Buffer

166

03:20:07

A simple buffer with 50kΩ input impedance -- audio line standard.

With a little help I got it right this time. :) (I tried something complex before this.) The help: http://sound.whsites.net/amp-basics2.htm#current

There was a small problem with my original component values: they made the input impedance much lower than I thought. I missed the implications of a line in that page I linked, "Bipolar transistors are inherently low impedance, and additional circuitry is needed to make them work satisfactorily in high impedance circuitry. Noise is also a problem when high impedance sources are used." I'm trying to turn a high impedance input to a low impedance output with just one transistor. 

I've actually succeeded, more or less, by tweaking resistor values. The impedance of the load affects the input impedance. With loads above 5kΩ, the difference is very small, which is good enough! I am, however, giving up my idea of driving old-school headphones (300Ω stereo, 600Ω mono,) from this one stage. :)

published 7 years ago

jason9

7 years ago

The 1kOhm resistor at the emitter draws current from the emitter, and if the voltage at the base rises or drops by 1V, then the current at the emitter rises or drops by 1mA, which makes the current at the base rise or drop by 10uA. 1V gives 10uA works out to 100kOhm (which is 1kOhm multiplied by 100, the gain of the transistor). That makes three 100kOhms at the input, or about 33.3kOhms. If you want 50kOhms, try increasing the 1kOhm resistor to 2.2kOhm and making the top and bottom resistors something like 120kOhm.

jason9

7 years ago

Actually, for an input impedance of 50.1kOhms, the emitter resistor should be 2.2kOhms and the top and bottom resistors should be 130kOhms. That doesn’t include the 1kOhm load resistor, but if the input impedance is 50kOhms shouldn’t that be the load too, especially since it’s just a buffer?

eekee

7 years ago

Hum! Thanks jason. I didn't realize the load matters too, in this configuration. I don't want to give it a high impedance load, because then why not just discard the circuit? :) The kind of buffer I meant is a voltage follower, current amplifier. Anyway, I'll try some things including your suggestions, because right now I calculate the input impedance to be only 25.25kΩ. That's from measuring the input current, 39.6μA @ 1V. I suspect it may not be possible to do it right in this configuration anyway. I'll post notes above.

eekee

7 years ago

Done. It's now a reasonable buffer when the load is 5k or above.

jason9

7 years ago

Making it a darlington will remove the load-reducing-input-impedance problem. Or attach the base of a PNP to the collector of the NPN for an equivalent effect.

eekee

7 years ago

Ah! That's a simple fix. Would the latter be a Sziklai pair? I should look those up again. This is good enough now, except for the high input resistors -- noisy. Reducing load effects will help reduce those.

eekee

7 years ago

http://everycircuit.com/circuit/6717742685356032

jason9

7 years ago

Yeah, that should be a Sziklai pair, only I can never remember what it’s called or how to spell it.

eekee

7 years ago

aha! ty

add comment in editor