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Voltage Regulators for Line Level Audio. Part IX : Two Transistor Series Regulator

Posted 21st February 2014 at 12:50 AM by rjm
Updated 4th March 2014 at 02:56 AM by rjm

Part of a series.

So you have a small handful of parts and want to build a (simple) discrete voltage regulator instead of using an IC. What to do?

For line-level audio circuits, especially op amp based (IC or discrete) preamps with high PSRR, something like the Z-reg is generally sufficient. Robust, works well, has enough ripple rejection to cut power line noise from the preamp output.

If you add just a couple more parts, however, you can add feedback to the Z-reg circuit, a simple error amplifier in the form of an additional transistor Q2, with the output-sampling voltage divider R1,R2.

The ripple rejection is not vastly superior to the circuit without the feedback unless some additional bypass capacitors are added as shown in the first version of the circuit below. The output impedance, however, improves from a few ohms to a few tenths of an ohm as a result of the feedback. Which could, in principle, be of use.

Note R3 controls the current through Q2 and D1, the smaller the value the better the performance, at the expense of extra power dissipation in Q2,D1 and overall the circuit efficiency goes down.

Finally, for you new people: this circuit has none of the niceties that come with an IC voltage reg, like protection against thermal overload, overcurrent, or short circuits.
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Total Comments 2


  1. Old Comment
    Calvin's Avatar

    instead of plotting V(out) in dB and calculating Z(out) from those values, plot V(out)/I(I1) instead and set the Y-axis to linear (placing the cursor over the axis .. it changes to a ruler ... click .. choose db, log, or linear).
    The Impedance will then be plotted as Bode plot with Z in Ohms + Phase in a more familiar way.

    Posted 7th March 2014 at 09:31 AM by Calvin Calvin is offline
  2. Old Comment
    rjm's Avatar
    Thanks for the tip.

    To plot V(vout)/I(I1), click "add traces" on the plot menu, and type the expression in the blank line at the bottom of the dialog box. Then change the axis to linear as Calvin says above.

    Since the small signal current I(I1) is set to "1" in the example files, however, dividing though by the current does not change the numerical answer. I agree its more comforting to see it displayed in the right units though.
    Posted 7th April 2014 at 07:55 AM by rjm rjm is offline

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