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Output Impedance
Since we seem to have swapped maximum output voltage swing for low output impedance, what is the output impedance of this circuit? Once again the method of forcing the output 1 volt positive by placing a 1 volt battery at the output and then determining the ensuing current flow from the output into the battery gives us the inverse of the output impedance. In much less tangled words: When a voltage is placed across a resistance, a current flows through the resistance. The lower the value of the resistance, the greater the current flow. One volt across one ohm equals one ampere of current. Two volts equals two amperes. If we place an unmarked resistor across a 1 volt battery and we have the means of measuring the current flow through the resistor, we then can determine the value of the resistor by working the math backwards, as resistance is equal to voltage divided by current. For example, if we measure 4 amps of current, then the resistor has a value of 0.25 ohms. In the same way we can calculate the output impedance of this circuit by finding the current flow out or into the output of this circuit and then dividing that current into the voltage that rise it.
Starting at the top triode, if the cathode is forced 1 volt positive the grid will effectively become more negative relative to the cathode and the current through the triode will decrease. How much less current? The transconductance of the cathode times the voltage yields the value of the current. The cathode's transconductance is equal to (mu + 1) / rp. The transconductance of the grid is less: mu / rp. The difference results from the difference in cathode-to-plate voltages; when the grid receives the drive voltage the cathode-to-plate voltage remains constant (we are now measuring the attributes of the triode statically with fixed voltages not amplifying signals in a circuit). When the cathode receives the drive voltage the cathode-to-plate voltage changes, which causes the rp to respond with greater or less conduction.
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