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What is the purpose of the Red Leds D1 and D2 in the attached schematic?

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  • What is the purpose of the Red Leds D1 and D2 in the attached schematic?

    Could someone share the purpose of the Red Leds D1 and D2 in the attached schematic? Is this some type of diode clipping circuit?

    Thank you in advance for any help,
    yeto
    Attached Files

  • #2
    Originally posted by yeto View Post
    Could someone share the purpose of the Red Leds D1 and D2 in the attached schematic? Is this some type of diode clipping circuit?

    Thank you in advance for any help,
    yeto

    It's a clipping circuit, of sorts. It will limit the signal from swinging higher than about 700mV or lower than about -700mV (so a bit more than 500mV rms). Anything beyond those limits will be clipped.

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    • #3
      Get back to work, wombat.

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      • #4
        Yes, it is a clipping circuit.

        Typical diode clipping scheme relies on forward conduction of diodes at high enough input signal amplitudes. Near the forward voltage threshold diode's resistance more or less gradually changes from very high (reverse bias) to very low (forward bias). This variable resistive element can then be harnessed in passive circuit (e.g shunt clipping) or active circuit (e.g. diodes in feedback loop).

        For example, in "shunt diode clipping" (clipping diodes from signal path to ground or other suitable current "sink", the most basic setup) clipping happens because of voltage divider formed by source resistance and dynamic diode resistance. When diodes "clip" they are forward biased and effectively present a very low "shunt" impedance. Source impedance becomes significantly higher than the dynamic diode impedance. In this state the divider presents maximum "attenuation", signal current gets largely "sinked" and the passing signal is therefore "clipped".

        --

        This scheme is "in reverse". The LED diodes are initially forward biased but high enough signal amplitudes will reverse bias them.

        End result is still the same. When forward biased, LEDs present a very low series impedance to the signal path and signal current can largely flow uninterrupted. When a high enough input signal amplitude reverse biases the diode its dynamic resistance ramps up and signal path effectively gets a very high series resistance. Signal current largely stops flowing and the passing signal is therefore "clipped".


        In practice the circuit - as is - is not exactly directly comparable to basic "shunt clipping" as the architecture that estasblishes the LED bias points will most definitely also affect LED currents and overall this will have some effect on conduction characteristics. But overall you should understand that this is simply a "reverse" alternative of the usual clipping scheme.

        The reverse scheme can naturally be used in "active" clipping circuits too but naturally one must acknowledge the "backwards" operation in dynamic characteristics. e.g. In feedback clipping you simply switch the clipping diodes to affect "that other" part of the voltage divider that effectively forms the negative feedback loop.

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        • #5
          Fully agree and add, let's put some Math into it:
          to begin with, yes, it's works "in reverse" ... but what does that mean?

          As explained above, usual way is that diodes clip voltage when threshold is reached, and that is a fixed value, such as around 700mV for Silicon, around 0.3V for Germanium, and 1.9V for Red Leds.
          Cool but not adjustable at will (we can only stack more or less diodes in series, a very coarse control)

          Here we clip current (which if passing through a resistor of course will drop a certain voltage) with the beauty that by careful choosing of current and load we can vary the clipping point as will.

          Not clear on first sight, but conceptually is similar to the VOX "limiter" circuit.

          How does it work?:

          1) I'm rounding off and simplifying some values, the idea is to explain the concept, not calculate within 1% , in the real life I'm happy with 5% or 10%

          2) let's think the 47k resistor ( I'll call it 50k) a constant current source and the 10k resistors the load.

          3) the diodes drop 1.9V , let's say 2V

          4) "constant current" will be (9000mV-2000mV)/55000 ohms=(7000/77000)mA=0.09mA
          It will split equally between both 10k resistors, which for this purpose are in parallel, so each will drop:

          5) (0.09mA/2)*10000 ohms=0.045mA*10000 ohms=450mV

          6) now, and here's the trick, when the input signal reaches extra 450mV across the first resistor (above the 450mV it was already passing at idle) , *it* will supply those 0.045mA , so those original 0.045mA coming through the diode are no longer needed, so diode current drops to 0 , diode opens and signal can not pass, so its clipped at that point.

          Notice that signal swing across the left 10k appears also at the right 10k, both are joined by the forward biased diodes and both are fed from the same constant current source, also total current through both is always the same, just they can share it equally (at idle) or not (high signal).

          A symmetrical clipping happens when the first signal swings 450mV the negative way, now the left 10k and diode suck the full constant current, 0.09mA and the right diode "dries up" and opens because voltage at the right side drops 450mV from idle 450mV (remember right and left track each other) , so there's net 0V across the right resistor and it passes 0 current.

          0 current in the resistor means 0 current in the right diode, so it opens.

          So signal clips on +/-450mV peaks, that's what we see, but what's actually clipping is current.
          Neat, huh?.

          So far it looks an unnecessarily complex way to clip, why not just use 2 antiparallel diodes and call it a day?

          Well, the beauty is that it's very very flexible:

          7) the designer set it for 450mV clipping ... you can set it to *any* value you wish
          You can mimic Germanium, Silicon, Schottky, Leds, Zeners ... you name it.

          8) it is **adjustable** , you replace the 47k resistor with a trimmer and you set clipping point at will

          In fact, VOX Limiter does roughly the same (analyzing it is where I understood how current based limiters work) , in that case they can be set to clip from roughly 1 diode drop (around 700mV) to roughly 2400/2700mV peak.

          I bet the bench Tech watched the power amp clip on a scope , it probably was an ugly signal (hum, sticking to rails, etc.) and then he set Limiter to clip cleanly *just* below that.

          Sticking to rails (very slow Power Transistors which do not turn off when needed) is a tried and true way to destroy transistors, doubly so if they are weak Germanium or poorly understood early Silicon .

          9) it can easily be unsymmetrical, simply by using varying load resistors, so you can mimic, say, 1n4007+1n34 clipping or whatever.

          10) HH and GK amplifiers also used a kind of current limited clipping , based on a Fet or starved transistor (quoting from memory).
          FWIW it took me a looooong time to analyze how they worked

          11) almost forgot: believe it or not clipping level does NOT depend on diode used , only on the resistor network and biasing.

          Only effect of diode choice is how much you must substract from Vcc .

          12) notice that it will clip even if fed from a low impedance source, such as an Op Amp or buffered output, does not need a series resistor like regular diode clippers.
          Juan Manuel Fahey

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          • #6
            Originally posted by teemuk View Post
            Yes, it is a clipping circuit.
            Thank you for taking time out of your day to write such a lengthy reply. This really helps.

            Again thank you,
            yeto

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            • #7
              Originally posted by J M Fahey View Post
              Fully agree and add, let's put some Math into it:
              Wow, this gives me a lot to study. Thank you for taking time to write such a valuable explanation. This is a circuit I am going to take time to fully understand.

              Thank you,
              yeto

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              • #8
                I have gone through your circuit diagram and also found the LEDs you mentioned. According to me these LEDs are there to work as a clipper.
                Because amplifier needs the limited signal level.
                So maintain the limited signal LEDs are connected to give some fixed voltage drop.
                [URL="http://www.7pcb.co.uk/prototype-PCB-assembly/"]pcb prototype assembly[/URL]

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