why is there 0.7V instead of 1.2V on the common emitter?


if you go through Q1 mesh : Ve = 0.5+Vbe = 1.2 instead of Ve = 0+Vbe = 0.7 from Q2.

  • \$\begingroup\$ Why do you think there should be 1.2 V? Please show your math for this. \$\endgroup\$ – AnalogKid Jun 15 at 0:22
  • 1
    \$\begingroup\$ Q1,2 are in parallel, not series \$\endgroup\$ – Sunnyskyguy EE75 Jun 15 at 0:48
  • \$\begingroup\$ if you go through Q1 mesh : Ve = 0.5+Vbe = 1.2 instead of Ve = 0+Vbe = 0.7 from Q2 \$\endgroup\$ – Thiago Baldino Jun 15 at 0:52
  • \$\begingroup\$ Will Q1 or Q2 determine the voltage at the common emitters? Either transistor can determine that voltage. Which one wins the fight? \$\endgroup\$ – Peter Bennett Jun 15 at 1:09
  • \$\begingroup\$ Starting with the voltage on the Q2 base, what is the voltage at the two emitters? Now, what is the voltage across the Q1 base-emitter junction? Pay attention to which way the emitter arrow is pointing. \$\endgroup\$ – AnalogKid Jun 15 at 1:18

Your circuit is:


simulate this circuit – Schematic created using CircuitLab

You know that the emitter voltage for both BJTs is the same (they are tied together.) It follows from the schematic that \$\mid\: V_{\text{BE}_2}\mid\:\:=\:\:\mid V_{\text{BE}_1}\mid+500\:\text{mV}\$. You should be able to easily see that fact, directly from the schematic: \$V_{\text{B}_1}=V_{\text{B}_2}+500\:\text{mV}\$ and \$V_{\text{E}_1}=V_{\text{E}_2}\$.

Since you also know that for every \$60\:\text{mV}\$ difference in \$V_\text{BE}\$ there will be about \$10\times\$ the collector current, it follows that \$\frac{I_{\text{C}_2}}{I_{\text{C}_1}}\approx e^{^\frac{500\:\text{mV}}{26\:\text{mV}}}\approx 225\times10^{6}\$. In other words, \$Q_2\$ hogs all available current in \$R_1\$. What remains for \$Q_1\$ is in the small "parts per billion." (In short, nothing at all.)

From here, you have to only decide the approximate collector current for \$Q_2\$. If you temporarily assume that \$V_{\text{BE}_2}=700\:\text{mV}\$ then you find that \$I_{\text{C}_2}=\frac{5\:\text{V}-700\:\text{mV}}{1\:\text{k}\Omega}\approx 4.3\:\text{mA}\$. Since this value is fairly consistent with \$V_{\text{BE}_2}\approx 700\:\text{mV}\$ for most small-signal BJTs, you can reasonably rest on this computed value.

This assumption confirmed, the result is easily worked out.


Redraw the diagram with Q1 and Q2 emitter- base paths replaced by diodes, D1 and D2. D1 has 0.5V applied to its cathode, and D2 has 0V applied to its cathode. A diode will drop 0.7V when it's on.

Given the above, only one on/off configuration of diode states, out of four possibilities, can occur.


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