1. 30 dbm corresponds to a power level of: 2. An amplifier has a power gain of 2

Question

1. 30 dbm corresponds to a power level of:


2. An amplifier has a power gain of 20 db. If the output power is doubled (for the same amount of input power), the power gain is:


4. If a diff amp were perfect, the CMRR would be:

15. The common-mode voltage gain of a diff amp is equal to RC divided by:




16. The power gain of an amplifier equals 20 db. If the input power equals 10 mW the output power equals:




17. A 2N5555 has IDSS =16 mA and VGS(off) = -2V. What is the pinchoff voltage for this JFET?



18. The voltage gain of a diff amp with a differential output is equal to RC divided by:



19. The voltage gain of an amplifier is 200. The decibel voltage gain is:



20. MOSFETs are often shipped with a wire ring around the leads to protect the MOSFETfrom static charges.
A) True
B) False

Explanation / Answer

1. 10^-3 * (10^(30/10)) = 1 W 2. Power gain is half due to ( 2Pout = ((100 or 20db ) * Pin) , Pout = (100/2)*Pin = 50 or16.99dB 4. the perfect differential amplifier's output voltage should hold absolutely steady (no change in output for any arbitrary change in common-mode input). This translates to a common-mode voltage gain of zero. 16. Pgain = 20dB or 100 Pin = 10mW Pout / Pin = Pgain Pout = Pgain * Pin Pout = (100) * (10m) = 1W 19. 20log(200) = 46.02 dBV 20.True

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.