4
\$\begingroup\$

circuit I was reverse engineering a power tool battery charger (21V SMPS) that derives its 5V rail from a simple shunt regulator, employing a TL431 programmable precision reference and a NPN transistor, as per my crude sketch (sorry I didn't have time to paint it). The programmable reference runs the gate of the NPN transistor, maintaining a stable 5 volts on the rail that powers the microcontroller, indicator LED. etc. (The reference is programmed to 5V by R1 and R2, but is actually running 0.6V higher, as it needs to overcome the additional Vbe of the transistor, so Vka is really 5.6V) The voltage measurements from the sketch were measured at the same time, in an idle, no load state. A SMPS 21V source feeds R4, dropping 0.577V, which per Ohm's law means about 0.0012A are flowing through R4. The transistor gate is biased by a 4.7kOhm resistor R3. The voltage drop on R3 also corresponds to Vcb, measuring 14.99V. Now I stop understanding things. With this voltage drop, R3 should be passing 0.0032A, but this is impossible since the "inrush limiting" resistor R4 is only passing 0.0012A! I also probed the points with my oscilloscope to make sure there wasn't some kind of weird wave action taking place that would mess up my multimeter reading but no, it showed a stable voltage. I understand that what is probably happening is some borderline open circuit condition, with the transistor just teetering on some kind of almost-off state 'commanded' by the reference, but this feels rather vague in my mind. Can somebody please explain what is happening, more to the point, what current can R3 be passing if its voltage drop measures 14.99V and Ohm's law would dictate it passing 3.2mA, but the circuit is only being fed a MAXIMUM of 1.2mA? Many thanks for any explanation! Cheers!

EDIT: as one commenter immediately mentioned, R4 I was reading as 470 Ohm was really 47 Ohm, so the currents all add up and all is good, as 12mA are coming into the circuit, and R3 is only shunting 3mA of those. Many thanks to the keenly eyed! I would still appreciate if anyone would be so kind as to describe "boundary" operation of the transistor, teetering on dropping 15.6V as maintained by the reference. Many thanks!

\$\endgroup\$
4
  • 3
    \$\begingroup\$ Please ask a new question for the new part about the "boundary" operation of the transistor. It's not fair to the users who already answered to have to amend their answers after you change your question. \$\endgroup\$ – Null Mar 24 at 20:34
  • 2
    \$\begingroup\$ Note this is not a shunt regulator, it is a series regulator (all load current passes through the transistor). \$\endgroup\$ – bobflux Mar 24 at 20:45
  • \$\begingroup\$ Thanks, I'm super new to the platform so was unaware about the etiquette. Thanks again for all the helpful advice! \$\endgroup\$ – parkside Mar 24 at 21:24
  • 2
    \$\begingroup\$ This is not a crude sketch, it is a very well done and detailed schematic. \$\endgroup\$ – Vladimir Cravero Mar 25 at 12:43
12
\$\begingroup\$

I am guessing that you are looking at an SMT resistor marked "470" and that it is actually 47 ohms.


As to operation-

The TL431 draws as much current as required (up to 100mA+) to reduce the voltage at the sense input to 2.495V nominally. As long as it gets > 1mA to use itself it will function properly (400uA typically). If we assume the circuit draws about 10mA the base current might be of the order of 100uA.

So the 4.7K resistor needs about 5.2V across it to function (so there is 1.1mA+ flowing). Add that to the 5.6V required at the base of the transistor.

So this regulator will work from about 11.5V up to well over 21V, taking into account the drop across the 47 ohm resistor.

As @bobflux rightly points out, this is a series regulator. The NPN transistor is the pass element.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Thanks mate spot on! It all fits now, always good to have someone double check lest one get lost down a rabbit hole... \$\endgroup\$ – parkside Mar 24 at 20:30
  • 3
    \$\begingroup\$ @parkside An assembly house made a mistake on a client's PCB many years ago and stuffed many resistor networks marked 510 on pcbs. They were supposed to be 510 ohms not 51 ohms. Hilarity ensued (not really). \$\endgroup\$ – Spehro Pefhany Mar 24 at 20:42
  • 3
    \$\begingroup\$ This reminds me of a time when an inspector on an SMT line rejected a batch of boards because they had 195R resistors fitted instead of 560R. I turned them around and gave them back to him :) \$\endgroup\$ – SiHa Mar 25 at 9:12
3
\$\begingroup\$

The beauty of R3 is it lowers the output impedance by negative feedback. When you choose Ic around 1mA, Vbe becomes around 600 mV so the rise in the precision Zener @ rated current @ 5600 mV +/-x% matches Vbe to result in 5.00 output while biasing the Zener at the same time. As Iz >> Ib the circuit is almost independent of hFE.

You Icc calculation must be low as it must be the sum of IR3 and Ic. As Spehro guessed correctly, Rc must be 47 Ohms which is a good thing because now you can drive ~50mA more than before than with 4k7 because you now have more NFB current.. But Pd max must be examined if you choose this.

Also

The same negative feedback ideal for linear H bias as well achieving exception THD with a compromise in Zin=Ri and Rfb / Rin about 1/2 of you achievable gain which increases with Ic.

Also C ought to be small enough to prevent large surge energy E= 1/CV^2 yet enough to reduce dV = Io dt / C for a small step current. As the NFB is very wide band and high feedback gain with 47 Ohms so it the output impedance is reduced significantly for any flyback +ve load . Rfb/hFE for positive dumps. And Zener Zo/hFE for negative step currents, unlike series LDO’s which only pullup. This is a low impedance Common Base with the NFB from a load regulation perspective. And emitter follower from a supply perspective.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks for the helpful reply, I wasn't aware of the answer/edit etiquette after my main question was sorted. \$\endgroup\$ – parkside Mar 24 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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