Open-circuit Output Voltage of Bipolar Output Stage


In TI's doc Designing With Logic, p. 16, it shows a method to calculate the internal resistance (\$R_{o}\$): using the open-circuit output voltage (\$V_{o}\$) and short-circuit current (\$I_{os}\$).

R_{o} = \frac{V_{o}}{I_{os}}

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I can understand the method to get \$R_{o}\$, but i can't understand how get the 'open-circuit output voltage', if it's output floating, ideally there will no current flow in Q1 and Q2, where the voltage come from?

Best Answer

Notice how at \$Io \gtrapprox 0\$, Vo drops sharply from Vcc down to 3.5V. This is the output voltage with just enough load to bias Q1+Q2 so \$ V_o = V_{cc} - R_1 \cdot I(R_1) - V_{be}(Q_1) - V_{be}(Q_2)\$ with \$I(R_1) = I_b(Q_1) \approx 0\$. That's a bit less than 5v - 0.7v - 0.7v = 3.6v.