--- title: "Double channel bolometer bridge method" subject: "RF and Microwave Engineering" chapter: 8 key: "semester_7/rf_and_microwave_engineering:double-channel-bolometer-bridge" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-channel-bolometer-bridge site: BCT Engineering --- # Double channel bolometer bridge method RF and Microwave Engineering · Chapter 8: RF and Microwave Measurements - Asked in: 2081 Baisakh, 4 marks; 2080 Bhadra, 8 marks (asked 2 times) - Video: [Bolometer | Power measurement | Microwave Engineering | Lec-118](https://www.youtube.com/watch?v=kCT40ZZ3jd0) — Education 4u - 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/double-channel-bolometer-bridge - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-8 ## Question Briefly describe power measurement and the working principle of double channel Bolometer Bridge method. ## Short answer The double channel bolometer bridge uses two matched thermistors in the same thermal block: one absorbs RF, the other compensates for ambient temperature. Self-balancing amplifiers keep both at resistance R, and P_RF = (V_c^2 - V_rf^2)/(4R). Temperature drift cancels, giving about 1% accuracy. ## Answer Microwave **power measurement** finds the average power delivered to a matched load. At microwave frequencies voltage and current are hard to define (especially in waveguides), so power is the basic quantity that is measured. #### Power measurement – overview Power ranges and usual sensors: | Range | Typical level | Sensor / method | |---|---|---| | Low power | < 10 mW | Bolometer (barretter, thermistor), diode detector | | Medium power | 10 mW – 10 W | Thermocouple, calorimeter, bolometer with attenuator | | High power | > 10 W | Calorimeter wattmeter, directional coupler + attenuator | A **bolometer** is a small resistor whose resistance changes with temperature when it absorbs microwave power: - **Barretter** – thin platinum wire, positive temperature coefficient (R rises with heat). - **Thermistor** – semiconductor bead, negative temperature coefficient (R falls with heat). The bolometer forms one arm of a bridge. In the **DC substitution** method, DC bias first balances the bridge; RF power then heats the bolometer, and the DC is reduced until the bridge balances again. The DC power removed equals the RF power absorbed: ```text P_RF = (I1² − I2²) · R ``` #### Double channel (dual) bolometer bridge A single bridge drifts because the thermistor also responds to **room temperature** changes. The double channel bridge (e.g. self-balancing thermistor mount, HP 432 type) removes this error. ```text RF in | +--v---------+ +------------+ | RF thermis-| | Compensat- | | tor bridge | | ing thermis| | (channel 1)| | bridge (ch2)| +-----+------+ +-----+------+ | V_rf | V_c self-balancing self-balancing amplifier amplifier | | +------> Meter <-------+ P = (V_c² − V_rf²)/(4R) ``` Working principle: 1. Two matched thermistors are mounted in the same thermal block, so both see the same ambient temperature. 2. **Channel 1 (RF bridge):** its thermistor absorbs the microwave power. A feedback amplifier automatically adjusts the bias voltage V_rf to keep the thermistor at resistance R (e.g. 200 Ω), so the bridge stays balanced. 3. **Channel 2 (compensating bridge):** its thermistor gets no RF; its bias V_c also keeps it at R. V_c changes only with ambient temperature. 4. With no RF, V_rf = V_c. When RF is applied, the RF bridge needs less bias, so V_rf drops. 5. The metering circuit computes the difference of the DC powers: ```text P_RF = (V_c² − V_rf²) / (4R) ``` Since temperature drift changes both channels equally, it cancels. **Advantages:** automatic balance, temperature compensation, accurate (about ±1 %), direct reading, good for 1 µW – 10 mW. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which pairing of bolometer and temperature coefficient is correct? A. Barretter: negative; thermistor: positive B. Both: positive C. Barretter: positive; thermistor: negative D. Both: negative Answer: C. A barretter (platinum wire) has a positive coefficient; a thermistor (semiconductor bead) has a negative one. (Key point: Bolometer types) 2. In DC substitution, how is the RF power obtained? A. It equals the DC power removed to restore bridge balance B. It equals the DC power added to the bridge C. It equals the galvanometer reading squared D. It equals the ambient temperature change Answer: A. The bolometer stays at the same resistance, so the DC power reduction equals the absorbed RF power, P = (I1^2 - I2^2)R. (Key point: DC substitution) 3. What problem does the double channel bridge solve? A. Low sensitivity of the thermistor B. Drift due to room temperature changes C. High power burnout D. Slotted line mismatch Answer: B. A single bridge drifts as the thermistor responds to ambient temperature; the compensating channel cancels this. (Key point: Why double channel) 4. When RF is applied in the double channel bridge, what happens to V_rf? A. It rises above V_c B. It stays equal to V_c C. It becomes zero D. It drops below V_c because less bias is needed Answer: D. RF supplies part of the heating, so the self-balancing amplifier lowers V_rf; P = (V_c^2 - V_rf^2)/(4R). 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