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While looking at a radio unit I was trying to repair, I found a 5V line at 1.2V. I started to reverse-engineer it, but was later able to obtain the schematic:

PS schematic

The right side is the input, and according to the service notes, it should show around 12V in power on state. For me it shows about 15V, which I'm assuming is because the radio won't power up (because of the lack of control 5V supply for the main chip.) I'm observing most of it dropping to around 1.5V after R174. The transistors are the usual suspects.

Instead of blindly ripping things out and replacing them, though, I'd like to understand how this thing was even supposed to operate. The three diodes in line with the backup battery seem to be disconnecting the battery when power is present (which would again, nominally be around 12V) and dropping the battery voltage to 5-and-half-ish V, which after Q110 should be right on point for the output. The circuit doesn't work when powered from the battery either.

I'm guessing that R169+R170/R171 form a voltage divider as a reference of sorts. That would make Q112 pass current from base of Q111 through R173 down, bringing the voltage there lower. This in turn would make Q111 block, lowering the Q110 EB current and thus perceived voltage at the output. The dropping voltage at the output would then feed back to Q112, closing it up, raising Q111 base voltage, and thus allowing more current through both Q110 and Q111. The final value at the output seems to depend on a very careful balance of that voltage divide and the transistors used.

Does that analysis make sense, or am I missing something important here?

Edit 1: As recommended in the comments, I've created a simulation. It seems to indeed stabilize at 5V:

simulation schematic

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    \$\begingroup\$ Why don't you simulate the circuit? \$\endgroup\$
    – Andy aka
    Jan 23, 2023 at 12:06
  • \$\begingroup\$ I'd be measuring V_BE on all three transistors, if any of them are ouside 0.6 to 0.7V that would be a red flag. \$\endgroup\$ Jan 23, 2023 at 12:10
  • \$\begingroup\$ @Andyaka I'm not very experienced in building simulations that work. Everything I've tried so far exhibited weird properties that didn't show practical effects i was looking at in real circuits. \$\endgroup\$ Jan 23, 2023 at 12:14
  • \$\begingroup\$ May I suggest that you simulate this circuit and, if it indicates weird properties, you post the schematic and results for analysis by the good folk on SE. \$\endgroup\$
    – Andy aka
    Jan 23, 2023 at 12:16
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    \$\begingroup\$ If you want to learn more about some function blocks in this design: Q112 is a Vbe multiplier a.k.a. rubber diode. Q110+Q111 are configured similarly to a Sziklai pair, but not exactly since the Q111’s emitter goes to ground and not to 5.2V. Overall, I’m a bit perplexed why they didn’t just use a 78L05. \$\endgroup\$ Jan 23, 2023 at 12:50

1 Answer 1

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Normally we expect to see the inputs coming from left-side and outputs going to right-side, so it looked a bit confusing at first.

Anyway. This is a series-pass linear regulator with BJT, Q110, as series element.

Your analysis is roughly correct. Some minor details need to be reviewed, though.


Here's a more detailed one for those who are interested:

Power Up: Q111's base current will come from the input through R174, R173 and D106. R174-C164 pair brings a crude soft start i.e. C164 will charge up through R174 and as it charges up the VBE of Q111 will rise, so will the base current, and therefore collector current. This will make Q110 turn on relatively slower (like a few milliseconds). As Q110 turns on more (by pulling more current from its base via Q111) the voltage across the output (right side) rises. The output voltage is divided and fed to the base of Q112. So the VBE of Q112 rises up. Once it approaches to somewhere around 0.5V the base will start to draw some current. Since the VBE of Q112 continues to rise the base draws more current.

Steady State: As the VBE of Q112 approaches to 0.6~0.7V the base will draw more current. So the collector current rises which means some of Q111's base current will be stolen. This will decrease the collector current of Q111 (and therefore base current of Q110), so Q110 starts to turn off (i.e. it's resistance increases) so the output voltage starts to drop.

As the output voltage drops the base current of Q112 drops, so it steals less current from Q111's base, therefore its collector. This results in higher base current for Q110, hence lower resistance. So the output starts to increase and the cycle starts afresh.

Apparently, the regulation can be obtained by keeping the VBE of Q112 at somewhere around 0.55~0.6V:

$$ \mathrm{V_O = V_{BE-111} \ \Big(1 + \frac{R169 \ || \ R170}{R171}\Big) \approx 0.6 \ (1+168.75/22)=5.2V} $$

So the output voltage heavily depends on the VBE of Q112 which may vary in a broad range.

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  • \$\begingroup\$ I agree that the orientation is somewhat weird, but this is the original diagram from the service manual, which i figured is a better resource than my own reverse-engineered version. \$\endgroup\$ Jan 23, 2023 at 12:40

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