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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)?

Differential Amp. with Active Load

Differential Amp. with Resistive Load

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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\$ – James Nov 8 '20 at 8:18
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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: -

enter image description here

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.

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