6
\$\begingroup\$

Okay guys, this one has me baffled. Take these two circuits:

enter image description here

Typical stuff; Power in goes to the + rail, GND to the - rail. This is a 19V power supply going in. It goes through the voltage regulator (rated for up to 35V, 1 A), output pin goes through the load and back to ground.

Now, In the circuit on the left, taking a multimeter and measuring voltage between the positive rail and negative rail (total voltage) gives 19V. Measuring from A to B (output pin of the voltage regulator to the negative terminal) gives 5V.

Now replace that resistor with a different load: The Raspberry Pi. It's wired correctly (i.e. converted to USB correctly, it's supposed to run on 5V).

In the circuit on the right, measuring voltage between the positive rail and negative rail gives 19V. Measuring from the A to B, output pin to end, gives 2.5V.

It just does. The Raspberry Pi doesn't even turn on. Take it back out and put a resistor back in, you measure 5V at the output pin.

Is this a dysfunctional regulator? Or am I not understanding something?

\$\endgroup\$
1
  • 1
    \$\begingroup\$ P = ( Vin - Vout ) * Iload = ( 19 - 5 ) * 0.5 = 7W (at least, assuming the Pi pulls 2.5W) \$\endgroup\$
    – user2497
    Commented Jul 13, 2017 at 7:31

2 Answers 2

Reset to default
6
\$\begingroup\$

First observation: Decoupling capacitors ought to be connected at both the input and output legs of the regulator, as close as possible to the pins, connecting to the ground rail. Without those, most likely the regulator is going into oscillation at the output.

Notice that your observed output voltage is exactly half the expected output - an indication that the output is oscillating between 5 and 0 Volts. A DC voltmeter won't catch the oscillation, just the resultant average voltage. An oscilloscope trace will show it.

Second observation: Even though the regulator is rated for 1 Ampere, and the load is probably rated for much lower, this does not take into account the surge current the Raspberry Pi board needs at start-up. That surge is most likely forcing the regulator into over-current protection, and that'd be one of the probable causes for the oscillation.

The resistor on the other hand is a passive load, which will carry essentially a constant current both at start-up and later. Hence no surprise current surge to destabilize regulation.

Adding a suitably large decoupling capacitor at the output leg of the regulator will also help damp the surge load on the regulator, hence it is quite feasible that the regulator and your Pi will stabilize. If not, you may need to use a higher rated regulator to cope with the surge and the operating current needs of your arrangement.

Final observation: Some linear regulators need a basic minimum load current for stable regulation. This can be achieved through a load resistance next to (parallel to) the output decoupling capacitor - the resistance to use would be calculated to just barely draw the required minimum load current. Though this is probably not the problem in your circuit, it is useful to know, and address, based on information in the regulator's datasheet.

\$\endgroup\$
9
  • \$\begingroup\$ Great answer. The highest power consumption on the RPi will most definitely be during startup, when the embedded OS boots up. If you're using the RPi Model B, the constant current can reach 750mA, but the peak will be significantly higher. There's a nice wiki entry on homemade RPi PSUs, if you want to take a look. Most of the successful ones are rated up to 1.5A or 2A. \$\endgroup\$
    – Polynomial
    Commented Nov 21, 2012 at 14:09
  • \$\begingroup\$ I don't think there's that big of a surge on startup, as many phone adapters that are only rated at 750mA work quite well with the Pi. Most likely it's more to do with the Pi acting like a constant-power load (drawing more current at lower voltage) instead of a constant-resistance load (like a resistor). \$\endgroup\$
    – Adam Lawrence
    Commented Nov 21, 2012 at 16:13
  • \$\begingroup\$ @Madmanguruman Phone adapters use switched mode regulation, often with cycle-by-cycle current limiting at switching frequency. Also, there would be output leg decoupling. The OP's schematic is a 3-terminal (presumably) linear regulator, and is missing the capacitors. \$\endgroup\$
    – Anindo Ghosh
    Commented Nov 21, 2012 at 16:23
  • \$\begingroup\$ @AnindoGhosh Look at my profile. I know what a switching regulator is. Inrush is generally generated by capacitor charging in the load or constant-power loads drawing high current when the input voltage is low. I doubt that the Pi draws significantly more than it's maximum steady-state current, which is the definition of 'surge'. \$\endgroup\$
    – Adam Lawrence
    Commented Nov 21, 2012 at 16:53
  • \$\begingroup\$ @Madmanguruman Nice profile. \$\endgroup\$
    – Anindo Ghosh
    Commented Nov 21, 2012 at 17:05
4
\$\begingroup\$

Besides the decoupling caps mentioned in another answer, do note that the regulator has to dissipate:

\$P = U × I = (19V - 5V) × 0.75A = \boldsymbol{18W}\$

A standard TO-220 7805 without a big heatsink will go into thermal shutdown.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Should check the datasheet for max. power dissipation, but I have to run. \$\endgroup\$
    – jippie
    Commented Nov 21, 2012 at 15:35
  • 3
    \$\begingroup\$ Yes. The poster does not want to be running a substantial fraction of an amp at 5v stepping down from 19v with a linear regulator. This is a job for a switching regulator. \$\endgroup\$
    – Chris Stratton
    Commented Nov 21, 2012 at 15:55
  • 1
    \$\begingroup\$ I think this was the issue. Trying this with a 9V supply instead of 19V does not produce the issue. Thanks a lot for your information; I did some research into switching regulators and now see why they are useful for this situation. \$\endgroup\$
    – cemulate
    Commented Nov 22, 2012 at 5:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.