2.1 What is the minimum number of pins required for a so-called dual-op-amp IC package, one containing two op amps? What is the
1 answer:
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Answer:
a) the output voltage is 50 V
b)
- the average inductor current is 10 A
- the maximum inductor current is 13 A
- the maximum inductor current is 7 A
c) the output voltage ripple is 0.006 or 0.6%V₀
d) the average current in the diode under ideal components is 4 A
Explanation:
Given the data in the question;
a) the output voltage
V₀ = V/( 1 - D )
given that; V = 20 V, D = 0.6
we substitute
V₀ = 20 / ( 1 - 0.6 )
V₀ = 20 / 0.4
V₀ = 50 V
Therefore, the output voltage is 50 V
b)
- the average inductor current
= V
/ ( 1 - D )²R
given that R = 12.5 Ω, V = 20 V, D = 0.6
we substitute
= 20 / (( 1 - 0.6 )² × 12.5)
= 20 / (( 0.4)² × 12.5)
= 20 / ( 0.16 × 12.5 )
= 20 / 2
= 10 A
Therefore, the average inductor current is 10 A
- the maximum inductor current
= [V
/ ( 1 - D )²R] + [ V
given that, R = 12.5 Ω, V = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz
we substitute
= [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]
= [20 / 2 ] + [ 60 / 20 ]
= 10 + 3
= 13 A
Therefore, the maximum inductor current is 13 A
- The minimum inductor current
= [V
/ ( 1 - D )²R] - [ V
given that, R = 12.5 Ω, V = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz
we substitute
= [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]
= [20 / 2 ] -[ 60 / 20 ]
= 10 - 3
= 7 A
Therefore, the maximum inductor current is 7 A
c) the output voltage ripple
ΔV₀/V₀ = D/RCf
given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz
we substitute
ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )
ΔV₀/V₀ = 0.6 / 100
ΔV₀/V₀ = 0.006 or 0.6%V₀
Therefore, the output voltage ripple is 0.006 or 0.6%V₀
d) the average current in the diode under ideal components;
under ideal components; diode current = output current
hence the diode current will be;
= V₀/R
as V₀ = 50 V and R = 12.5 Ω
we substitute
= 50 / 12.5
= 4 A
Therefore, the average current in the diode under ideal components is 4 A
Answer:
a) P ≥ 22.164 Kips
b) Q = 5.4 Kips
Explanation:
GIven
W = 18 Kips
μ₁ = 0.30
μ₂ = 0.60
a) P = ?
We get F₁ and F₂ as follows:
F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips
F₂ = μ₂*Nef = 0.6*Nef
Then, we apply
∑Fy = 0 (+↑)
Nef*Cos 12º - F₂*Sin 12º = W
⇒ Nef*Cos 12º - (0.6*Nef)*Sin 12º = 18
⇒ Nef = 21.09 Kips
Wedge moves if
P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º
⇒ P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º
⇒ P ≥ 22.164 Kips
b) For the static equilibrium of base plate
Q = F₁ = 5.4 Kips
We can see the pic shown in order to understand the question.
Answer:
Explanation:
For pressure gage we can determine this by saying:
The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.
We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h