Help me understand this input power supply circuitry

[alexv1n]

Hi all, I'm new here, so please take it easy on me

I'm trying to repair an Android TV box. It has no power. And I'm a bit perplexed by its input power circuitry. An external power brick supplies +8V through a barrel power connector. There are two input decoupling capacitors as expected. Then there is some circuitry that I'm having hard time understanding its purpose. I tried to recreate the schematic (see attached).

So, there is a voltage reference U1 (TL431). Its cathode is connected to the base of an NPN BJT transistor Q1 (2F) with collector output used as a gate of the MOSFET Q2. I don't understand two things: first, the purpose of U1 and its specific application. Normally, if TL431 is used as a voltage reference, the R1/R2 divider needs to be connected to its cathode (pin 2) - typical shunt regulator application. However, here R1 is connected to VIN. The second thing is, while the voltage on U1's cathode is +1.86V, how could the voltage on the Q1's base be +7.3V? This results in Q1 being open and its collector voltage is almost VIN, which keeps Q2 closed and there is no voltage on VOUT.

If I just disconnect the resistor R4, Q1 shuts off which opens up Q2 and the device powers on. So, where does the voltage on the base of Q1 come from? And in general, what is this circuit trying to do? It looks like some sort of a protection circuit, but I fail to understand its exact purpose. Any help?

P.S. I thought that there might be a leak current between Q1's emitter and base, so I replaced the transistor. This didn't help. I also replaced U1 to no avail.

 

[barry]

According to the TL431 data sheet, the voltage on the cathode should be about 5.8 volts by my calculation.

I don’t see why you have a problem with the cathode voltage being 1.8V and the base being 7.3V. Obviously theres current flowing through the resistor. Which is why Q1 is on and Q2 is off.

To be honest, I don’t see how Q2 would ever turn on. Maybe you’ve drawn it wrong?

 

[alexv1n]

According to the TL431 data sheet, the voltage on the cathode should be about 5.8 volts by my calculation.

I don’t see why you have a problem with the cathode voltage being 1.8V and the base being 7.3V. Obviously theres current flowing through the resistor. Which is why Q1 is on and Q2 is off.

To be honest, I don’t see how Q2 would ever turn on. Maybe you’ve drawn it wrong?

How did you come up with the conclusion of 5.8V expected on the cathode? If my understanding is correct, TL437 tries to maintain Vref of about 2.5V on its REF pin (2) by opening up or closing its anode-cathode junction. If the voltage divider R1/R2 is connected between the cathode and ground, this would vary the cathode voltage until Vref reaches the desired 2.5V. Hence the voltage on cathode should be 2.5*(1 + R1/R2) and in my case that would have been 5.8V. But that's only in the case of the top of R1 being connected to the cathode, below R3. However, R1 is connected to VIN. Which is why I'm confused about what this is trying to achieve.

I'm still wondering what is this circuit is supposed to do? And I'm fairly positive that I drew it correctly. I spent too much time tracing and retracing it, and measuring continuity between different nodes to confirm that it is wired like that. Obviously, there is always a chance of a mistake somewhere, but so far I think I did it correctly.

 

[alexv1n]

After looking at it some more, I think this might be some sort of overvoltage protection circuit. TL431 opens up or closes down depending on VREF voltage. If it is greater than 2.5V, then the diode breaks down and current start flowing from cathode to anode (towards the ground). If the voltage is lower, the junction closes. Since the R1/R2 voltage divider is not connected to the cathode, VREF does not change with the diode opening up or closing and as a result it will be either fully open or fully closed (if VREF is greater than 2.5V, U1 will essentially short to the ground and if it is lower, then U1 will be effectively an open circuit). This in turn connects the base of Q1 to the ground or VCC (through the resistors R3 and R4). If the input voltage is too high, U1 shorts to the ground, resulting in Q1 opening and Q2 closing, cutting the power to the rest of the circuit. Does this assessment look plausible?

If this is the case, the cutoff voltage seems to be around 5.8V. And as such, the power supply of 8V that the owner brought with the device probably does not belong to it and the device needs likely a 5V PSU instead.

 

[alexv1n]

I just confirmed that my theory is exactly right. As soon as input voltage crosses 5.8V threshold, Q2 cuts off the power to the rest of the circuit! Problem solved. Thanks for listening

 

[KlausST]

Hi,

the circuits function is:
* Output OFF for input voltage > 5.8V
* Output ON for input voltage < 5.8V

==> So in my eyes the circuit is some kind of overvoltage protection for a 5V supply. (Maybe there should never be 8V at the input)

In so far everything is OK: (The circuit works as expected)
* R1/R2 voltage
* U1 voltages
* R4 voltages
* Q1 emitter, Q1 base voltages
* Q1 collector voltage
* Q2 gate
* Q2 is OFF
* Output is OFF

--> try to connect 5V at the input ,, then the output should be 5V as well.

Klaus

ADDED: Interesting, when I started to write my post .. there were no post#4 and #5 (although it was clearly after their timestamp).
I guess they were in the moderation queue ...

 

[FvM]

Yes, two posts approved while you are writing.

I guess, you'll find a +5V supply input specification on the device if you look sharp.

 

[alexv1n]

Thanks, Klaus, for confirming. I'm glad I was able to figure out this myself but also for your help confirming my findings.

 

[barry]

How did you come up with the conclusion of 5.8V expected on the cathode? If my understanding is correct, TL437 tries to maintain Vref of about 2.5V on its REF pin (2) by opening up or closing its anode-cathode junction. If the voltage divider R1/R2 is connected between the cathode and ground, this would vary the cathode voltage until Vref reaches the desired 2.5V. Hence the voltage on cathode should be 2.5*(1 + R1/R2) and in my case that would have been 5.8V. But that's only in the case of the top of R1 being connected to the cathode, below R3. However, R1 is connected to VIN. Which is why I'm confused about what this is trying to achieve.

I'm still wondering what is this circuit is supposed to do? And I'm fairly positive that I drew it correctly. I spent too much time tracing and retracing it, and measuring continuity between different nodes to confirm that it is wired like that. Obviously, there is always a chance of a mistake somewhere, but so far I think I did it correctly.

You’re right about the cathode voltage of the TL431, i missed that. Still, you presented the circuit as having 8V as VIN which means Q2 would always be off.
I agree with the others, it’s an OVP circuit.