Hight input impedance DC buffer ?

[Externet]

Hi.
I believe am loading my circuit too much and collapses. Am I seeing it wrong ?

A 7555 IC as astable has 100KΩ as 10μF capacitor charge path and 100KΩ at its discharge path for ~ 0.2 Hz
I want to use the oscillation at the capacitor, that goes from ~1VDC to ~2VDC (1/3 to 2/3Vcc) triangle waveform to drive a 5KΩ load.
Using a transistor common emitter as amplifier, the triangle wave is overloaded and stops working.
What is a convenient single 3V supply circuit that may work ? I guess with a >1MΩ input resistance.

Like this ? How is Re set ?

 

[Harald Kapp]

From Wikipedia: R_in ~ ß₀*R_E
Knowing ß₀ and the expected R_in allows calcuation of R_E.

 

[BobK]

If you really want an exact copy of the input, a rail-to-rail opamp would do the trick.

Bob

 

[Externet]

The voltage follower works well. Re=100K yields 15MΩ loading to Vin with a 150beta 2N2222A. No more overloading of the triangle wave. Output is near desired range.
What would be the effect of adding some resistance to the collector ? Would it produce less or more attenuation at Vout ?

 

[AnalogKid]

As long as you are taking your output from the emitter, adding *reasonable* resistance (not 1 ohm, not 1 Meg) to the collector leg will not affect the output. It does shift some heat from the transistor to the resistor.

Note that your circuit has a *very* asymmetrical output impedance. Pulling the output up is at a low impedance. Pulling the output low is the 100K resistor.

It sounds like the -0.6 V shift in the output waveform is not a problem for your application. If it is, there is a relatively simple way around it.

ak

 

[duke37]

Note that your circuit has a *very* asymmetrical output impedance. Pulling the output up is at a low impedance. Pulling the output low is the 100K resistor.

The output impedance will only become asymmetrical when the transistor current has reduced to zero so will depend on signal amplitude and load.