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I want to build an input buffer for audio purposes. This buffer should do 3 things:

  1. have high input / low output impedance.
  2. divide the signal by 2.
  3. low pass filter the signal (cut freq. = 15KHz).

How about the circuit below?? Any other suggestion? Please let me know if you have better ideas.

Would it be ok to use a MOSFET like the BS270 instead of the bjt?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ One challenge is that the BC547C is obsolete. Can you be more specific than 'high' input impedance? Do you need 100k such as in a typical guitar amp input? \$\endgroup\$ Jan 14, 2017 at 19:17

2 Answers 2

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You are over-thinking this. Here is all you need:

This satisfies all your specs:

have high input / low output impedance.

The input impedance is 2.2 kΩ and the output impedance 550 Ω.

divide the signal by 2.

It does that, since it is basically a resistor divider with two equal resistors.

low pass filter the signal (cut freq. = 15KHz).

Assuming you really mean kilo-Hz, not Kelvin-Hz as you wrote it, this is accomplished by C1 working against R2. The rolloff frequency is

  1 / (2 Π R C) = 14.5 kHz

which is within 4% of the target, and considerably less than the tolerance of ordinary parts.

This circuit has additional advantages over something like what you show:

  1. The gain is more accurate since there is no additional attenuation thru a emitter follower.

  2. It doesn't require power.

  3. It is more linear. The B-E voltage of your circuit isn't completely fixed or a linear function of the output voltage. As a result, the emitter follower is slightly non-linear.

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DC bias must always be a smaller R than load.

ie. R3 must be << AC coupled load e.g. <1/2 load R to avoid large swing current starving Ie.

Then Rb must be biased to Vcc/2 such that Req<< hFE* Re for hFE=150 Re/Rb<100

Thus ...

R4 = 100 Ohm is your load with \$R4C1=1/(2\pi f)\$
This needs a large C1. A bipolar supply avoids this.

schematic

You can divide signal /2 easily but you forgot to ensure DC bias >> load AC current.

BS270 needs a Vgs of >3V for a drain follower so no good. Vgs(th)=1V ok.

Define Zin , Zout and Vin pp , f range and Av better next time.

This one has Zin=1k.

Bootstrapping can raise Zin with extra cap to split bias input, if needed, but I do not recommend as this limits BW severely in this low voltage, low impedance.

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  • \$\begingroup\$ Recommendation: Use bootstrapping for drastically increasing the input resistance (bootstrapping: positive feedback between emitter and base). Start a google search for "emitter follower bootstrap methods". \$\endgroup\$
    – LvW
    Jan 14, 2017 at 17:34
  • \$\begingroup\$ Maybe but not needed. also, No results found for "emitter follower bootstrap methods". \$\endgroup\$ Jan 14, 2017 at 18:00
  • \$\begingroup\$ Thank you for your reply. Can you please tell me what is the best way to add low pass filtering in this buffer stage? Would it be ok to add the capacitor at the base of the bjt? \$\endgroup\$
    – Dukenukem
    Jan 14, 2017 at 20:02
  • \$\begingroup\$ Base coupling capacitors make a highpass function. \$\endgroup\$
    – LvW
    Jan 15, 2017 at 9:31
  • \$\begingroup\$ Tony Stewart - I cannot believe that you did not found anything about bootstrap for emitter followers via google. \$\endgroup\$
    – LvW
    Jan 15, 2017 at 9:33

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