what is this transistor circuit?

Thread Starter

onlyvinod56

Joined Oct 14, 2008
363
Hello,
i found a circuit (soldered PCB) in my college. I have drawn the circuit and attached.

The circuit consists of a 230/12v T/F, rectifier, 7812, 7809, 7806 and 7805.
all the outputs of regulators are connected to a rotary dialing switch. and a single output is taken out. By rotating the switch, we can select the desired voltage.

That output connection is given to a transistor circuit. what is that circuit meant for? Is it a overload protection circuit?

thankyou
 

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cravenhaven

Joined Nov 17, 2011
34
I'm thinking it is an overvoltage protection circuit, but I cant find what the SK100E device is?.
When the output voltage rises above the input, it turns on the BC548 which then presumably turns off the SK100E?.
 

thatoneguy

Joined Feb 19, 2009
6,359
Can't find a datasheet for an SK100E

I'd guess some sort of over-voltage cutoff circuit in the event a regulator fails. Assuming emitter is at bottom of center transistor, and it is NPN.

SK Reverse Lookup Shows it is a universal replacement for 2N396 or 2N317A PNP transistor

Don't see many SK/ECG/NTE parts around anymore, they are "universal replacements" for the most common transistors. I used a few 20 years ago, but have stuck with same part since.
 
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debjit625

Joined Apr 17, 2010
790
I don’t think its a over voltage protection or cut off circuit, rather its a over load protection circuit.... as far I can see the circuit it should work like this....

The 230VAC is stepped down into 12VAC and then converted into DC with using rectifier, then for linear voltage regulation the linear voltage regulators are used 78xx.
The desire voltage is fed into the circuit, now in the circuit the first transistor BC548 is biased properly so that the base of PNP transistor SK100 is at near supply voltage, you might have noticed that SK100 is in voltage follower or common collector configuration i.e... it have a voltage gain of almost 1 and it have a current gain of \(\beta + 1\) ,these circuit are typically used in voltage regulators.

Now the interesting part is a buzzer and a led is connected in series from the emitter of the SK100 to the base of BC548, at normal load conditions the circuit will work normally but if you increase the load i.e.. decreasing the load resistance at output ,their will be a minimal voltage drop across the load or if the load is too low for example 0 ohms (short circuit) then the emitter of SK100 will be at 0V ,as a result current will start flowing from negative side of the power supply through the load to red LED and buzzer to 1.5K biasing resistor of BC548 ,the buzzer will beep and red LED will lit up indicating the output is connected to a very low load, as this happens their will be also a significant voltage drop across the 1.5K biasing resistor of BC548 and as a result the biasing of BC548 will be affected so the biasing current at base of SK100 will be also effected as a result SK100 will not conduct properly.

In simple words, if their is any low load condition at the output the circuit will decrease its current supply and the red LED will glow and the buzzed will beep. You may use a low load to verify this.

Good Luck
 

Adjuster

Joined Dec 26, 2010
2,148
I would say that whatever the detail of how this circuit was intended to work, its sensitivity will depend on transistor Ic vs Vbe characteristics, which vary with temperature and are not identical for different devices.

If anything more than a very rough test is required, a set-up using something like a level comparator with a zener diode or a band-gap device would be better.
 
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