Help !!!! Query on PNP voltage regulator on a HK subwoofer

[rfengg]

Hi there,

Was trouble shooting a HK subwoofer and found a broken MMBT 4403 BJT which caused the subwoofer to not produce any sound......changed the transistor (B-E junction was open) and she sings again.
I had access to the subwoofer service manual but cannot get my head around the implementation the simple circuit which I have attached.

-7VSB is fed to the collector of the MMBT 4403 PNP transistor and between the base and collector there is a 1K resistor. There is a 7.5V Zener to ground and as the voltage is only -7V, the zener is cutoff.
In this scenario , how does -7VSB appear at the emitter? What exactly is the purpose of this implementation?

thanks

[Chris56000]

That circuit is a negative mirror–image of the common NPN emitter follower regulator used with positive supplies!

Even if the supply voltage is too low to enable the zener diode to conduct, it will still produce an output, because the base voltage of the transistor will only be about 0.1 to 0.2 V above the collector, relative to positive earth, and the emitter/base junction of the transistor will be forward biassed, because the cathode of the emitter/base diode goes to a point more NEGATIVE than its emitter, so you will get the forward emitter voltage drop of a PNP transistor, giving an emitter voltage of about –6.4 V.

Think of it as you would with the usual NPN emitter follower stabiliser, but with positive voltages replaced by negative ones!

Chris Williams

[wasedadoc]

Hi there,

Was trouble shooting a HK subwoofer and found a broken MMBT 4403 BJT which caused the subwoofer to not produce any sound......changed the transistor (B-E junction was open) and she sings again.
I had access to the subwoofer service manual but cannot get my head around the implementation the simple circuit which I have attached.

-7VSB is fed to the collector of the MMBT 4403 PNP transistor and between the base and collector there is a 1K resistor. There is a 7.5V Zener to ground and as the voltage is only -7V, the zener is cutoff.
In this scenario , how does -7VSB appear at the emitter? What exactly is the purpose of this implementation?

thanks

(Attachment Link)

Zener diode followed by emitter follower.  What don't you understand?

The input on the left side is labelled -7 Volts.  That does not necessarily mean it is -7 Volts.  It means that it is the raw input which will be used to provide the -7 Volts on the output side.

[Audiorepair]

Either the -7vSB on the left side is a typo or  is misleading at best.

[rfengg]

Thanks Chris......I tried this on the breadboard using a PNP BJT after ensuring that the -7VSB was indeed only -7V by measurements taken on the subwoofer.
So the zener is indeed cutoff completely but I failed to recognise that the emitter is pulled to a higher voltage thru the load connected to ground which indeed forward biases the BE junction.

Thanks for your inputs