What is the gain gain of a common emitter with split supply?

Thread Starter

Santa klaus

Joined Nov 16, 2014
32
So the gain of a common emitter stage should roughly Rc/Re .it works fine in simulation . However, when i use emitter biais (use split supply on the transistor ) there is no gain. Am i doing something wrong !
upload_2015-9-28_19-24-58.png
 

Jony130

Joined Feb 17, 2009
5,163
In first circuit (split supply) transistor is in saturation region and this is why the gain is one. Notice that Ie = (15V - 0.7V)/1K =14.3mA.
and VRc = 14.3mA * 2K = 28.6V. So this clearly show that BJT is in saturation.
 

Thread Starter

Santa klaus

Joined Nov 16, 2014
32
In first circuit (split supply) transistor is in saturation region and this is why the gain is one. Notice that Ie = (15V - 0.7V)/1K =14.3mA.
and VRc = 14.3mA * 2K = 28.6V. So this clearly show that BJT is in saturation.
Isn't the collector voltage Vrc = Vcc-RI so 15-28.6 = - 13.8 ? what about the positive supply potential
 

Jony130

Joined Feb 17, 2009
5,163
Isn't the collector voltage Vrc = Vcc-RI so 15-28.6 = - 13.8 ? what about the positive supply potential
No, the voltage at collector cannot be lower than -0.7V. Notice that in saturation NPN transistor behavior just like two diodes.
https://upload.wikimedia.org/wikipedia/commons/c/cf/NPN_Transistor_scetch.svg
How do i make a common emitter in split supply with predictable gain?
Use a differential amplifier or increase Re for example to 7.5kΩ and add additional Re1 (1.538kΩ) resistor together with a series Ce capacitor.
 

Jony130

Joined Feb 17, 2009
5,163
I forgot to mention that this additional Re1 resistor must be in parallel with Re.
1.PNG

And the gain is Av ≈ (Rc||RL)/( re + Re||Re1) ≈ R4/(R5||R7)
where re = 26mV/Ie
 

Russmax

Joined Sep 3, 2015
82
How do i make a common emitter in split supply with predictable gain?
Santa Klaus, the mistake you've made is in failing to re-reference the input signal to the negative supply. In circuit one, you have a 3V DC supply in series with your AC signal. This 3V gives your transistor the correct DC operating point. When you went to split supply, your base voltage is way too high, which causes your emitter voltage to be way too high, which causes your emitter and collector currents to be way too high, which causes the voltage across Rc to be way too high, which causes Vcollector to be way too low--biasing the transistor out of the active region and into saturation.

To get the split supply circuit to work like the single supply, you need to do a couple of things.
1) The total supply voltage should be the same. VCC-VEE = 15V. So use +/- 7.5 V supplies.
2) The single supply circuit has the base 3V above GND. The split supply circuit also needs the base 3V above VEE (-7.5V). So you need to set it up so VB = VEE + 3V.

Hope this helps,
Regards
 

Thread Starter

Santa klaus

Joined Nov 16, 2014
32
Santa Klaus, the mistake you've made is in failing to re-reference the input signal to the negative supply. In circuit one, you have a 3V DC supply in series with your AC signal. This 3V gives your transistor the correct DC operating point. When you went to split supply, your base voltage is way too high, which causes your emitter voltage to be way too high, which causes your emitter and collector currents to be way too high, which causes the voltage across Rc to be way too high, which causes Vcollector to be way too low--biasing the transistor out of the active region and into saturation.

To get the split supply circuit to work like the single supply, you need to do a couple of things.
1) The total supply voltage should be the same. VCC-VEE = 15V. So use +/- 7.5 V supplies.
2) The single supply circuit has the base 3V above GND. The split supply circuit also needs the base 3V above VEE (-7.5V). So you need to set it up so VB = VEE + 3V.

Hope this helps,
Regards

Thanks, you have been really helpful. The base voltage Vb was way too big causing a big voltage drop across Re and too much current.
 
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