Hello everyone, I'm Electrical Appliance! The differential amplifier Re with both or single-ended inputs has a smaller resistance value, so the common-mode rejection ratio is smaller. As mentioned earlier, the larger the common-mode rejection ratio, the better. In order to improve the common-mode rejection ratio, the resistance value of Re should be increased, but in order to ensure that the operating point is unchanged, the negative power supply must be increased, which is a very uneconomical means. At this time, we can also think of the way is to use the constant current source T3 to replace Re, the constant current source dynamic resistance can improve the common mode rejection ratio, while the constant current source of the tube voltage drop is only a few volts, can not increase the value of the negative power supply. This circuit is called a constant current source differential amplification circuit, as shown in the figure
The circuit below the middle two emitters can be seen that the constant current source is composed of a voltage regulator V., the upper resistors R, T3 and Re, which are not re, V can also be replaced by a resistor with the same temperature coefficient as R. In fact, many circuits use mirror current sources. So why can a current source replace Re?
This is because a double-ended input differential amplifier can replace a constant current source when the current of Re is constant when the input differential mode signal is unchanged.
So what happens if a single-ended input differential amplifier inputs a common-mode signal? Single-ended input single-ended output differential amplification harm common mode magnification is equal to β (Rc∥RL)/[Rb+rbe+2(1+β) Re] The larger the Re, the smaller the common-mode magnification, which is exactly what we need If the common-mode magnification is zero, it is better. The current source is usually larger than a resistor value, so common-mode rejection is also very good.
In fact, the two transistors inside the integrated circuit are not resistors, and they are all current sources, which we should pay attention to! At this time, how to calculate the value of the constant current source? Look at the picture
IE3Q＝（Ⅴz－VBE3）／Re
IC3Q≈IE3Q
IC3Q/2 is the emitter currents IEQ1 and IEQ2 in the T1T2 tube
Then divide (1+β) by IEQ1 and IEQ2 at the same time to get the Ib1Q Ib2Q
Multiply β by IbQ1 IbQ2 to get IcQ1 and IcQ2, and then UcQ1 and UcQ2 can be calculated.
When calculating the quiescent, calculate how much the current of the current source Ic3 is and then calculate it later. For AC calculations, the single-ended input of the two emitters is floating ground and the double-ended input is ground. The calculation results are the same as single-ended and double-ended inputs. Not here. In addition, although the above is called a constant current source but not a constant current, the current source is relatively accurate
Let's take a look at the constant current source differential amplification circuit diagram with a zero-modulated potentiometer
RW is a zero-to-zero potentiometer whose purpose is to adjust the asymmetry of T1 and T2.
Avd=βRc/〔Rb+rbe+(1+b)Rw/2〕(1)
(1) It is a case without load
Replace Rc with Rc∥ 1/2RL when under load
Ro＝2Rc
The following is a brief introduction to the FET differential amplification circuit, as shown in the figure
The insulated gate type field should be connected to the current source of the tube and the other and the triode connection method is basically the same
Differential mode voltage gain Ad=-gmRd when not loaded
Load Connection Ad = one gm(Rd∥1/2RL)
Ri=∞ Ro=2Rd
FET differential amplifiers perform better than transistors.
Finally, thank you very much for watching!