ! why wasn't I taught this ? lol Great chip of the day ! in case you didn't know I love chip of the day !......cheers.

  • @goosgitaar
    @goosgitaar 3 месяца назад +1

    ❤ Jfet s they Sound good to

  • @reedreamer9518
    @reedreamer9518 3 месяца назад +2

    Interesting

  • @MsFireboy2
    @MsFireboy2 3 месяца назад +1

    Awe another idea for the old differential amplifier. Your thoughts on this?

  • @rtybn2012
    @rtybn2012 3 месяца назад +1

    I was watching one of your videos and you said at one time that you worked at HP. I have acquired an HP 8601A sweep generator for a rather low cost for an unknown problem. It turns out that the -75V regulator does not work. I have found a replacement for one of the transistors but not the other one. Its HP part number is 1854-0037 in is a PNP and used a current source for the error amp. Any help would most appreciated.

    • @IMSAIGuy
      @IMSAIGuy  3 месяца назад

      1854 is NPN hparchive.com/Bench_Briefs/HP-Bench-Briefs-1974-01-02.pdf
      www.qsl.net/n9zia/hp/HP_Cross_Reference.pdf
      hparchive.com/PARTS/300-hpxref.pdf

    • @rtybn2012
      @rtybn2012 3 месяца назад

      @@IMSAIGuy I have the schematic and shows it is a PNP. The 1854-0037 is not listed. How would it effect the the circuit if I used a common type of a PNP transistor.

    • @markusberg2770
      @markusberg2770 3 месяца назад

      @@rtybn2012Interesting, I also read the HP schematic as PNP: 1854-0037 PNP Current Transistor. Unfortunately, after searching a digitized schematic for over 20 minutes, I did not find a parts list either. You’ll notice, though, some of the itemized parts offer a substitution.

    • @markusberg2770
      @markusberg2770 3 месяца назад

      2N3251A could be an appropriate alternative?

    • @rtybn2012
      @rtybn2012 3 месяца назад +1

      @@markusberg2770 Thanks for the info. I checked the specs, they looked good.I ordered some from E-Bay.

  • @SkippiiKai
    @SkippiiKai 3 месяца назад +1

    I think you either drew the potentiometer wrong in the schematic, or it's not doing what you said it's doing. The JFET is still 100% on, since the gate is still grounded. The current only changes because it's going through the resistor. I think you meant to have the pot change the current to the base to change how turned on the transistor is to vary the current.

    • @misterhat5823
      @misterhat5823 3 месяца назад +2

      Nope. The current through the pot develops a negative gate voltage, with respect to the source, and that lowers the current to below Idss.

    • @windshield11
      @windshield11 3 месяца назад

      Similar things are done with vacuum tubes, you add a cathode resistor so that the grid is "pushed up" and you get less amplification. In reality the reference point of the grid goes up so the total voltage difference grid to cathode decreases. You'd usually bypass this with a cap so that you get more bass.

  • @pault6533
    @pault6533 3 месяца назад +7

    Thanks for doing this! (It was my request). As interesting as I expected. The variable version is a bonus! PS Would be interested in your take on the varactor diode too🥸

    • @IMSAIGuy
      @IMSAIGuy  3 месяца назад +1

      ruclips.net/video/Sz0-cUxC2NI/видео.html
      ruclips.net/video/4vPPA85BmT0/видео.html

    • @pault6533
      @pault6533 3 месяца назад

      @@IMSAIGuy Thanks for taking the time to find these for me!

  • @paulpaulzadeh6172
    @paulpaulzadeh6172 3 месяца назад +1

    JFET you connected is current sink , NOT current source .

  • @byterock
    @byterock 3 месяца назад +1

    'Acrossed' you giving away your Upper Midwest upbringing ;)

    • @joeteejoetee
      @joeteejoetee 3 месяца назад

      Down South people pronounce the word HeigHT, as HeigTH.

  • @qutube100
    @qutube100 3 месяца назад +1

    OHHHhM curve tracer of loveliness design me a constant current circuit that operates variably to two amps ohhhhhmmmmm!

  • @lo2740
    @lo2740 Месяц назад

    not rare nor expensive, a semitec S-822T cost 0.1USD in large quantity, and provides 8.2mA at up to 80V, with a max dissipation of 500mW. This is often the best option when PCB realestate is at a premium (standard SOD123) and usually the most effective solution in term of cost / parts count, as long as the current is below about 35mA.

  • @AndyFletcherX31
    @AndyFletcherX31 3 месяца назад +1

    Those expensive constant current diodes don't even have a good stability or accuracy. I can't imagine why anyone would use them when there are better and cheaper alternatives.

  • @robinbrowne5419
    @robinbrowne5419 3 месяца назад +1

    Cool 👍 I like the fact that a simple pot makes a variable current source. Also the fact that it is a 2 wire device which should be able to go almost anywhere in the circuit where you need a current source.

  • @beakytwitch7905
    @beakytwitch7905 2 месяца назад

    The 78xx devices can also be used as constant current sinks, or the 79xx constant current sources.
    Basically you put a load across the regulated voltage output, and the constant current is what flows from the "unregulated" supply through the regulator.
    Since the breakdown voltage can be as high as 40 V and the current can be from milliamps to an amp, they could be used in vacuum tube circuits.

  • @ГеоргиХристов-г6р
    @ГеоргиХристов-г6р 3 месяца назад

    And here are some notes on using a JFET for a current regulator:
    1. Select a transistor with a minimum Vth for low supply voltage operation.
    2. Select a transistor with a Idss slightly greater than, or equal to the desired output current.
    3. The transfer characteristics of the JFET have a thermally stable point. If you establish the current at this point, it will not depend on temperature.
    4. Be mindful of power dissipation and thermal effects on drain current.
    CRDs are very useful parts, but they are rare and cost like a wing of Jumbo Jet! For this reason they never got popular or mass produced. Also it is much cheaper to take two(or more) BJTs and make the same thing.

  • @p_mouse8676
    @p_mouse8676 3 месяца назад +1

    These are great little devices, but are becoming kinda absolute these days unfortunately.

  • @hobbified
    @hobbified 3 месяца назад +1

    Why aren't these more common for LED drivers? Cost?

    • @VEC7ORlt
      @VEC7ORlt 3 месяца назад +2

      At low power resistors are cheaper, at higher powers they are pointless efficiency wise.

    • @SeanBZA
      @SeanBZA 3 месяца назад

      @@VEC7ORlt Yes mostly cost, as driving a LED a resistor is very cheap, under a cent cost wise, and with a limited supply voltage the brightness change is not an issue. Having a device where a resistor can replace it for 5% of the cost means it will never be used. The constant current diode is really a specialist device, mostly used where you need constant current over a wide input voltage range.

    • @SkippiiKai
      @SkippiiKai 3 месяца назад

      They kinda are, or at least were. I remember buying a pack from Mouser, probably 12 years ago, specially made for driving LEDs. I think they were only like 50 cents each, but being only 20mA at a max of 90v, they have a pretty narrow usefulness.

    • @SkippiiKai
      @SkippiiKai 3 месяца назад +1

      The TL431 can perform a similar function, is way more versatile, and cheaper. Even the tiny half watt LEDs will want over 100mA, so a 1 to 35ma supply isn't helpful without an amplifier - in which case there are better ways to do it. Even the little through-hole 5mm LEDs will take 20mA, but those aren't going to be used in a circuit likely to have wide voltage swings, which is the application where these are useful.

  • @mr1enrollment
    @mr1enrollment 3 месяца назад

    cute - never heard of that B4