Im studying realimentation in electronic circuits, I know that for have transconductance gain the circuit have to be of serie-serie type....I have V in the entry and I in the out of it.
To have a transfer function of it I divided in two parts: A=>the amp-op and the part of diodes and transistors and 2) Beta=>the two Resistances of 1kohm.
This way I think I can calculate the transfer function of the circuit to a form A/(1+Beta*A)
In the diodes and transistor part the gain it
To have a transfer function of it I divided in two parts: A=>the amp-op and the part of diodes and transistors and 2) Beta=>the two Resistances of 1kohm.
This way I think I can calculate the transfer function of the circuit to a form A/(1+Beta*A)
In the diodes and transistor part the gain it
audioguru2
- Apr 6, 2004
- 12,026
- Joined
- Apr 6, 2004
- Messages
- 12,026
The schematic is small and is a very fuzzy JPG file type. It would be much clearer if it was made as a GIF or PNG file type.
I can hardly see the part numbers. I cannot see the inverting and non-inverting symbols on the opamp.
Why did you post the schematic "over there"? Please attach a clear schematic to your reply here.
I can hardly see the part numbers. I cannot see the inverting and non-inverting symbols on the opamp.
Why did you post the schematic "over there"? Please attach a clear schematic to your reply here.
audioguru2
- Apr 6, 2004
- 12,026
- Joined
- Apr 6, 2004
- Messages
- 12,026
Thank-you, Upsurt.
Your new schematic is very clear as a GIF file type. But it was huge so I reduced its size, cropped it and made some of its dotted lines solid lines.
I don't think the resistor in series with the load is needed. The voltage gain of the amplifier will be 1 when the load resistance is very low.
I also don't think the capacitor is needed because the output transistors are emitter-followers with a very wide bandwidth, very low phase shift and no voltage gain.
I changed the 47k feedback capacitor to 100k so that the output impedance of the amplifier can be higher.
Your new schematic is very clear as a GIF file type. But it was huge so I reduced its size, cropped it and made some of its dotted lines solid lines.
I don't think the resistor in series with the load is needed. The voltage gain of the amplifier will be 1 when the load resistance is very low.
I also don't think the capacitor is needed because the output transistors are emitter-followers with a very wide bandwidth, very low phase shift and no voltage gain.
I changed the 47k feedback capacitor to 100k so that the output impedance of the amplifier can be higher.
audioguru2
- Apr 6, 2004
- 12,026
- Joined
- Apr 6, 2004
- Messages
- 12,026
The opamp has an extremely high input resistance because its input transistors are FETs. It has an open-loop voltage gain of typically 200,000 so the resistors set the closed-loop gain with their negative feedback.
I made the 1k resistor on the inverting input R3 because you have multiple R2's
From writing the nodal equations for the opamp and solving:
Without the capacitor across the feedback resistor G(s) = (R2 + R3)/R3 = 48
With the capacitor added G(s) = (R2 + R3 + R3.C1.R2.s)/(R3 + R3.C1.R2.s)
Imput impedance 1k, output impedance ~ 24m Ohm
From writing the nodal equations for the opamp and solving:
Without the capacitor across the feedback resistor G(s) = (R2 + R3)/R3 = 48
With the capacitor added G(s) = (R2 + R3 + R3.C1.R2.s)/(R3 + R3.C1.R2.s)
Imput impedance 1k, output impedance ~ 24m Ohm
Last edited by a moderator:
