Search Engine

Power Amplifier Attenuation

Add Question

10 Threads found on Power Amplifier Attenuation
And is 10DB equal to 30 dBm?? ???? Minimum attenuation is 20 dB. (30dBm - 20dB = 10 dBm). You need another attenuator. dB = attenuation or gain dBm = power
That's a lot of power. And more than that, it's an unusual low impedance, that you surely won't find with any wideband RF amplifier. At least frequencies above 1 MHz can't be feed through ususal 50 ohm cables without considerable attenuation. I guess, you have to design a wideband amplifier on your own.
Hi edaboard members! I am currently designing a Class AB power amplifier (in 90nm cmos process) operating at 5.8GHz. In my initial design (which consists of transistor, rf inductor choke and dc block capacitor at the output), I obtained a power gain of 12.26dB. However, when i implemented the input matching network (circuit that matches
Spectrum analyzer, power sensor or power meter, all are fine to measure the gain power, from the moment they are well calibrated. For spectrum analyzer setup beware choosing the right RBW and the right front-end attenuation. DC power can be measured using a simple current meter, or a DC (...)
Hi, I have a 30 dBm power amplifier. If I use 20 dB directional coupler with the power amplifier, how much power does coupled port give? Regards.
Passive pre-amp can be audio pre-amp consisting of connectors, switch and potentiometer. Audio power amplifier with enough gain (giving max power at nominal audio level) does not need any amplification in pre-amp, only attenuation (potentiometer). Passive op-amp is nonsense.
High power S11 and S21 you can measure theoretically without problems using a network analyzer. Beware that you need a high power attenuator at the output do don?t damage the NA. SR706 can deliver 300W (+55dBm), so you need at least 40dB attenuation after the amplifier. To measure the proper output impedance at high (...)
"If the power amplifier generates 1W, then an attenuation of 2dB in the filter translates to 370mW loss of power" I think 10*log(0.37/1)=-4.3dB , not -2dB. Am I right?
Check out this: Input power = 1W Output power = 1W - 0.37W = 0.63W 10 x log (0.63/1) = -2dB Regards, IanP
I am design a blutooth PA with output power 4dBm. But the output power must be controled from -30dBm to 4dBm. 1. How can power control be realized? Someome says that using an voltage attenuator to attenuate the input rf signal is an good method. How can an attenuator be realized in cmos or bipolar porocess, could you give the schematic? (...)