What do these capacitors do?

Discussion in 'General Electronics Chat' started by markbr, Mar 21, 2010.

  1. markbr

    Thread Starter New Member

    Mar 21, 2010
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    Hey guys,

    This is the Danelectro Honeytone circuit which i traced out but im not sure what these capacitors do:

    1)C10 & C3

    2) C5 & C14

    3)C13

    Thank you.
     
  2. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
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    R4 and R3 form a divider to give a bias voltage near half the supply voltage.

    R12, C10 and C3 are decoupling or noise filtering to prevent any variations in the supply voltage (due to the current drawn by the rest of the circuit) affecting the bias, which could cause distortion or feedback.

    C5 and C14 provide an effective ground for the audio signal, while allowing the DC level on the opamp inputs to be held at the bias voltage level by R2 and R9.

    C13 looks to be a bit of HF cut or hiss reduction, or possibly correction for the tone circuit. It could also be for RF protection to reduce pickup of stray signals.
     
  3. BillB3857

    Senior Member

    Feb 28, 2009
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    It looks like C10 and c3 are used to insure a well filtered bias to the op amps. The bias will cause the outputs of the op amps to be near 50% of the supply voltage, thus allowing AC (audio) to be reproduced without clipping. C5 and C14 keep the inverting input of their respective op amps at AC ground while allowing the negative input to be at a voltage near 50% of the supply. C13 appears to be a high frequency filter. Have you tried running this circuit in a simulator?

    Others on this site could give a more exact or detailed answer but I'm sure this is the general idea.
     
  4. BillB3857

    Senior Member

    Feb 28, 2009
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    Looks like great minds run alike, Bob. :) I took a short break while typing.
     
  5. markbr

    Thread Starter New Member

    Mar 21, 2010
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    Thanks Robert, I was wondering if i supply -9V now to the -Vcc pin of theTL072. At the moment, it's only receiving +9V with -Vcc grounded. If i give +9V and -9V, i should then remove biasing resistors R2 & R9? Would i then have to remove C5 and C14 too?

    Also does C1 and C2 make sense? Shouldn't there be a resistor in front of C1?
     
  6. markbr

    Thread Starter New Member

    Mar 21, 2010
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    Thanks Bill. Do you mean the bias will cause the outputs OF the opamps or TO the opams? As far as i know, R12, R4 & R3 roughly split the supply voltage in half to about +4.8V. This should be the voltage at the point between R2 and R9. But im wondering why R2 and R9 are needed...possibly to reduce the current supplied to the non-inverting inputs?

    Could you explain this a little more please? What would happen if i removed them?
     
  7. BillB3857

    Senior Member

    Feb 28, 2009
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    As you have already noted, the op amps in the original drawing are operated from a single polarity supply with one power lead grounded. Since audio is essentially AC, in order to reproduce it, the no signal output of the amplifier must be lifted above ground. The output of the op amp will then vary positive and negative around THAT point. The coupling capacitors (C6 and C12) between stages removes the DC element and allows the effect of the AC to be coupled through.

    As for R2 and R9, they allow the input audio (pin 5 TL072(1)) and the inter-stage audio (pin 3 TL072(2)) to be developed. If they didn't exist and pins 5 and 3 were tied together, the circuit would only take the outputs of each stage to a fixed DC level. You calculated 4.8VDC. I estimated it to be "near 50% of supply".

    As for C5 and C14.... Op amps operate with an effective virtual short between the inverting and non-inverting inputs. (They operate off the CURRENT developed between the two). The caps prevent any DC current from the Bias line from going through R5 and R10 while allowing fluctuations in the voltage to generate fluctuations in current based upon the caps charging and discharging. The feedback resistors (overdrive and R11) maintain the proper DC relationships while controlling gain.

    If you are curious about D1 and D2, they will limit the voltage fed on to the tone control and ultimately the 2nd stage. If the input level is too high or the first stage gain is set too high (limited to about 100 by ratio of Overdrive and R5), then clipping at +/- 0.7 volts will result at the right side of R6.
     
  8. markbr

    Thread Starter New Member

    Mar 21, 2010
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    Does this mean D1 & D2 are soft clippers? Or are they the same as limiters?
     
  9. BillB3857

    Senior Member

    Feb 28, 2009
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    In my simple mind, a clipper and a limiter are one in the same. They limit or clip at a particular value. The limiting/clipping function is the result of current to flow through the diodes any time the voltage at the anode/cathode exceeds the breakdown voltage of the PN junction. Others on the site may have a better definition and be able to offer the difference between the two.
     
  10. Darren Holdstock

    Active Member

    Feb 10, 2009
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    I'd say D1 and D2 are clipping pretty hard. Strictly speaking they won't suddenly limit to an absolute voltage, as plot Vf against log(If) for a silicon diode and you'll get a straight line (albeit a bit droopy at the higher currents due to ESR), but at room temperature and moderate currents 0.6 V is a good rule of thumb (the 1N4148 typically has a Vf of 0.4 V at 10 uA and 1.1 V at 100 mA).

    You can play around with these diodes quite easily. Germanium point-contact diodes (such as the once-popular OA91) have a different characteristic to silicon diodes, so try these in the D1/D2 position for softer clipping. Adding a bit of +/- asymmetry produces a higher proportion of even harmonics, and you can do this by making D1 and D2 different (e.g. one silicon, one germanium), or LEDs of different colours (an old Marshall trick), or use the same diodes but put 3 in series for D1 and 2 in series for D2. Or whatever - have a play. Watch out for DC offsets if you're going a bit asymmetric, these can make pots crackle horribly, though C9, C7 and C8 will prevent this here.

    If you want to be a bit hardcore, put the diodes in the feedback loop of an op-amp circuit, but be warned that it will need some taming to prevent instability due to the heavy non-linearities therein. A small resistor in series adds a bit of linearity, and a pot in parallel doubles as a gain/linearity control. A small cap between op-amp output and - input helps stability, but is no guarantee.
     
  11. Mike33

    AAC Fanatic!

    Feb 4, 2005
    349
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    Diodes in the feedback loop will get you something like a Tubescreamer/Bluesbreaker; set up the way this one is, it's more like an MXR Distortion + or Marshall Guv'nor.

    Germanium diodes such as 1N34A, 1N60....will clip earlier, giving a harder-sounding output (fuzzier), altho a little lower in level (stack them to get more level as mentioned above). The 1N4148's will sound a bit more 'modern' and crunchy, with LEDs being crunchiest (and loudest, all things being equal) of all!
    These are great distortion circuits to mess around with, you can create a lot of custom sounds, and place different diodes on switches, etc.
     
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