Input Z Local Feedback Triode

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5v333

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Jun 30, 2013
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An amp with a series Rin and a Rfeedback around it has an input Z of about Rin. At least this is how i have understood it.

Although if we have Rin and Rf around a triode with DC potential around its Anode would the input Z depend if the Rf is AC or DC coupled?
See the attached pic.

 

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5v333 said:
An amp with a series Rin and a Rfeedback around it has an input Z of about Rin. At least this is how i have understood it.
That is correct. How close it is to Rin depends on the open loop gain, which for a triode is not exactly huge.
Although if we have Rin and Rf around a triode with DC potential around its Anode would the input Z depend if the Rf is AC or DC coupled?
See the attached pic.
The reason the input impedance is usually close to Rin is because the NFB creates a virtual earth at the grid. In other words, for ac the grid is at 0V so the input impedance must be Rin. In reality, the input impedance is Rin plus the virtual earth. In op amps the open loop gain is so huge it is safe to assume the virtual the is zero ohms.

In both tube schematic cases, the total feedback impedance between anode and grid divided by the stage gain of the triode determines how good the virtual earth is. Its value therefore depends on the stage gain, and the total impedance between anode and grid including that of the capacitors. Provided the caps are sized to have a small impedance compared to Rf or Rin at the lowest frequency of interest then they will not matter.

Cheers

Ian
 
Will the virtual ground collapse if the closedloop gain(set by Rf and Rin) is higher than the initial openloop gain?
 
When you say feedback impedance do you mean Rf/closedloop gain?
For ex
Rin = 100k
Rf = 2M
Triode OLG = 10

CLG = 2M/100K = 20
Feedback Z = 2M/20 = 100K
Virtual ground = 100K/OLG = 10K

 
5v333 said:
When you say feedback impedance do you mean Rf/closedloop gain?
For ex
Rin = 100k
Rf = 2M
Triode OLG = 10

CLG = 2M/100K = 20
Feedback Z = 2M/20 = 100K
Virtual ground = 100K/OLG = 10K

In your example, virtual earth = Rf/OLG = Rf/10 = 200K.

Th big problem is that NFB theory is usually simplified by assuming OLG is very much greater than desired closed loop gain but in you example this is clearly not the case so the simplified theory lets you down.  Instead you have to use the complete equation which is:

Vout/Vin = A/(1 + B*A) where:

Vout/Vin = CLG
A = OLG
B = feed back fraction in this case Rin/Rf

If A is large then this simplifies to CLG = 1/B = Rf/Rin

but with A only equal to 10 you get

CLG = 10/(1 + 10*Rin/Rf) and with the values you gave this becomes:

CLG = 10/(1 + 10/20) = 10/ (1.5) = 6.67

Cheers

Ian
 
5v333 said:
So in this ex, virtual ground is 100k / 6.67 ?
The virtual ground is always OLG/Rf or in the ex 2M/10 =200K

As you can see it is not really a virtual ground because OLG is so small.

Cheers

Ian
 
I would like to make a circuit with in Z of about 330K.
CLG not less than 10.
My OLG is 20.

Am i right that this is achievable with
Rf 4.7M and Rin 100K
 
5v333 said:
I forgot to say that source Z is about 50K.
I guess this matters!

If OLG is 20 and required CLG is 10 and you want Rin to be 330K then

CLG = OLG/(1 + OLG*Rin/Rf)

10 = 20/(1 + 20*330K/Rf)

10 + 10*20*330K/Rf = 20

10*20*330K/Rf = 10

Rf = 10*20*330K/10 = 20 *330K = 6.6Meg

Rin in includes any source impedance, in this case 50K so you can subtract that from thr 330K total and make the actual Rin 280K.

You have very little feedback here. Feedback equals open loop gain divided by closed loop gain, in this case 2 or 6dB.

Cheers

Ian
 
So with that setup, the source will see a 330k load?

I need to brush up my skills in equation solving with paranthesis!

Im in the middle of experimenting. Ill have to come back later if i get some good results.
 
I have playd around with the  closed loop gain formula for a while and got it to stick to my brain!

Now i wonder, if we remove Rin and only have Rf and Rsource, how do i know what load the source is seeing?

For ex
Source is 50k
Rf is 6.8M
OLG is 20
And lets say there is a 1M from grid to ground.

Is it (Rf/CLG)//1M
 
Strike that...

You said Rf becomes Rf/OLG

That would suggest that the source sees (Rf/OLG)//1M

And if theres an Rin, we just add this to the formula above.
 

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