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I am attempting to make a voltage regulator. When I make the circuit below but without anything to the left of the dotted line I get the expected voltage; however when I make the full circuit the voltage is much lower than expected. Also the led is a bright for about a quarter second then dims, I can only assume that the voltage across the led is initially as expected but then drops to 2.7v (assume as I cannot get the multimeter ready in time to test initial voltage).

Can someone please help me understand why this voltage drop occurs?

Many thanks.

schematic1

edit: I realise I have drawn the battery upside down.

edit2: the purpose of the left hand side is to ensure there is enough voltage available to split it into two 4.5v outputs (9v required)

edit3: typical input voltage is between 9 and 12 volts.

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  • \$\begingroup\$ What is the purpose of the stuff to the left of the dotted line? What is the Zener supposed to do in that position? \$\endgroup\$ Apr 4, 2020 at 10:54
  • \$\begingroup\$ I think you need to swap the zener and resistor in the thing on the far left... also your battery is either backwards or you've just drawn it backwards. \$\endgroup\$
    – Hearth
    Apr 4, 2020 at 10:58
  • \$\begingroup\$ if voltage below 9v then block all flow \$\endgroup\$
    – MrOtto
    Apr 4, 2020 at 10:58
  • \$\begingroup\$ IMO, you should use feedback from output to regulate it, not from source. \$\endgroup\$ Apr 4, 2020 at 10:58
  • \$\begingroup\$ top of the battery is +, sorry if i got it wrong \$\endgroup\$
    – MrOtto
    Apr 4, 2020 at 10:58

1 Answer 1

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The circuit to the left of the dotted line drops approximately 7.7 volts across the transistor. Note that, because of the zener diode, the voltage at the base of the transistor is going to be approximately 7.1 volts less than the input voltage. The output voltage then is a further diode drop of about 0.6 volts less than this, because of the base-emitter diode. So with 9 V input, you'll be seeing about 1.3 V output from that part of the circuit.

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  • \$\begingroup\$ thanks for your answer. \$\endgroup\$
    – MrOtto
    Apr 4, 2020 at 11:17

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