01:26 - Intro and Review 04:50 - Finding the Feedback Factor K 09:36 - Complete Analysis of Previous Example: Voltage-Voltage Feedback 18:14 - Complete Analysis of Previous Example: Voltage-Current Feedback 28:46 - Complete Analysis of Previous Example: Voltage-Voltage Example with Heavy Attenuation because of Impedance Dissemblance 44:15 - Example: TIA
For anyone wondering why current I2 is going inside the output port (at 7:20 ) , it's because we are subtracting a current from Iin, in the closed loop system.
26:16 "For an open-loop system, loop gain should be positive", basically, we have inverted the output current as that we get positive gain for a negative amplification across the loop, we are intentionally inverting the output current :) Normally if we would look for trans-conductance gain of K by itself, we would have measured for output current from output port to the ground across a short. :)
At 9:12, do those two tables or this technique have a name? This is so if I want to use this method, I don't have to try to find this video. EDIT: After poking around a while, I think the name of this tune is select the correct form of 2 port (Z, Y, g etc.) for the feedback and it all takes care of itself.
01:16 Analyzing feedback circuits accurately by considering loading effects 04:06 Finding the feedback factor and loop gain in feedback circuits. 10:11 Using transistor as a subtractor for accurate feedback circuit analysis. 13:27 Analysis of Feedback Circuits and Calculation of Key Parameters 20:07 Analysis of open-loop circuit parameters and gain calculation 23:01 Calculating feedback factor K for a trans impedance amplifier 29:04 Using a buffer to reduce heavy attenuation in feedback circuits 31:21 Feedback circuit analysis with detailed steps 37:02 Impedance analysis of feedback circuits in bipolar transistors 41:08 Feedback circuits affect voltage gain and input impedance. 46:53 The problem of instability in feedback systems Crafted by Merlin AI.
At 46:30 I calculated the gain to be Vout/I in = (((Rc//r π2)*Rf) / ( Rf+r π1 ))* 1. ( I approximated the gain of the follower as 1 ). Input current splits to two, the one flowing thru Q1 (Iq) is ( Rf/Rf+r π1)*Iin. this current flows thru (Rc1//r π2) = Vx. Vx/Iq = ((Rc//r π2)*Rf) / ( Rf+r π1 ) multiply this gain by the gain of the follower. Correct ?
41:30 I think (β+1)/gm_F should "approximately" equal to rπ_2, because β=gmrπ. (gm=I_C/VT and rπ =VT/I_B in BJT Charac.) [BTW I usually approximate r_e = 1/gm in BJT Transistor. Don't know if this approximation is valid in general or not?]
If I include QF in feedback network, can I say the topology is a voltage-current feedback instead of voltage-voltage feedback? Is this generally true that according to how we choose our feedforward circuit and feedback circuit the topology changes yet gives the almost similar results?
Current input is a current divider. the collector of Q1 will see Rc in parallel with r π2+(β+1)*Rf. and the gain of the emitter follower is Rf／(Rf + 1/gm2) (All neglect Early effect)
@@許祐嘉-u1cIs the voltage at node x going to be Rc in parallel with r π2+(β+1)*Rf multiplied by [β.(Rf/(Rf+rπ1) *Iin with a minus sign in front of the result. the total gain is then all that multiplied by the follower gain that is approx =1
At 26:00 k should be (-1/Rf)
Yes, because the loop gain A*K has to be adimensional Rf*(1/Rf) so A=Rf and K=-1/Rf
01:26 - Intro and Review
04:50 - Finding the Feedback Factor K
09:36 - Complete Analysis of Previous Example: Voltage-Voltage Feedback
18:14 - Complete Analysis of Previous Example: Voltage-Current Feedback
28:46 - Complete Analysis of Previous Example: Voltage-Voltage Example with Heavy Attenuation because of Impedance Dissemblance
44:15 - Example: TIA
For anyone wondering why current I2 is going inside the output port (at 7:20 ) , it's because we are subtracting a current from Iin, in the closed loop system.
Even I am wondering, I have this doubt from earlier lectures also.
26:16 "For an open-loop system, loop gain should be positive", basically, we have inverted the output current as that we get positive gain for a negative amplification across the loop, we are intentionally inverting the output current :) Normally if we would look for trans-conductance gain of K by itself, we would have measured for output current from output port to the ground across a short. :)
At 9:12, do those two tables or this technique have a name?
This is so if I want to use this method, I don't have to try to find this video.
EDIT: After poking around a while, I think the name of this tune is select the correct form of 2 port (Z, Y, g etc.) for the feedback and it all takes care of itself.
01:16 Analyzing feedback circuits accurately by considering loading effects
04:06 Finding the feedback factor and loop gain in feedback circuits.
10:11 Using transistor as a subtractor for accurate feedback circuit analysis.
13:27 Analysis of Feedback Circuits and Calculation of Key Parameters
20:07 Analysis of open-loop circuit parameters and gain calculation
23:01 Calculating feedback factor K for a trans impedance amplifier
29:04 Using a buffer to reduce heavy attenuation in feedback circuits
31:21 Feedback circuit analysis with detailed steps
37:02 Impedance analysis of feedback circuits in bipolar transistors
41:08 Feedback circuits affect voltage gain and input impedance.
46:53 The problem of instability in feedback systems
Crafted by Merlin AI.
25:50 k=I2/V1 I2=-V1/Rf so substituting I2 in k we get -1/Rf.
At 46:30 I calculated the gain to be Vout/I in = (((Rc//r π2)*Rf) / ( Rf+r π1 ))* 1. ( I approximated the gain of the follower as 1 ). Input current splits to two, the one flowing thru Q1 (Iq) is
( Rf/Rf+r π1)*Iin. this current flows thru (Rc1//r π2) = Vx. Vx/Iq = ((Rc//r π2)*Rf) / ( Rf+r π1 ) multiply this gain by the gain of the follower. Correct ?
41:30 I think (β+1)/gm_F should "approximately" equal to rπ_2, because β=gmrπ. (gm=I_C/VT and rπ =VT/I_B in BJT Charac.)
[BTW I usually approximate r_e = 1/gm in BJT Transistor. Don't know if this approximation is valid in general or not?]
thank you Sir 👍
Simple coepect 😊
If I include QF in feedback network, can I say the topology is a voltage-current feedback instead of voltage-voltage feedback? Is this generally true that according to how we choose our feedforward circuit and feedback circuit the topology changes yet gives the almost similar results?
L
It will be voltage-voltage in both cases
Input is voltage so it can't be voltage-current
46:37 i calculated the gain to be : r π1 / (Rf + r πι1) *(- gm1 * Rc ) , is that right ?
You'll also have r π2 in parallel with Rc
Current input is a current divider.
the collector of Q1 will see Rc in parallel with r π2+(β+1)*Rf.
and the gain of the emitter follower is Rf／(Rf + 1/gm2)
(All neglect Early effect)
@@許祐嘉-u1c you're right....that's accurate... thanks
@@許祐嘉-u1cIs the voltage at node x going to be Rc in parallel with r π2+(β+1)*Rf multiplied by [β.(Rf/(Rf+rπ1) *Iin with a minus sign in front of the result.
the total gain is then all that multiplied by the follower gain that is approx =1
How can I calculate accurately using direct loop gain instead of multiplying K and open loop gain?
you mean finding KA by braking the circuit and putting Vtest on one side and Vf and the other?
Use either -Vf/Vtest or -If/Itest (for current-current) feedback
Thank You