4
\$\begingroup\$

enter image description here

This is from the datasheet of LM2931.

What baffles me is how this actually works. The LM2931 drops the voltage to 5 V and I assume the 68ohm resistor is there to drop the voltage around 0.7 V to open the PNP.

Basically I can’t understand how the heck the PNP let’s through current at only 5 V if Vin was let’s say 12 V.

Granted, I also have trouble understanding PNPs, so it may be related to that as well. Thanks!

\$\endgroup\$

2 Answers 2

6
\$\begingroup\$

The voltage across R is also the voltage between the base and emitter of the transistor. If the output current is small enough, all of the output current passes through R and the transistor is off. The circuit behaves essentially as it would if the transistor were not even present. The voltage regulator would simply output its regulated voltage.

However if the output current rises to a sufficient level, the voltage across R becomes sufficient to start conduction in the transistor. This aids in providing output current.

If, for any reason, the output voltage should rise above the regulator's programed output voltage, the regulator will reduce its current, thus reducing the voltage across R, thus reducing the collector current of the transistor. So, adding the transistor does not prevent the regulator from doing its job regulating the output voltage. It only provides extra current when needed.

In a way, looking at the voltage regulator "backwards" it acts like a voltage detector, and adjusts the current on its input pin, according to the voltage on its output pin. It thus regulates the transistor to output the proper voltage.

Basically I can’t understand how the heck the PNP let’s through current at only 5 V if Vin was let’s say 12 V.

For "normal" operation of a PNP transistor, the emitter more positive than either the base or collector. For there to be collector current, the base should be a least 600 mV or so less positive than the emitter. The collector can be less positive than the emitter by anywhere from the saturation voltage, which could be as small as 200 mV, up to the breakdown voltage of the transistor. In this circuit, the emitter is always (about) 7 volts more positive than the collector, which, which puts the transistor in it's linear operating region. In linear operation, a transistor acts more like a variable resistance, than as a switch.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ To put some numbers on it, the LDO provides all output current up to about 10 mA. That particular transistor only has a current gain of about 20 (worst case), so for every additional 21 mA drawn by the load above 10 mA, 20 mA flows through the transistor collector and 1 more mA flows through the LDO and the base of the transistor. At that point, the voltage across (and current through) the resistor changes only slightly. \$\endgroup\$
    – Dave Tweed
    May 14 at 20:41
  • 1
    \$\begingroup\$ I think the questioner also has another hole in their understand based up this: "I can’t understand how the heck the PNP let’s through current at only 5 V if Vin was let’s say 12 V". They probably need a little more explanation about bipolar collector behaviors, in particular. \$\endgroup\$ May 14 at 21:37
  • \$\begingroup\$ Likely so. Can you point me to a good source for study? \$\endgroup\$ May 15 at 9:23
0
\$\begingroup\$

It may be easier to understand this circuit by first considering it without the transistor. In that scenario, all the current to the output must flow through the 68R resistor and the LM2931. If a circuit connected to the output is drawing 1 mA, then you will have 68mV dropped across the resistor and the rest of the remaining voltage between the input and output dropped across the LM2931. For example, with a 10V input, you will have (10 - 5 - 0.068) = 4.932 volts dropped across the LM2931.

If the circuit connected to the output draws 10 mA, then you will have 680mV dropped across the resistor and, again with a 10V input, 4.32 volts dropped across the LM2931.

If the circuit connected to the output tries to draw 100mA, then you will 6.8v dropped across the resistor, leaving only 3.2v for both the LM2931 and the output. At this point, you've overloaded the circuit.

If you repeat this with the transistor in place, what happens?

For a 1mA load, you have 0.068v dropped across the 68R resistor, and therefore 0.068v across the base-emitter junction on the transistor. The transistor is off and no current flows through it. The LM2931 is delivering all the current to the load.

For a 10mA load, you have 0.68v dropped across the 68R resistor, and therefore 0.68v across the base-emitter junction on the transistor. At this point, the transistor is starting to "turn on" (for a very simplistic transistor analysis). The LM2931 is delivering most of the current to the load, but a little bit is starting to flow through the transistor in parallel with the LM2931. Note that the LM2931 is controlling the output voltage.

As the load increases above 10mA, the LM2931 will deliver more current. The current delivered by the LM2931 will mostly flow through the 68R but a little bit will flow through the base of the transistor. As the voltage across the 68R resistor increases above 0.7v, the transistor will turn on. More current will flow through its base and more current will flow through its collector, bypassing the LM2931 and being delivered directly into the load.

At this point, the transistor is acting as a current multiplier. The more current the load tries to draw, the LM2931 will deliver a bit part of the load, increasing the current across the 68R resistor, and therefore increasing the base-emitter voltage on the transistor. This turns the transistor on "harder" and therefore the transistor also supplies more current to the load.

For any load that is significantly above the 10mA threshold that starts to turn the transistor on, the LM2931 will deliver approximately 1/(1+hfe) of the transistor and the transistor will deliver the rest of the current.

[To all transistor experts out there, the above is very deliberately simplified to (hopefully) make the general behaviour more understandable. I know that it's not in any way accurate.]

\$\endgroup\$

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.