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Regulator for HV Drive of IC Voltage RegulatorDescriptionThe intent of this circuit is to regulate a high voltage source which, unregulated, is unsuitable to drive an IC voltage regulator. In this circuit, V+=75V. Referring to the datasheet for the LM7812, we see that the maximum input voltage is 35V. We also see that given an ideal heat sink, the power dissipation of the device is limited to 20W. If we apply the unregulated 75V to the input of U1, then at .5A,required by our design, the device will be required to dissipate P=VI = (75-12)*.5 = 37.5W. From the datasheet, this is clearly not possible. Consequently, we need to limit the input voltage to U1; thereby, limiting the required power dissipation when driven at our upper current limit of .5A. Given that we intend to operate the device at room temperature with an adequate heat sink, from the datasheet, we require the power dissipation of U1 to be less than approximately 7.5W. We select 6W to allow a margin of error. Therefore, our target input voltage to U1 is given by Vin-Vout=P/I = 6/.5=12V. Since Vout=12V, Vin=24V. Design Details -- How it WorksR14 and R13 provide a voltage divider biasing Q6 at 75*R13/(R13+R14) = 25.33V. Assuming the forward diode drops across Q6 and Q7 to be .7V, the voltage at the emitter of Q7 is: 25.33V-1.4V=23.93V. At .5A, this requires U1 to dissipate (23.93-12)*.5=5.97W which is below our upper limit of 7.5W. (Note that this is an upper limit on the power dissipation requirements of U1 since we designed the input voltage under a no-load assumption.) Load AnalysisFirst, assume we omit Q6 from the design and drive the base of Q7 directly from the voltage divider. Referring to the datasheet for the 2N3055, the worst-case beta for this device is 20. Since the collector current, IC, equals the product of the device beta and the base current, IB, and since the emitter current equals the base current plus the collector current, driving Q7 directly from our voltage divider would require -- using the worst case beta of 20 -- a base current of: .5/(20+1)=23.81mA. Specifically, we have IE = IB + IC = IB + Beta*IB = IB*(Beta+1). Passing 23.81mA through R14 would result in a voltage drop of R14*23.81=238V. Clearly this is not possible. To overcome this problem, we introduce Q6 in a Darlington pair configuration with Q7. This configuration results in a net increase of the input beta equal to the product of the individual betas of each transistor in the Darlington pair. Referring to the datasheet for the 2N4923, the worst-case beta for this device is 40. Therefore, our Darlington pair provides a worst-case beta of 40*20=800. Therefore, for a load current of .5A, we require a base current into Q6 of: IB=.5/801 = 624uA. To ensure an adequate input voltage for U1, we will assume that the 4.96mA current through R14 does not change under the .5A load condition. Given this, the voltage drop across R13 becomes (4.96mA-624uA)*5.1K=22.11V. Therefore, under these assumptions, the input voltage to U1 becomes 22.11-1.4 = 20.7V. According to the datasheet for the 7812, this voltage is sufficient. In reality, the current through R14 will increase under load, as loading the circuit will effectively place a resistance in parallel with R13. Therefore, the current through R14, say I, will be 4.96mA < I < 5.56mA. Using 5.56mA, we have the power dissipation required by R14 given by: 5.56mA*5.56mA*10K=.311W; therefore, we specify a .5W power rating for R14. |
________________________________________ Fine print: Note that this site is intended as an information and education repository. Readers assume all risks and liabilities associated with the use of any information contained herein. The author assumes no liability of any kind for errors, omissions, or claims against any individual or corporation for use of the information contained herein. |