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I asked similar question recently, which also referred to differential transistor pair, but this one is different.

I made this circuit (in real world):

schematic

simulate this circuit – Schematic created using CircuitLab

I was only experimenting with DC (no signal applied). Potentiometer was used to adjust I_e1 to equal I_e2. Beta of transistors were around 300 and both of I_e were set to be exactly 1.1 mA.

After everything was connected as shown, I started to observe the I_e of each transistor when one transistor gets heated up. So, I "touched" the Q1 with two fingers for a few seconds - but nothing happened (no changes in current), same happened with the Q2. It seemed a bit strange that nothing happens. Then I took my lighter and heated one transistor to temperature about 100 degree Celsius - the current through both of transistor didn't even blink! At the end I really really heated it up - and only at this level of heating, the currents moved up/down the scale for 4-6 uA (micro amps)!

  • How is this even possible? When lighter was usually approached, the I_e changed on the scale of mA, if not tens of mA.

I even replaced transistors - same results!

I also reconnected the whole circuit again - same results!

At the end I used different multimeter (in case one wouldn't work properly) - same results!

*This thing could be possible with current mirrors at collectors and constant current source at emitters, but I only used usual passive resistors.

Comparison of multimeter values:

enter image description here enter image description here

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    \$\begingroup\$ The voltage at the Q2 base is around 11V. And if Ic1 and Ic2 is 1.1mA and Rc are 10k your BJT's are at the edge of saturation or in a saturation region. So, try to change Rc resistors value to 4.7k \$\endgroup\$ – G36 Oct 31 '17 at 19:14
  • \$\begingroup\$ @G36 Yes, you are right. But I don't understand why were both BJT's in saturation.. Aren't the Rc resistors calculated as V_Rc/I_e and then Re is calculated as V_Re/2*I_e ? \$\endgroup\$ – Keno Oct 31 '17 at 19:51
  • \$\begingroup\$ @G36 Oh sorry, I messed it up in the voltage divider - Voltage across R1 should be only around 6V; according to calculations, and R2 should be 10.7V. Actually R1 should be 19.3V instead of 6V. \$\endgroup\$ – Keno Oct 31 '17 at 19:59
  • \$\begingroup\$ Yep, your base voltage is too high or you have chosen the too large V_Rc value in your calculations. For Rc = 10k and Ic = 1.1mA you will get V_Rc = 10.1V. And the base voltage is around 11V so the BJT is clearly in the saturation region. \$\endgroup\$ – G36 Oct 31 '17 at 20:07
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    \$\begingroup\$ @Keno It's because it acts more like a current source (which is what you really want there) with more compliance voltage available to it. The more, the better current source it is. To see why, imagine you have a load that varies from \$1-1000\:\Omega\$ and you hook it up to a "current source" made out of \$1\:\textrm{MV}\$ (mega) voltage source and a \$1\:\textrm{G}\Omega\$ series resistor to your grounded load. What's the current variation over your load variation? Not much. Same situation now, except \$10\:\textrm{V}\$ source and \$10\:\textrm{k}\Omega\$ series resistor. Now how good is it? \$\endgroup\$ – jonk Nov 1 '17 at 0:00

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