Recent content by Rumination

It's up to you, if you want to help. I can't force you. :wink:

I know, I should not have used the symbol ϒ. re is better. The emitter resistor and the load resistor form a voltage divider, so the output peak negative swing will be: -21V* 49/(R6 + 49) = -8.8V and from there I found R6 = 62Ω. And Ie4 = 21V/62Ω = 0.34A. What to do next?

Bipolar transistors have a small internal resistance built into their Emitter region called Re. I have found R6 to be: -21V* 49/(R6 + 49) = -8.8V R6 = 62 Ω. And Ie4 = 21V/62Ω = 0.34 A I have been given the value of β = 400 for all transistors in the assignment.

ϒe = re Gain A1: ϒe1 = ϒe2 = VT / IE = 0,025/0,001825 = 13,7 Av1= (R3 II Rib3) / (ϒe1 + ϒe2 + R1 + R2) The gain of the second stage A2: ϒe4= VT/Ie4 = 0,025 / 0,34 = 0,339 Ri4= (β4+1) ( ϒe4+R6) = 401*(0,339+62) = 25 KΩ Av2 = (R5 II Ri4) / (ϒe3+R4) = ((R5*25 KΩ) / (R5+25 kΩ)) / (25 + R4)...

ϒe is: VT / IE = 25 Ω Av1= Rc / 2* ϒe = 2 / 2* 25 = 40. ϒe4= VT/Ie4 = 0,025 / 0,01 = 2,5 Ω I have found R6 to: (-21*49)/(R6+49) = -8,8 ⇒ R6 = 67,9. Ri4= (β4+1) ( ϒe4+R6) = 401 (2,5Ω+67,9) = 28,2 KΩ Av2 = (R5 II Ri4) / (ϒe3+R4) = ((R5*28,2 KΩ) / (R5+28,2 kΩ)) / (25 + R4) Is this correct?

I give up. It's too difficult for me to solve. But thanks for your help:frown:

I can solve two equations with two unknowns. But my problem is, what those two equations are? That's what I'm confused about. Are these the two equations?: (17.35+0.7)V=21-IC3R4 +0.7V=-21+IC3R5 But that's two equations with three unknowns Ic3, R4 and R5

Okay, but when I solve for the current in one equation and inserting it into the other one. I get: (17.35+0.7)V=21 V -IC3R4 (2,95 / R4 ) = IC3 21,7 V = (2,95 / R4 ) * R5

Oh okay. So we get this by eliminating Ic3: 18,05 = 21 - R4 R4 = 2,95 0,7 + 21 = R5 R5 = 21,7 Does it mean, that we have found the values of R4 and R5 now? And now we can find the value of Ic3?

I don't understand this - "Another relation between R4 and R5 can be derived from Q3 and the corresponding DC quiescent voltages/current".

You can see the current I1 under the first set of amplifiers.

My English is not so good. I have found this out: Vcm = 0 , Vin_D = 0 VB1 = VB2 = 0 V. Ic1 = IE1 = I/2 = 3,65 / 2 = 1,825 mA VB3 = 21 V - R3 * Ic1 = 21 V - 2kΩ * 1,825 mA = 17,35 V. Ic3 = IE3 = (24 V - VB3 - 0,7 V) / R4 But I don't know the value of R4?

The figure shows a differential amplifier with a differential input vid = (vin_d_pos - vin_d_neg) between the bases of Q1 and Q2, and an output (out) at the emitter of Q4. I have to Determine R1, R2, R4, R5 and R6 so that the following requirements are met: R 1 = R 2 Av = vout / vid = 100 [V /...