--- title: "Device to measure 7.5 mW microwave power" subject: "RF and Microwave Engineering" chapter: 8 key: "semester_7/rf_and_microwave_engineering:measure-7-5mw-power" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/measure-7-5mw-power site: BCT Engineering --- # Device to measure 7.5 mW microwave power RF and Microwave Engineering · Chapter 8: RF and Microwave Measurements - Asked in: 2072 Magh, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/measure-7-5mw-power - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-8 ## Question You are supposed to measure about 7.5 mW of microwave power. Choose a proper power measuring device and explain its working principle. ## Short answer Use a thermistor mount with a self-balancing double bolometer bridge. RF heats the thermistor; DC bias is reduced to restore balance, and P_RF = (V_c² − V_rf²)/(4R). A diode sensor or calorimeter is unsuitable. ## Answer 7.5 mW is a **low microwave power**: ```text P (dBm) = 10 log10(7.5 mW / 1 mW) = 8.75 dBm ``` This is inside the range of a **thermistor (bolometer) power meter**, about 1 µW–10 mW. A diode sensor is not suitable, because it is square law only below about −20 dBm, and a calorimeter is too slow and insensitive at mW level. So the proper device is a **thermistor mount with a self-balancing (double) bolometer bridge** – the DC substitution bolometer method. #### Bolometer element - **Thermistor:** semiconductor bead, negative temperature coefficient, robust – chosen here. - **Barretter:** fine platinum wire, positive temperature coefficient, more delicate. The thermistor sits in a matched waveguide/coax **mount** that absorbs all the incident power. #### Basic bridge (DC substitution) ```text +---R1---+---R2---+ | | | DC bias -+ G +- | | | +---R3---+--Rth---+ ^ 7.5 mW RF ``` 1. With no RF, DC bias current I1 heats the thermistor until it reaches the balance resistance R (e.g. 200 Ω); G reads zero. 2. RF power heats the thermistor more; R changes and the bridge unbalances. 3. DC is reduced to I2 to restore balance. The thermistor is at the same temperature, so ```text P_RF = (I1² − I2²) · R ``` #### Practical self-balancing double bridge To avoid drift with room temperature and manual balancing: 1. Two matched thermistors share one thermal block. One gets RF (RF bridge), the other only sees ambient temperature (compensating bridge). 2. Feedback amplifiers keep both bridges balanced automatically, with bridge voltages V_rf and V_c. 3. The meter shows ```text P_RF = (V_c² − V_rf²) / (4R) ``` Temperature drift changes both channels equally and cancels. #### Practical precautions for 7.5 mW - Zero the meter with RF off. - Use a well-matched mount (low VSWR) to avoid mismatch error, and apply the mount calibration factor. - 7.5 mW is close to the 10 mW upper limit; if the level may rise, add a 10 dB attenuator (reading becomes 0.75 mW) and add 10 dB back. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is 7.5 mW in dBm? A. 7.5 dBm B. 8.75 dBm C. 75 dBm D. 0.75 dBm Answer: B. 10 log10(7.5) = 8.75 dBm, a low power within thermistor range. (Key point: Power level classification) 2. Why is a thermistor power meter chosen rather than a diode sensor? A. Diodes are heavier B. Thermistors work only above 1 W C. A diode sensor is square law only below about -20 dBm, and 8.75 dBm is far above that D. Diodes measure only frequency Answer: C. At 8.75 dBm the diode is out of its square-law region, while a thermistor covers 1 µW to 10 mW. (Key point: Choice of device) 3. What does the self-balancing double bridge avoid? A. Drift with room temperature and manual balancing B. Need for a mount C. Need for RF power D. Mismatch of the source Answer: A. A compensating thermistor in the same block cancels ambient drift, and feedback amplifiers balance automatically. (Key point: Why double bridge) 4. 7.5 mW is close to 10 mW. What is the suggested precaution if the level may rise? A. Remove the mount B. Use a calorimeter C. Increase the DC bias D. Add a 10 dB attenuator and add 10 dB back to the reading Answer: D. A 10 dB attenuator reduces power to 0.75 mW for safe measurement; the 10 dB is added back. 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