The Hf/Hr formulation

Thread Starter

The Electrician

Joined Oct 9, 2007
2,970
In another thread, LvW says:

The closed-loop gain of an idealized opamp (Aol infinite) with feedback can be expressed as Acl=Hf/Hr
with Hf=(Vp-Vn)/Vin for Vout=0
and Hr=(Vp-Vn)/Vout for Vin=0.
This a powerful method to analyze opamp circuits, but I think there's a sign error somewhere.

Let's use it to analyze this simple opamp circuit:



All we have to do is find 4 voltage divider ratios. I get this result:



This circuit is so simple that it can be analyzed visually. Assuming an ideal opamp, the voltage across R2 is zero, so the current in R2 is zero. Therefore the current in R1 and R3 is also zero, so the voltage across R1 and R3 is zero. Thus, Vp = Vn = Vin = Vout, so Acl = 1, not -1.

LvW, can you explain where the sign error comes from? Is there a mistake in the algebra?
 

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Tesla23

Joined May 10, 2009
542
This is an application of superposition, the circuit has two voltage sources, Vin and the voltage source in the ideal op amp which we may as well call Vout. Even though Vout is a dependent source, we can still evaluate the network responses using superposition, so denoting

Vp - Vn = Hf * Vin for Vout = 0

and

Vp - Vn = Hr * Vout for Vin = 0

then Vp - Vn = 0 when both sources are active if the contributions from the two sources cancel,

i.e. if
Hf * Vin = - Hr * Vout

or Vout / Vin = -Hf/Hr


That's where your minus sign comes from
 

LvW

Joined Jun 13, 2013
1,759
Hi Electrician,
sorry for my late answer - I was absent for 10 days.

As you probably know - the simple expression HF/Hr rsults from Black´s famous formula
Acl=Ao/(1-Ao*Hr)
with Hr being negative for negative feedback.
However, this formula applies to the classical feedback model where the input signal is directly superimposed to the feedback signal.
If the input signal is attenuated by some circuitry in front of the opamp input nodes (example: inverting opamp operation) the above formula contains an additional term Hf.
Hence, we have:
Acl=Hf*Ao/(1-Ao*Hr)=Hf/(1/Ao-Hr).
In case Ao infinite, this results in
AcL=-Hf/Hr

Therefore, you are right. In my first posting I forgot to include the minus sign.
Thank you for correction.
 
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