question with gain bandwidth product

[Jenifer_gao]

Hi All:

If an amplifer operates in the close loop, is the gain bandwidth product still equal to that in the open loop operation?

J

 

[flatulent]

Generally yes. Think of operational amplifiers. When the gain is reduced by the feedback resistors the 3 dB bandwidth increases.

For the case of single transistor amplifiers the product declines with lower gains. This is due to the parasitic resistances staying constant. For bipolar amplifiers, using emitter resistor feedback will mitigate this shrinkage.

 

[noiseless]

Actually, it really depends. you can try use caps as feedback and use close loop do the analysis, you will find the GBP is changing.

 

[Btrend]

Actually, it really depends. you can try use caps as feedback and use close loop do the analysis, you will find the GBP is changing.

the GBP should be no change if and only if ur OP's dominant pole (assume single pole OP) is not affected by the loading cap. if u found the GBP changing with feedback caps, u should check ur dominant pole of ur OP , it must be shift away.
usually ω0=gm/Cload, if ur Cload change, ur ω0 change.

 

[Jenifer_gao]

Thanks Btrend, your explaination makes more sense. I have another question for you with the characteristics analysis on the closed loop amplifier. In the figure 1 in my attachment, it is a general testing setup for an open loop amplifier in frequency response analysis. If I want to test a closed loop amplifer with the configuration shown in figure 2, I couldn't use the same testing setup, because the Hspice complain that there is not path from node X to ground if I apply a Vbias to Vin node.

Do you have any experience with testing the closed loop amplifier, please share your knowledge with you. Thanks.

J

 

[Btrend]

u can connect a high resistance (e.g. 1GΩ) from X to GND.

 

[ezt]

Hi.
If the frequency range of your experiment is known, you can implement a big resistor in parallel with C2 (the magnitude should be carefully chosen in order to not interfere in the circuit operation.)
However, there may be some other useful techniques.

Regards,
EZT

 

[VVV]

The GBW is exactly what the name implies, gain times bandwidth. So if you increase the gain, you decrease the bandwidth.
When operating in open loop, the opamp has the GBW you see in the caalog. That does not change, nor does the internal opamp pole shift in frequency.

However, when you operate in closed loop, it is the components that determine the GBW of the circuit, in case they are not pure resistors.

Look at the picture. The red line is the charactersitic of the opamp, the green one, that of the circuit. The GBW is the bandwidth for a gain of 1 (0dB). Clearly the bandwidth of the circuit can be affected by the external components. But this is what interests you, since you will never use the opamp in open-loop.

If the network only contains pure resistors, the bandwidth is found at the intersection of the gain (horizontal dotted line) with the red line. If using reactive elements, then the GBW is narrower, given by the green line. The bandwidth is determined in the same manner, but using the green line. Thus, it will be narrower.

 

[Jenifer_gao]

Thanks for everybody to reply me

u can connect a high resistance (e.g. 1GΩ) from X to GND.

Btrend, once I connect a high resistance from X to GND, can I do the simulation like the following setup: 1. apply a sinusoidal signal to the positive input node; 2.apply a specific bias voltage to the Vin node (like the way I did in my open loop simulation)

J

 

[Btrend]

However, when you operate in closed loop, it is the components that determine the GBW of the circuit, in case they are not pure resistors.

I don't understand this,
1. as I know, in a sigle pole amplifier A(s)=A0/(s/ω0+1) where ω0 is the pole.
|A(jω)|=A0/sqrt((ω/ω0)^2+1),
at ω=0, |A(0)|=A0/sqrt((0/ω0)^2+1)=A0............. this is DC gain
at ω >> ω0 , |A(ω)|=A0/sqrt((ω/ω0)^2+1) ~= A0/(ω/ω0)~=A0*ω0/ω
let |A(ω)|=1 ===> we get ωu=A0*ω0 .............. this is the Gain-Bandwidth-Product (or called as UGB Unity Gain Bandwidth) of the amplifier

2. if there exist a feedback network with transfer function as β(s), then we should calculate the loop gain (LG=A(s)β(s)) as what we do in step 1.

3. if β(s) is pure resistive, i.e. β(s)=b ...... which is constant and b <1
the LG(s)=A(s)*b
the GBP = A0*ω0*b < A0*ω0
so the GBP of "loop gain" LG is smaller than the GBP of amplifier.
in VVV's drawing, the green line is GBP of LG = A(s)β
and the red line is GBP of amplifier A(s)
which two are different.

4. if β(s) is pure capacitance as in the case we discussed, it is Cf/(Ci+Cf), and this is still a constant < 1. and the result is the same as in step 3.

5. if β(s) is not constant, such as β0*(s/z0+1)/(s/ω1+1), then
the LG(s)=A(s)*β(s) = A0β0*(s/z0+1)/[(s/ω0+1)*(s/ω1+1)],
which is not a single pole system, and the above GBP derivation can not be
easily applied.

6. After calculation of open loop gain, LG(s), we can find the closed loop transfer function as H(s)=LG(s)/β/(1+LG(s))

7. the adding of cap after amplifier can affect ur dominant pole if the amplifier has no buffer stage behind the gain stage. if ur amplifier contains a buffer stage which has a low output resistance, then the load cap simply add a high frequency pole after the UGB, so it won't change anything before the unity gain frequency .

8. the importance is "open loop gain" of amplifier is not equal to "open loop gain" of "amplifier+feedback network" except feedback =1.

hope this make things more clear!

Added after 38 minutes:

Thanks for everybody to reply me

u can connect a high resistance (e.g. 1GΩ) from X to GND.

Btrend, once I connect a high resistance from X to GND, can I do the simulation like the following setup: 1. apply a sinusoidal signal to the positive input node; 2.apply a specific bias voltage to the Vin node (like the way I did in my open loop simulation)

J

1. the GND should be signal gnd, i.e. the DC level of Vin
2. use feedback resistor as stated in ezt's post
3. u can do what u had done before , as long as u can make sure ur virtual short is established, and so ur OP will function as an amplifier not a comparator.

 

[Jenifer_gao]

[quote="Btrend
[/quote]
I don't understand this,
3. if β(s) is pure resistive, i.e. β(s)=b ...... which is constant and b <1
the LG(s)=A(s)*b
the GBP = A0*ω0*b < A0*ω0
so the GBP of "loop gain" LG is smaller than the GBP of amplifier.
in VVV's drawing, the green line is GBP of LG = A(s)β
and the red line is GBP of amplifier A(s)
which two are different.

[/quote]

I don't quite agree with you, when you derive the GBP for the amplifier with a pure resistor feedback network, you missed something. For the system with the feedback network, the transfer function should be:
H(s) = A(s)/[1+A(s)B(s)], if we assme the amplifier is a single pole system, and
A(s) = Ao/sqrt((ω/ωo)^2+1), if B(s) is the constant, like b, we can get
H(s) = [Ao/(1+Aob)]/[1+s/((1+Aob)ωo))]
now the dc gain, Adc, becomes Ao/(1+Aob), and the cutoff frequency, ωc, becomes (1+Aob)ωo, so the GBP is Adc*ωc = Aoωo, keeps the same as the one in the open loop structure.

[/quote]
4. if β(s) is pure capacitance as in the case we discussed, it is Cf/(Ci+Cf), and this is still a constant < 1. and the result is the same as in step 3.
[/quote]

There is a trick in the pure capacitance feedback, because B(s) is called feedback transfer function, which depends on the feedback form. The B(s) is a constant only when the feedback is a series-shunt form. In this case the closed loop GBP is same as the open loop one. But if it is in the shunt-shunt form, the story becomes different.

 

[noiseless]

I agree with Gao.

Actually, when the feedback is cap load, it changes the effective load of the OTA, then change the GBP of the feedback loop.

 

[Btrend]

I think I misleading the concept of gain-bandwidth-product!
in ther previous derivation, I mistaken that gain of LG is not the gain in GBP, the gain is defined by Vo/Vi, and the bandwidth is also defined by Vo/Vi.
Vo/Vi=LG/β/(1+LG) when the loop are closed.

There is a trick in the pure capacitance feedback, because B(s) is called feedback transfer function, which depends on the feedback form. The B(s) is a constant only when the feedback is a series-shunt form. In this case the closed loop GBP is same as the open loop one. But if it is in the shunt-shunt form, the story becomes different

I don't understant these words.
Honestly speaking, I still don't understand clearly the 4-type feedback forms after so many years.
especially those series-series, series-shunt,shunt-shunt, shunt-series, what's main difference?
and what's the purpose to classfy them?
and why feedback of a series cap. is called series-shunt ? why ?

 

[Jenifer_gao]

There is a trick in the pure capacitance feedback, because B(s) is called feedback transfer function, which depends on the feedback form. The B(s) is a constant only when the feedback is a series-shunt form. In this case the closed loop GBP is same as the open loop one. But if it is in the shunt-shunt form, the story becomes different

I don't understant these words.
Honestly speaking, I still don't understand clearly the 4-type feedback forms after so many years.
especially those series-series, series-shunt,shunt-shunt, shunt-series, what's main difference?
and what's the purpose to classfy them?
and why feedback of a series cap. is called series-shunt ? why ?

If the feedback is a pure capacitive network, and we use OPAMP as the amplifier, we can get two types of feedback, one is shown in figure 1(series-shunt), another is figure 2(shunt-shunt). In the series-shunt version,
the B(s)= Vfb/Vout=C1/(C1+C2), it is a constant. But in the shunt-shunt version,
B(s) = Ifb/Vout = -sC2, and then you can derive H(s) = Vout(s)/Iin(s) = A(s)/(1+A(s)B(s)), but if you want to get voltage gain transfer function, you have to convert it to: H'(s) = Vout(s)/Vin(s) = Vout(s)/Iin(s) * Iins/Vin(s)=Vout(s)/Iin(s)*sC1, and you can prove this transfer function introduce a zero.

For the 4-type feedback forms, the first term of the name referes the feedback, and the second term of the name refers to the output sensing. For example, in figure 1. the feedback is Vfb, which is in series with Vin, but in figure 2, the feedback is Ifb, which is shunt with Iin. At the output node, figure 1 and figure 2 all sensing output voltage, so it's called shunt at the output. There is a simple rule to help you to identify the feedback form: if the feedback is voltage, it is called series at the feedback node, otherwise is called shunt; if the sensing is directly connected to the output node, which means sensing voltage, and it is called shunt; if the sensing is indirectly connected to the output node, which means sensing current, and it is called series.

In many cases, feedback is used to solve stability problem, and it can be used to modify the input and output impedence. Hope it can help.

 

[qslazio]

can anybody tell me why the close loop GBW is different with GBW in open loop version with a capacitor feedback?

I can not find the answer in the above discussion.
thanks