Voltage Regulator current flow

Discussion in 'Homework Help' started by jwarankie, Nov 24, 2015.

  1. jwarankie

    Thread Starter New Member

    Nov 24, 2015
    Can you please explain what path the current flows in the attached schematic?
    1. I would like to know for positive and negative cycles.

    2. I would like to know the input voltage at the transformer and the output voltage at the input of the regulator during the full A/C period if possible. ( I have seen examples online of a bridge rectifier showing how the input A/C signal waveform is and what the output after it has been rectified which would help me understand better )

    3. I am not sure why C1 and C2 have such a large capacitance value as my searching online show that the input and output filter capacitance is generally .1uF for a 7805 Regulator? ( my diagram is showing 22uF capacitor at the output and 4000uF at the input it seems to me? )

    4. I think it would make more sense to use 4 diodes as a bridge rectifier, then a filter capacitor for the input to the regulator.... is this circuit a cheaply made one or is this common? )

    This is from an older 1980's hand spin wheel balancer and I wanted to know if I needed to change out the capacitors because of their age, but the output to the regulator is showing 5.06 Vdc so it looks to be working ok?

    I did notice that there was a .037 Vac ripple measured with a multimeter when I place the negative lead to a metal part on the machine and to a ground pin on an IC. ( i.e. the Ground pin on the IC and "ground" on the machine )

    I also noticed that there was a .037 Vac ripple measured on the IC's positive pin. ( is this ripple insignificant and or normal or is this a result of a bad capacitor not filtering out the A/C ? )

    Sorry for all the questions, I appreciate any help you can lend me.

    Thank You for your time.

    - joe
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
    D1 and C2 are not needed to make the +5V at the regulator output. They would be used only if you needed an unregulated negative output voltage.

    Since the rectification for the 7805 is only half-wave, it will not supply much current..., likely much less than its rated 1A.

    C1 is the filter capacitor. It needs to be as big as possible because it holds up the supply current for 16ms until the next half-wave pulse from the transformer/rectifier.

    If you need both positive and negative outputs from a non-center-tapped transformer, then you cannot use a four-rectifier full-wave bridge.
  3. jwarankie

    Thread Starter New Member

    Nov 24, 2015
    Thank You for your help.
    I was suspecting that this was a half wave rectifier like you suggested but was wondering why D1 and C2 were put in place because usually this would be a waste of money for the company to use something that doesn't get used?


    - joe
  4. crutschow


    Mar 14, 2008
    If you don't need a negative voltage, it would be much better to add two more diodes and make a full-wave bridge to generate the plus voltage.
  5. jwarankie

    Thread Starter New Member

    Nov 24, 2015
    Thank You for your reply.

    If I simply use a 9volt battery instead of the 8.5Vac stepdown transformer to power the circuit, would this be a good substitution?

    My thinking is that the 9Volt battery will provide the proper voltage to the regulator and there would be no difference except that there would be 0 ripple voltage and I could have a portable wheel balancer.

    If I look at the diagram, I can see that the positive voltage from the battery would pass through the diode, it would bring the voltage down to 8.3Vdc which is plenty for the Regulator, the capacitor will charge up to 9V and sit there doing nothing, ( maybe I would need to replace the 4000uF capacitor with a resistor value instead) the regulator common lead will return the current from the battery to the negative battery lead.

    What do you think?
  6. MikeML

    AAC Fanatic!

    Oct 2, 2009
    What is the ultimate load?

    A 9V battery is suitable only for load that draws a few tens of mA.


    If the load is 1/4A, the 9V battery will be dead in minutes.

    Just the quiescent current into the 7805 will kill the battery in a day or two...
  7. Russmax


    Sep 3, 2015
    1. For a positive cycle, D2 conducts and D1 doesn't. For a negative cycle, D1 conducts and D2 doesn't.

    2. Given just the diagram, we don't know the transformer's input voltage, but I'm assuming 120V AC rms if you're in America.

    3. C1 and C2 need to be large for high ripple, as MikeML said. For C3, beware of moderate value capacitors on the regulator's output. There is stability with very large C or very small C, but often not with "medium" C. I'd stay with the values that you know are working if you replace caps. Which is a good idea, btw.

    4. Since the circuit is creating a +5V DC regulated and -11V DC unregulated output from a single transformer winding, you have to stay with 1/2 wave rectifiers.

    Other questions: The regulator appears to be working fine, but the AC voltage seems a bit high. I'd replace caps. It may not make a difference. This circuit will not work from a battery because it will not create the -11V DC on C2.