EMRFD Message Archive 1619

1619 2008-04-28 13:16:59 Ashhar Farhan Fig 8.22 - Modular Direct Conversion Receiver of EMRFD Message Date From Subject I was explaining the bias arrangement of the audio pre-amp of the modular direct conversion receiver of EMRFD to someone .. and i got confused myself. here is why ... 1. the common base stage is used so that i can provide a 50 ohms input impedance to the audio diplexer and detector. the Re should be 50 => Ie should be 0.5mA. 2. at 0.5mA, the emitter voltage across the 2.7K resistor between the emitter and the ground should be 1.3V. Accordingly, the voltage in on the base would be 1.8V. 3. The base is biased with a divider of 5.6K and 3.3K, implying that the decoupled power supply voltage to this stage is +5V. 4. If this were true, then the drop across the 10K collector load at 0.5mA would be 5V. However, the emitter is at 1.3V and the power supply is at 5V. Hence, the transistor is running saturated and fully sinking all the current from the load. On the other hand, I can see that the low pass filter is also transforming the impedance. Let's imagine that the impedance driving the audio preamp is 25 ohms. 1. The transistor now needs 1 mA current. giving 2.7 Volts at the emitter and 3.4V at the base. 2. back calculating the biasing dividers gives us a supply voltage of 9 volts. 3. 1mA across the 10K will result in a 10 volts drop. The transistor will still saturate. Am I missing something here? On a side-note, the active decoupler has a biasing arrangement that is crucially dependent upon the transistor's current gain. Hence, when I use a transistor like the BC147 or BC457 (standard in my junk box), their typically high gain (300 typ.) and large variation makes it difficult to repeat the performance and each requires choosing a biasing arrangement indivdually. - farhan I was explaining the bias arrangement of the audio pre-amp of the modular direct conversion receiver of EMRFD to someone .. and i got confused myself. here is why ... 1. the common base stage is used so that i can provide a 50 ohms input impedance to the audio diplexer and detector. the Re should be 50 => Ie should be 0.5mA. 2. at 0.5mA, the emitter voltage across the 2.7K resistor between the emitter and the ground should be 1.3V. Accordingly, the voltage in on the base would be 1.8V. 3. The base is biased with a divider of 5.6K and 3.3K, implying that the decoupled power supply voltage to this stage is +5V. 4. If this were true, then the drop across the 10K collector load at 0.5mA would be 5V. However, the emitter is at 1.3V and the power supply is at 5V. Hence, the transistor is running saturated and fully sinking all the current from the load. On the other hand, I can see that the low pass filter is also transforming the impedance. Let's imagine that the impedance driving the audio preamp is 25 ohms. 1. The transistor now needs 1 mA current. giving 2.7 Volts at the emitter and 3.4V at the base. 2. back calculating the biasing dividers gives us a supply voltage of 9 volts. 3. 1mA across the 10K will result in a 10 volts drop. The transistor will still saturate. Am I missing something here? On a side-note, the active decoupler has a biasing arrangement that is crucially dependent upon the transistor's current gain. Hence, when I use a transistor like the BC147 or BC457 (standard in my junk box), their typically high gain (300 typ.) and large variation makes it difficult to repeat the performance and each requires choosing a biasing arrangement indivdually. - farhan