#1886

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  • @Manf-ft6zk
    @Manf-ft6zk 20 днів тому +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 20 днів тому +5

    This is news to me. Great work.

  • @andymouse
    @andymouse 20 днів тому +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.

  • @robinbrowne5419
    @robinbrowne5419 20 днів тому +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.

  • @MrSlehofer
    @MrSlehofer 20 днів тому +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.

    • @user-zf5rz2iy7o
      @user-zf5rz2iy7o 12 днів тому +1

      Diotec parts are bipolar, they are for much higher currents, and are marketed as LED drivers.

  • @user-zf5rz2iy7o
    @user-zf5rz2iy7o 12 днів тому

    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.

  • @pault6533
    @pault6533 20 днів тому +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  18 днів тому +1

      ua-cam.com/video/Sz0-cUxC2NI/v-deo.html
      ua-cam.com/video/4vPPA85BmT0/v-deo.html

    • @pault6533
      @pault6533 17 днів тому

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

  • @KarlAdamsAudio
    @KarlAdamsAudio 20 днів тому +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 20 днів тому +1

      What does that achieve?

    • @KarlAdamsAudio
      @KarlAdamsAudio 20 днів тому

      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 20 днів тому +2

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

    • @KarlAdamsAudio
      @KarlAdamsAudio 20 днів тому +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.

  • @PTigris
    @PTigris 20 днів тому +1

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

  • @stamasd8500
    @stamasd8500 14 днів тому

    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.

  • @SeanBZA
    @SeanBZA 20 днів тому +2

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

  • @ats89117
    @ats89117 20 днів тому +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 20 днів тому

      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.

  • @qutube100
    @qutube100 20 днів тому +1

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

  • @byterock
    @byterock 19 днів тому +1

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

    • @joeteejoetee
      @joeteejoetee 18 днів тому

      Down South people pronounce the word HeigHT, as HeigTH.

  • @markp5726
    @markp5726 20 днів тому +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.

  • @goosgitaar
    @goosgitaar 20 днів тому +1

    ❤ Jfet s they Sound good to

  • @MsFireboy2
    @MsFireboy2 20 днів тому +1

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

  • @reedreamer9518
    @reedreamer9518 20 днів тому +2

    Interesting

  • @paulpaulzadeh6172
    @paulpaulzadeh6172 19 днів тому +1

    JFET you connected is current sink , NOT current source .

  • @p_mouse8676
    @p_mouse8676 20 днів тому +1

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

  • @mr1enrollment
    @mr1enrollment 20 днів тому

    cute - never heard of that B4

  • @massriver
    @massriver 20 днів тому +1

    Doesn't start doing till 2v, too $$

  • @AndyFletcherX31
    @AndyFletcherX31 20 днів тому +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.

  • @rtybn2012
    @rtybn2012 20 днів тому +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  20 днів тому

      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 20 днів тому

      @@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 20 днів тому

      @@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 19 днів тому

      2N3251A could be an appropriate alternative?

    • @rtybn2012
      @rtybn2012 19 днів тому +1

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

  • @hobbified
    @hobbified 20 днів тому +1

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

    • @VEC7ORlt
      @VEC7ORlt 20 днів тому +2

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

    • @SeanBZA
      @SeanBZA 20 днів тому

      @@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 20 днів тому

      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 20 днів тому +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.

  • @SkippiiKai
    @SkippiiKai 20 днів тому +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 20 днів тому +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 20 днів тому

      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.