A question about OTAs

Hello All,

I have a not so difficult question that have never quite got a
satisfactory answer for. Simply put why is it, at least to a first
order approximation, that the unity gain frequency for an OTA
regardless of the topology is (in radians) simply gm/Cl. Where gm is
the transconductance of the input differential pair and Cl is the load
capacitor. In every textbook/paper this is always the case from
topologies ranging from the simple Common Source amplifier to folded
cascodes to 2 stage amplifiers, for example a telescopic followed by a
common source, or a current mirror one. It doesn't matter if the output
is differential or single ended, or if it's ClassAB or ClassA, the
unity gain frequency is gm/CL. How does one intuitively explain that?

Regards,
Sanjay

 

[Phil Hobbs]

svinayag@gmail.com wrote:

Hello All,

I have a not so difficult question that have never quite got a
satisfactory answer for. Simply put why is it, at least to a first
order approximation, that the unity gain frequency for an OTA
regardless of the topology is (in radians) simply gm/Cl. Where gm is
the transconductance of the input differential pair and Cl is the load
capacitor. In every textbook/paper this is always the case from
topologies ranging from the simple Common Source amplifier to folded
cascodes to 2 stage amplifiers, for example a telescopic followed by a
common source, or a current mirror one. It doesn't matter if the output
is differential or single ended, or if it's ClassAB or ClassA, the
unity gain frequency is gm/CL. How does one intuitively explain that?

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

 

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

 

[Phil Hobbs]

svinayag@gmail.com wrote:

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

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