# Help please with this simple transistor circuit

#### bryankerr

Joined Apr 1, 2015
15
Hello all and thanks for taking a look at my thread.

I was hoping you could help me understand this circuit. But how is this transistor doing that exactly? Can you help me understand the circuit both intuitively and mathematically? I have never seen a transistor circuit like this. Thank you!

Last edited:

#### blocco a spirale

Joined Jun 18, 2008
1,546
The problem is, the circuit has been drawn incorrectly and won't work. Neither transistor can function.

#### bryankerr

Joined Apr 1, 2015
15
I copied it correctly. The only thing I left out is that the incoming voltage is actually from an AC rectifier bridge.

#### blocco a spirale

Joined Jun 18, 2008
1,546
Ok, then the source you copied it from was wrong.

#### ErnieM

Joined Apr 24, 2011
8,126
An NPN transistor needs current to flow from base to emitter to get current to flow from collector to emitter.

Both devices as drawn have a short across base to emitter. Thus no current, no function.

#### blocco a spirale

Joined Jun 18, 2008
1,546
I'm wondering whether with an AC input it might actually flash the LED. On the positive cycles the capacitor will charge and the transistor will be off but when the polarity reverses on the negative cycle the transistor's base would be positive wrt its emitter and would turn on and flash the LED.

I assume there should also be a resistor to limit base current.

Last edited:

#### bryankerr

Joined Apr 1, 2015
15
But there is no negative side of the voltage since its been rectified? This circuit indeed works. I checked the output pulses and they are perfect square waves 2.5 volts @ 120 hertz.

#### Dodgydave

Joined Jun 22, 2012
9,923
think your drawing is a Zero Crossing Detector, using an opto coupler and transistor driving it, the led lights when the voltage is near to zero ,(prob 3v), but it doesn't look right...

Last edited:

#### bryankerr

Joined Apr 1, 2015
15
Dodgydave your exactly right about it being a Zero Crossing Detector. What doesn't look right about it?

#### Dodgydave

Joined Jun 22, 2012
9,923
the transistor driving the led, has its base lower than the emitter, unless its being used as a zener.

#### bryankerr

Joined Apr 1, 2015
15
If I take a multimeter and put the positive on the base, and negative on the emitter I get -.3 dc and switching on/off .42 ac

#### bryankerr

Joined Apr 1, 2015
15
oops sorry, the .42 ac is not switching on/off

#### bryankerr

Joined Apr 1, 2015
15
just checked the base/emitter on the oscope and it shows square waves turning the resistor on/off at 120 hertz

#### Dodgydave

Joined Jun 22, 2012
9,923
so your using a 60Hz supply

#### bryankerr

Joined Apr 1, 2015
15
yeah 120 volts 60hz dropped to 3.5ac right before a rectifier bridge, then that is the circuit.

#### Dodgydave

Joined Jun 22, 2012
9,923
the transistor must be being used as a Zener diode, being reversed biased, strange design but it works.

#### bryankerr

Joined Apr 1, 2015
15
There is 3.5 volt ac applied to a bridge rectifier, and if I am careful not to leave anything out, there is also a resistor in parrallel with that whole circuit above. but I assume it has no function in the opto coupler function

#### blocco a spirale

Joined Jun 18, 2008
1,546
I know what's happening here; The capacitor charges as the input voltage rises (the diode is forward biased, transistor and LED are off), when the voltage starts to fall, the lower end of the capacitor tracks the input voltage and drops below 0V (the diode is reverse biased) taking the transistor emitter with it. The transistor base current comes from the 0V rail as it is now at a higher potential wrt the emitter, the transistor turns on and flashes the LED.

The LED will flash on the falling edges of the incoming waveform.

Last edited:

#### bryankerr

Joined Apr 1, 2015
15
blocco, I think you're on to something, because the base has somewhat square waves on the oscope at 120Hz, and I believe the peaks are at the zero crossing of the input. But can you explain why the capacitor voltage drops below 0V?

#### blocco a spirale

Joined Jun 18, 2008
1,546
As the input voltage rises, the capacitor will charge to the peak value of the incoming supply (minus the diode drop), let's say 5V. As the voltage falls, the capacitor will maintain this potential across its terminals (it can't discharge because the diode is now reverse-biased). So, when, for example, the input falls to 4V, the top end of the capacitor will also be at 4V wrt 0V but there will still be 5V across the capacitor which means that the lower end will be at -1V wrt 0V. The most you will see on this circuit is around -0.65V equal to, and limited by, the transistor's Vbe .

Last edited: