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| Hi, at first sorry english is not my native language ,but i have a question: someone could tell me how it works : http://www.designnotes.com/CIRCUITS/varps.htm I cannot understood how obtained 0V at output in this design ... ;( thanks in advance | |
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| It appears that the NPN pass transistor is also used as a switch. The only path for current is thru the transistor, and thru the IC. I can't find the datasheet for this IC on the internet. Chipdocs.com has it, but they charge a price to register to view their datasheets. If the IC is configured properly, it appears that there can be zero volts output on this variable power supply. The current thru the transistor can easily be turned off by turning off the voltage to the base. When there are zero volts at the base, there will be no current thru the transistor, and no voltage from it, the transistor is turned off, and it is like an open switch. The only other output path is thru pins 1, 2 and 10 on the IC, and it depends on how the IC is configured. It appears that 1, 2, and 10 are some type of sensing feedback paths. It also looks like pin 6 is the voltage output, and pin 6 is connected to the zener, which goes to the base of the transistor. The zener probably drops the voltage output below the normal minimum output down to zero volts, which would make the circuit capable of a zero volt output. But without the datasheet for the CA723CT, this cannot be verified. You could probably email the author for more info. Since there is a link to his e-mail address, he probably wouldn't mind.
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Here is the datasheet. http://xgistor.ath.cx/files/Electron...alog/LM723.pdf IC723 is wired as high-voltage, high-current regulator.
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| I've made some clarify on sematic, and copyed the metal-can pinout.(the numbers of sematic for metal-can package, the numbers is different as DIL pack and no contain a zener-diode for voltage shift.) Between 1-10 pins the IC contains an NPN transistor E-B and with external Rsc working as current limiter.The red line show the output current flow. On the non-inverting input is a divided reference voltage. The output voltage feedback coming via R3 to inverting input, and here compared with control voltage.(coming via R4 from poti.) The 7.5V zener is a level-shifter: allow control the voltage to zero without negative slave-supply. | |
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| maby something to with the ZENER DIODE? if the voltage does not exceede 7.5 volts it will not aloow voltage to the transistor, and therfore not allowing the NPN to flow current?
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| Yes, but that doesn't mean that it is operating as switch. For regulating the voltage, the pass transistor needs to be operated in active region. You might be thingking that without proper biasing how is the transistor kept in active region? This is due to the negative feedback used by the IC which constantly adjusts the output to keep it in active region. Regarding the 0V output, according to datasheet the IC can provide 2V-37V regulated output for a input of max. 40V. It has 2 operating modes namely the low voltage (2-7V) and high voltage(7-37V) modes. The above circuit gives 0V due to the use of zener diode.
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