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Modding a circuit

Discussion in 'General Electronics Discussion' started by electronoobz, Dec 28, 2012.

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  1. electronoobz


    Jan 14, 2012
    this is my my circuit

    i also did this:


    I removed this I'll just use trimpot instead

    Adding a "softness" or clipping threshold control

    You can put a pot in series between the signal and diode pair to adjust the clipping threshold of the diodes. This acts like a "softness" control. I saw this described by R.G.Keen.


    This works well. You can also put the bounding resistor between the diodes and ground.

    instead i put DPDT switch on pin 1

    -0-0-0- (-pair of diode - pin 1 - 2nd pair of diode-)
    -0-0-0- (-diode's ground - ground - 2nd pair of diode's ground)

    on this part of the first link i gave is where i have a problem

    Gain Control
    The transistor's gain can be increased by a lot by bypassing the emitter resistor with a fairly high-value capacitor (4.7uF+). Or, a gain control can be added by replacing the 680Ω resistor with a 1k pot going to ground, with the middle lug connected to a capacitor as before. By using a lower value bypass capacitor (0.1uF-1uF), a treble boost can be added, although I find it more effective to just add a high-value bypass capacitor and reduce the value of the input capacitor for a treble boost.

    i know how to put bypass cap in a fixed resistor, for what i know you'll put it parallel with the resistor positive side on emitter pin..

    but in this case.. im using a rheostat..

    which one is right?

    pin 3 positive side and pin 1 negative side or pin 2 positive side and pin 3 negative side?

    and i heard about making log pots out of lin pots.. i need it coz we dont have log pots here

    i heard that you need to put 10% or 20% value of pot resistor in pin 1 and pin 2.. what if i'll use it as a rheostat? what if i jumpered pin 1 and 2 will it work still?

    Last edited: Dec 29, 2012
  2. electronoobz


    Jan 14, 2012
    Bump.. Anyone here?
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    a rheostat? Do you mean a large wirewound variable resistor with a huge knurled knob on the top, or at you talking about just using a normal potentiometer as a variable resistor (as opposed to a variable potential divider)?

    Can you draw circuit diagrams of what you're asking questions about?

    The lack of clarity in your questions may be the reason for the lack of answers.
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Yes. Most of your question is too vague for me to know how to answer you.

    Regarding your last question, click here.
  5. electronoobz


    Jan 14, 2012
    so this is my circuit..


    i'll be using a pots as a rheostat (lin converted to pots since we dont have that here) as an exchange for the emitter resistor so i can control the gain..

    my doubt is the lin to log conversion may not work if i'll use it as a rheostat.. im doubting cause when pot is used as rheostat either 1 or 3 is connected to the wiper right? and the resistor to be used to convert the lin to log pot may not work.. coz i think the resistor that is used mwill just be connected in series.. do yo get it??

    another problem is.. adding a bypass cap on a pot..

    as to what i know adding bypass cap on the emitter is connected parallel with the emitter resistor..

    but this time it is a pot.. at first i thought i'll just connect the cap on pin 3 and 1

    but the site says:

    i need to put the cap in the middle.. now im confused...

    Attached Files:

    Last edited by a moderator: Dec 29, 2012
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    You can vary the stage gain by putting a variable resistor (a potentiometer with only two terminals used) and a capacitor in series with it, all in parallel with the emitter resistor. In this case it's probably best to use a logarithmic potentiometer. And you're probably right that the conversion from a linear to a logarithmic potentiometer won't work when it's used as a variable resistor aka rheostat.

    You can also vary the gain by putting a potentiometer in series with the input signal. Again this probably should be a logarithmic potentiometer.

    Can you scrounge a log pot from a broken amplifier or something? They shouldn't be hard to find.
  7. electronoobz


    Jan 14, 2012
    now you see what my problem is
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Relayer, your gain adjustment in the emitter leg is wrong. You should use a fixed resistor to set the DC conditions, and a potentiometer and capacitor in series, in parallel with the fixed resistor. The way you have it connected will mean that the gain will not vary much, since the capacitor across the trimpot provides a low-impedance path for AC, effectively shorting out the potentiometer (well, nearly) at audio frequencies.

    Also, the potentiometer in series with the base adjusts the gain, although it's labelled "distortion". This is simply because the more gain you have, the earlier the distortion will set in. So, adjusting the stage gain with the trimpot in the emitter path will have the same effect as adjusting the input potentiometer; you don't need both.

    Finally, that circuit doesn't have the proper DC conditions set up. Actually I would recommend a JFET-based gain stage because they have a much higher input impedance and are simpler to bias. Using a bipolar transistor with a single resistor (especially such a high value - 1000 times higher than the collector load resistor) from collector to base does not stabilise the DC conditions in the stage. You will probably find that the collector voltage sits very close to the positive power supply rail, and the stage will clip on one half of the signal with or without the diodes on the output. As I said, a JFET stage would be a lot better. I realise this isn't your fault; that problem was present with the original circuit.
  9. duke37


    Jan 9, 2011
    Q1 is a virtual earth amplifier. The DC conditions are set by the emitter and collector resistors in series and by the feedback resistor. The emitter resistor is only 1/20 of the collector resistor and should be eliminated.

    The necessary feedback resistor depends on the gain of the transistor and should be set to get half the supply voltage at the collector. The stage gain will be Rfeedback/Rin.
    With 1mA through the transistor, the base current will be 1/gain mA, say 50uA. At 4V, Rfeedback = 4/50Mohm = 80k, this agrees with Kris that the collector voltage will be too high to amplify properly in the circuit as shown.

    Connect the emitter to ground and initially try 100k feedback resistor.
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    If you want the amplifier stage to amplify without clipping, so the diodes are the only clipping element, you should use a voltage divider to provide the base bias. I would recommend using high resistances in the voltage divider, and a Darlington transistor, to give high input impedance so you don't load down the guitar pickup.

    The DC conditions in the circuit will be determined by the two base resistors, the collector resistor, and the emitter resistor. That is, the resistor from the emitter to ground, not the resistor in the series resistor-capacitor network that is connected from the emitter to ground.

    The gain is determined by the ratio between the collector resistor and the resistor in the series resistor-capacitor network from the emitter to ground.

    Have a look at I think there are explanations in there, although they are written in mathematical form and don't really explain things very well.

    If you want a more detailed explanation I will provide one, but not now.

    As I said, I would use a JFET amplifier instead of a transistor amplifier in any case.
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