gain bandwidth product for error amplifier

Thread Starter

anhnha

Joined Apr 19, 2012
880
I am design an error amplifier for LDO. However, I don't know why we need a high gain bandwidth product. It seems to me that we are working with low frequency signals and so the bandwidth is not a problem.
What I am wrong here?
 

Jony130

Joined Feb 17, 2009
5,022
It seems to me that we are working with low frequency signals and so the bandwidth is not a problem.
Are you sure that we we are working with low frequency signal?.
What about digital circuit, RF circuit ect. Thous circuit are also supply from voltage regulators. But as with all digital circuits, the supply current is drawn in very short spikes on the clock edges, and if I/O lines are switching, the spikes will be even higher. So digital circuit will cause current pulses is drawn from our LDO reg. And this is why our LDO need to be fast.
 

Thread Starter

anhnha

Joined Apr 19, 2012
880
Using a low bandwidth amp leads to poor transient response plus high frequency noise on the output.
I also read that a wide bandwidth will cause more noise generated by the op-amp since this noise is integrated overall its bandwidth.
It seems to contradict with the above.
 

Thread Starter

anhnha

Joined Apr 19, 2012
880
Are you sure that we we are working with low frequency signal?.
What about digital circuit, RF circuit ect. Thous circuit are also supply from voltage regulators. But as with all digital circuits, the supply current is drawn in very short spikes on the clock edges, and if I/O lines are switching, the spikes will be even higher. So digital circuit will cause current pulses is drawn from our LDO reg. And this is why our LDO need to be fast.
I only know that my input voltage is from 2 to 3.3V. How can I know that if there are high and low frequency components in that?
 

Thread Starter

anhnha

Joined Apr 19, 2012
880
I still can't see the relation between slew rate and gain bandwidth product. They are two different concepts and seems there is no relation.
 

Thread Starter

anhnha

Joined Apr 19, 2012
880
From the first link:
Vout = Iout1/(2πCc*f)

How can you know that the voltage at the output of transconductance stage is 0V?
Also from that link:

Another indicator of op amp speed is the small-signal slew rate. This parameter is tied to the frequency corner where the closed-loop gain meets the amplifier open-loop gain curve. This point is shown in Figure 2.
How to know that the intersection point of closed-loop gain and open-loop gain curves is f-3dB?
 
Last edited:

Jony130

Joined Feb 17, 2009
5,022
From the first link:
Vout = Iout1/(2πCc*f)

How can you know that the voltage at the output of transconductance stage is 0V?
But Vout is not 0V. Vout = Iout * Zload

Also from that link:
Another indicator of op amp speed is the small-signal slew rate. This parameter is tied to the frequency corner where the closed-loop gain meets the amplifier open-loop gain curve. This point is shown in Figure 2.
How to know that the intersection point of closed-loop gain and open-loop gain curves is f-3dB?
Because this is how we define our corner frequency.
Corner frequency is defined as the frequency at which the gain drops to 0.707 corresponds approximately to half power.
Fc = Vout/Vin = gain = 1/√2 = 0.707 = -3.01dB
So in figure 2 Fc is at frequency when the gain is equal to 37dB
 

Jony130

Joined Feb 17, 2009
5,022
Q1 - Va is 0V because the next stage is a transimpedance stage (current in, voltage out).
And as you should now, the Zin for transimpedance stage is 0 Ohms.

We simply have a ideal voltage amplifier with gain equal to A = -10V/V
Next we connect a resistor R = 10Ω between the input and the output of the amplifier.


Now let as try to find a input resistance.

Rin = Vin/Iin

In = (Vin - Vout)/R = (Vin - A*Vin)/R = Vin * (1 - A)/R

Rin = Vin/Iin = R/(1 - A)

Iin = (1V - (-10V))/10Ω = 1.1A

So Rin = 1V/1.1A = 0.909Ω

So as you can see our Rin resistance is (1 - A) smaller then R if we have inverting amplifier.
So if A ---> -∞ then Rin ---> 0Ω. And this is why no voltage can exist at the input of this transimpedance stage. Only current can flow.

As for the Q2.
Do you know what Open loop gain vs frequency represent? And what Closed loop gain is ?
 

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Jony130

Joined Feb 17, 2009
5,022
Ok, so you indeed understand what open loop gain and closed loop gain is.
Let as exam this circuit



As you can see we have a op amp circuit without negative feedback. So we have open loop circuit.
And the Aol - Open loop gain is shown in figure 2.
Aol_max = 120dB = 1 000 000V/V and GBW - Gain bandwidth product = 1 000 000 * 10Hz = 10Mhz
Now we closed the loop by connecting point A with point B.
And the Acl - Closed loop gain = Aol/(1 + Aol *β)

β - feedback factor ( feedback gain ) = R2/(R1 + R2)

So our Acl for low frequency is equal to

Acl = Aol/(1 + Aol *β) ≈ 1/β ≈ 1 + R1/R2 = 100V/V = 40dB

But as signal frequency increase, the open loop gain start to drops.
And at frequency when Aol < Acl the closed loop gain also start to drop.
Because op amp don't have don't have enough open loop gain to keep the Acl at 40dB. So the corner frequency is when Aol drops to the value when Acl = 37dB = 70.7V/V.
See this
http://users.ece.gatech.edu/~alan/ECE3040/Lectures/Lecture29-OP Amp Frequency Response.pdf (page 15 and 16).
Also check the simulation file.
 

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Thread Starter

anhnha

Joined Apr 19, 2012
880
Thanks. I think the picture in the post #13 is a bit misleading.
fc = -3dB point should be the point in which closed-loop gain drops to 37dB not 40dB as in the figure.
 

Jony130

Joined Feb 17, 2009
5,022
Thanks. I think the picture in the post #13 is a bit misleading.
fc = -3dB point should be the point in which closed-loop gain drops to 37dB not 40dB as in the figure.
Misleading? Maybe for the beginners. Because we always using asymptotic approximation when we draw gain vs frequency.
 
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