What is transistor saturation?

For the circuit below, VDD = 10V and VOUT = 2V. Assume the transistor is in saturation, VTN = 1.2V, VDSQ = 5?

• For the circuit below, VDD = 10V and VOUT = 2V. Assume the transistor is in saturation, VTN = 1.2V, VDSQ = 5V, and VGS ≈ ID*RS. The current across the bias resistors (I1) is equal to .05*ID, and R1 || R2 = 96. Find R1, R2, RD, and RS. I need to know how to do like this problem. Please help me Here is the circuit: http://www7.0zz0.com/2012/03/26/01/464318029.gif

• Answer:

  Okay, finally this is posed well enough to solve. Prior attempts weren't so good. Although I'm sure you realize that for a p-channel mosfet Vto=-1.2V, not +1.2V, and it must be that Vdd=-10V, not +10V. I'll assume you are just speaking of magnitudes and work from there. Your statements, as I interpret them, are: 1.      Vdd = -10V 2.      Vs = Vout = -2V 3.      Vds(quiescent) = -5V 4.      Vto = -1.2V 5.      I₁ = I₂ = Id ⁄ 20 = Vdd ⁄ (R₁ + R₂) 6.      (R₁ ∥ R₂) = 96Ω I can easily add a few things: 7.       Vd = Vs + Vds = -7V 8.       Vg = Vs + Vto = -3.2V 9.       Id = (Vdd - Vds) ⁄ (Rd + Rs) Vg will need to be adjusted slightly in the real world because the channel won't be completely open with Vto met. And none of this takes into account Ron because you didn't provide it. Just keep those limitations in mind while moving forward below. From (9) and (4), we can find: 10.     (R₁ + R₂) ⁄ (Rd + Rs) = (20⋅Vdd) ⁄ (Vdd - Vds) = 40 11.     Rd + Rs = (R₁ + R₂) ⁄ 40 So, first thing is to work out R₁ and R₂. You should already know (12) and be able to work out (13) and (14) from basic algebra: 12.     Vg = Vdd ⋅ R₂ ⁄ (R₁ + R₂) 13.     R₁ =(Vdd ⁄ Vg) ⋅ (R₁ ∥ R₂) = 300Ω 14.     R₂ = (Vdd ⁄ [Vdd - Vg]) ⋅ (R₁ ∥ R₂) ≈ 141.18Ω From (11) you know that: 15.     Rd + Rs = (R₁ + R₂) ⁄ 40 ≈ 11.03Ω So, from (9) above: 16.     Id = (Vdd - Vds) ⁄ (Rd + Rs) ≈ -453.3mA And from basic Ohm's law: 17.     Rs = Vs ⁄ Is ≈ Vs ⁄ Id ≈ 4.41Ω 18.     Rd = (Vdd - Vd) ⁄ Id ≈ 6.62Ω I think you will find that (17) and (18) added together match with (15). The above won't be right, of course. You need a Vgs that is, in magnitude, a little larger than Vto to completely open the channel. But this will be close if the pmosfet has an appropriately low Ron and you can easily redo all the above calculations by using a slightly different Vg (moved less than a tenth of a volt towards Vdd, for example) and get closer, if you try and simulate with spice. You could also take into account Ron. But the above is a general approach. Simulation (.op) with the above calculated values and using a Siliconix Si7137DP (which does have Vto=-1.2V and Rs=2mΩ) yields: Vdd = -10 Vd = -7.09481 Vg = -3.20005 Vs = -1.93533 I(R1) = -0.0226665 I(R2) = -0.0226665 I(Rd) = -0.438851 I(Rs) = -0.438851 Vs is a little lower than as-designed, but it was to be expected. On the order of 60-70mV, which is probably about which you'd need to increase Vto (redo the above calculations with this change.) I'd call this good enough for homework.