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I was reading the datasheet of the LM324 from TI (http://www.ti.com/lit/ds/symlink/lm324.pdf).

They specify a generic GBW of 1.2MHz. This is usually done with no output applied, from what I understand. Then there is the \$ A_{vd} \$ which is max 100K (100 V/mV). This is the open-loop gain measured with a load, so taking into account loading effects, if I get it right.

Is this gain really reducing by a factor of 10 with few \$ k \Omega \$ on the output?

To understand this better, I read the SLOA11 from TI. They mention \$ B_1 \$ to be the bandwidth at unity-gain with a load specified under the \$ A_{vd} \$ measurement, which should result in a BW of 100kHz. Making it different from the GBP measurement with no load (resulting in unity BW of 1.2 MHz). But in the LM324 datasheet, \$ B_1 = 1.2MHz\$ instead of \$ 100kHz \$. So I am not sure here what are the correct terms and how to use them.

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    \$\begingroup\$ "A few \$k\Omega\$" equates to a few mA, which definitely hinders the performance of a decades-old low-power op-amp. \$\endgroup\$ – DSWG Oct 25 '18 at 22:18
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    \$\begingroup\$ Do you dimensional analysis. 1.2MHz / 100K does not equal 12 -- it equals 12Hz. GBW is gain bandwidth product. In theory a vendor could try to pass off a device with a gain of 100 and a 3db frequency of 1MHz as an op-amp with a GBW of 100MHz (and, in fact, I recall seeing a data sheet for one about 20 years ago -- so maybe it's not just theory!) \$\endgroup\$ – TimWescott Oct 25 '18 at 23:14
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I would reckon that the 100k figure is the gain at DC. Now if the dominant pole is at 12 Hz, then the GBW product above this frequency will be approximately 1.2 MHz. The output impedance of an opamp is normally of the order of 100 ohm so a load of a few k is not significant.

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  • \$\begingroup\$ A little more detail on this. As to how to use the GBW figure, when resistive feedback is applied to the opamp to obtain a finite gain, the small signal bandwidth of the amplifier multiplied with its gain will be approximately GBW. So if a gain of 10 is implemented, the bandwidth will be approx 120 kHz. 12 kHz for a gain of 100. \$\endgroup\$ – Steve Hubbard Oct 26 '18 at 1:04
  • \$\begingroup\$ Regarding the output resistance, I simulated (frequency domain) an inverting amplifier with a gain of 100 and a supply of 15 V using the Pspice model from the TI website. I then drove 1 A AC into its output. At high frequencies the output voltage tends to 50 V indicating an output resistance of 50 ohm. Of course, one should always be wary of the accuracy of Spice models. \$\endgroup\$ – Steve Hubbard Oct 26 '18 at 1:04
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They specify a generic GBW of 1.2MHz. This is usually done with no output applied, from what I understand. Then there is the Avd which is max 100K (100 V/mV). This is the open-loop gain measured with a load, so taking into account loading effects, if I get it right

Having a GBW product of 1.2 MHz doesn't mean that there is any frequency where the gain is 1.2 million. Having an open-loop DC gain of 100 V/mV doesn't mean that the gain must drop to 0 dB at or below 100 kHz.

Assuming this is an old-fashioned dominant-pole-compensated design, you could imagine that the DC gain is 100 V/mV and the dominant pole in the open loop response is at about 3-4 Hz, so that the response falls of at 20 dB/decade to reach 0 dB at 1.2 MHz.

enter image description here

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  • \$\begingroup\$ The only inconsistency is tolerances. The breakpoint is always 45 deg open loop or 10 Hz in your graph and with 100dB DC gain results in only 1MHz small signal BW rather than 1.2MHz. The large signal BW will be limited by current and slew rate, will of course be less. \$\endgroup\$ – Sunnyskyguy EE75 Oct 26 '18 at 0:48

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