A question about OTAs

Hello,

Thanks for the quick reply. I agree with your above explanation that if Iout = g_m*Vin then it follows by your explanation that the unity gain frequency is g_m/C_L. However, is that the same for a two stage OTA as well? For example the classic case of the 5 transistor differential input stage whose output is the input for a common source (CS)? Here, I_out is determined by in the input g_m as well as the CS g_m right? Or am I missing something? Nevertheless the unity gain is primarily determined by in the input diff pair g_m and not the CS g_m.

Thanks again for your time,

Regards, Sanjay

OTAs are current-mode devices, so the voltage swings on their internal nodes are small. This means that the internal capacitances are not generally the main bandwidth limiters, and so the calculation is simple.

If you put an input voltage Vin on the OTA, it produces an output current Iout = g_m Vin. The transfer function is therefore H = g_m*Z_L.

Assuming that Z_L = R_L || 1/(j omega C_L) and g_m*R_L >> 1, the unity gain cross will occur when g_m/(omega C_L) = 1 ==> omega = g_m/C_L.

Cheers,

Phil Hobbs

You're making life too complicated. OTA transconductance is defined as d(I_out)/d(V_in). All it is is a diff pair with a bunch of current mirrors, so the input stage transconductance is what sets the total transconductance--all the other stages have current gains of almost exactly 1.0.

Cheers,

Phil Hobbs