How to make the output of a transistor the same as the input?
It's not clear to me where you are headed, but your statement:


  I want the transistor to output the same voltage as the input is.


Suggests a thought to me. If it is totally NOT what you want, no harm done. At worst, we're talking at cross-purposes and I'll remove by answer.







simulate this circuit &ndash; Schematic created using CircuitLab

The output will follow the input pretty closely, but within some reasonable limits with respect to the power supply. (This topology will have a very familiar look to some audio amplifier aficionados where it is one part of a so-called diamond buffer.)

In general, \$V^-\lt V_\text{IN}\lt \left(V^+-1\:\text{V}\right)\$ to avoid saturation. But the output load, in combination with the values of \$R_1\$ and \$R_2\$, matters. (Not shown in the diagram.) So the maximum input voltage will be:

$$V_\text{IN}\le \frac{\beta_1\:R_1\:R_\text{LOAD}\left(V^+-V_\text{BE}\right) + R_2\:V^-\left(R_1+R_\text{LOAD}\right)}{R_1\:R_2 + R_\text{LOAD}\left(\beta_1\:R_1+R_2\right)}$$

For a single rail supply, where also \$R=R_1=R_2\$, this becomes:

$$\begin{align*}V_\text{IN}&amp;\le \frac{\beta_1\:R_\text{LOAD}\left(V^+-V_\text{BE}\right)}{R + R_\text{LOAD}\left(\beta_1+1\right)}\\\\&amp;=\frac{V^+-V_\text{BE}}{\frac{\beta_1+1}{\beta_1}+\frac{R}{\beta_1\:R_\text{LOAD}}}\end{align*}$$

And from this, you can see that you want \$R\ll \beta_1\:R_\text{LOAD}\$ in order to support input voltages that are within a diode drop of the positive voltage rail.



See following simulation: