In the BJT case, why is it that the upper 3-dB frequency of a differential amplifier with a current-mirror load (1st picture) is lower than that of a differential amplifier with a purely resistive load (2nd picture)?
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\$\begingroup\$ An important thing to realize is that a power amplifier which has a differential input stage with a current mirror active load will achieve twice the open loop gain and twice the slew rate compared to that when the differential input stage is a well balanced resistive load. \$\endgroup\$– user173271Nov 8, 2020 at 8:18
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2 Answers
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Resistors as loads immediately reduce the potential DC gain of your circuit compared to using a current source. But, using resistors also means that the 3 dB point is higher compared to using a current source load. However, if you looked at both circuits side-by-side and compared gain numbers at the 3 dB point of the resistor loaded circuit, you will find that their absolute gain values are pretty similar or slightly favouring the current source loaded amplifier.
It's like comparing an op-amp circuit with a gain of ten versus an op-amp circuit with a gain of 100. both run out of steam at the same high frequency point but the higher-gain circuit has a lower frequency 3 dB point: -
As you can see, when configured as an amplifier with a DC gain of 100, it will have a 3 dB point around 100 kHz whereas when configured as a gain of ten amplifier, the 3 dB point is around 1 MHz - same amplifier but different resistor values.
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Q4 will have Miller Effect multiplication of the Cob of Q4.
You might also consider the Source resistance, that drives the bases of the two differential transistors.
answered Nov 8, 2020 at 4:27