Thanks @ericgibbs ...hi,
On your opening circuit its Q1 and on the LTS circuit its Q3.
E
Already corrected the .asc file to match the picture!
Thanks @ericgibbs ...hi,
On your opening circuit its Q1 and on the LTS circuit its Q3.
E
hi,
hi,
Thats good.
The LTS sim should now give you the correct plots, so that you can check out your calculations.
Regards
Eric
I simply change the MOS simulation model to get Vt =1V and K= 0.5 mA/V² and I did the same think with the BJT.Yes, but only if I use @Jony130 .asc file, because he changed some parameters that I didn't know how to do it!
K = 0.5mA/V² is that Kp parameter you changed to 1m???I simply change the MOS simulation model to get Vt =1V and K= 0.5 mA/V² and I did the same think with the BJT.
Because LTspice is using this formula for Id in saturation region, Id = 0.5*Kp (Vgs - Vt)², and this is why I had to use Kp = 1m instead of 0.5m.K = 0.5mA/V² is that Kp parameter you changed to 1m???
You already forget how to do it ? All supply Voltage are shorted to gnd. And you add one additional voltage source Vx between Q2 collector and ground, and find Zo = Vx/Ix.I mean, in the other side I have +Vcc and R3. After replacing +Vcc by it's internal impedance, typically a short-cut, R3 will be connected to GND... My question is what do I place at Q2 collector terminal? Is it a voltage supply of -3 V???
I haven't forgot that part... I was just trying to understand what is placed at Q2 collector terminal, between the original circuit and the hybrid model. We use to draw an intermediate step. I'll try to draw it!Because LTspice is using this formula for Id in saturation region, Id = 0.5*Kp (Vgs - Vt)², and this is why I had to use Kp = 1m instead of 0.5m.
You already forget how to do it ? All supply Voltage are shorted to gnd. And you add one additional voltage source Vx between Q2 collector and ground, and find Zo = Vx/Ix.
Because Ib =0ASo, the input impedance of the circuit above is 1/hoe because
Your task is to find a gain Vout/Vin not Vout/Vsig. Do you see the difference?I didn't understood why you said that I have Vin connected to Rs.
Yes , you right. But Vid = Vg1 - Vg2 or Vid = Vgs1 + Vsg2we have an example (made with MOSFETs) where from the Gate to the ground we have a voltage source of value of vid/2 on both sides of the Diff amplifier.
Thread starter | Similar threads | Forum | Replies | Date |
---|---|---|---|---|
Exam question on Ohm's law. | Homework Help | 22 | ||
U | Help understanding Electrical FE Exam Question | Math | 2 | |
B | Stuck on this problem on my final exam review guide! | Homework Help | 18 | |
M | Exam question superposition problem | Homework Help | 3 | |
H | Difficult Diode Exam Problem | Homework Help | 6 |
by Jake Hertz
by Jake Hertz
by Jake Hertz