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I'm wondering what the 1.2k resistor does in the following circuit series pass voltage regulator. I should also mention, \$Q1\$ is a Darlington transistor.

ShuntRegulator

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Interesting circuit - the zener could be biased from the incoming power but instead it's biased from the output of the regulator and the reason is this: -

  • it protects Q2's base-emitter junction from reverse bias when the power is applied or if there is an output short or temporary overload.

For a typical NPN transistor such as the BC547 the max reverse voltage that can be applied to the B-E junction is 6V and if the 1k2 fed the zener from the incoming power supply the emitter would always be at about 6.2V - if the output were at zero volts you've exceeded the max limits of the device.

However, with the 1k2 being where it is, exceeding the max limit is impossible because the output level sources both base voltage and emitter voltage.

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  • \$\begingroup\$ Although the 1k2 (why don't people use reference designators in their schematics???) will degrade the regulator's performance slightly, by providing positive feedback instead of negative feedback. If the output voltage rises, then the Zener voltage will also rise slightly, since the Zener diode has a nonzero resistance. \$\endgroup\$ – Dave Tweed Oct 3 '13 at 14:39
  • \$\begingroup\$ @DaveTweed it certainly will but it's unlikely to be a serious regulation showstopper. \$\endgroup\$ – Andy aka Oct 3 '13 at 14:42
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It supplies current to the zener diode which acts as a reference voltage to the emitter of Q2. The 470R and 56OR form a voltage divider that produces a signal at the base of Q2 (6.9V). This is compared with the zener voltage (6.2V) which is lower by the base -emitter voltage of Q2 (0.7V). When this condition is met the output will be 12.7 volts. If the output is pulled lower (large load current) more current is allowed to go into the base of Q1 (less through Q2) and raise the output voltage to the desired level (12.7V)

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  • \$\begingroup\$ I'm still not too sure of why the 1.2k is there. What happens if I open circuit it? \$\endgroup\$ – user968243 Oct 3 '13 at 14:15
  • \$\begingroup\$ @user968243 Not a lot. The zener would still receive current through Q2 and the 680R resistor (5.67mA) and so be at 6V2. The 1.2k does provide an extra 5.4mA for the zener to play with and so could provide more base current to Q1 if needed but as Q1 is a darlington (I suspect the original circuit wasn't) this isn't necessary. \$\endgroup\$ – JIm Dearden Oct 3 '13 at 15:11

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