1
\$\begingroup\$

Typical application of the AL1663R IC I use is below.

My question is: when I was trying to touch the feedback resistors' (R5 - R6) output with a wire, I saw the LEDs at the output went off.

The wire I used doesn't even have a connection. It isn't grounded, so the circuit cannot be completed through the wire. How is it possible it makes the LEDs turn off?

img

\$\endgroup\$
3
  • \$\begingroup\$ Probably parasitic capacitance/inductance. Don't know too much about this circuit, but it's conceivable that the feedback signal could be very sensitive, especially if R5 and R6 are very large, or the wire is very long. \$\endgroup\$ Aug 31, 2022 at 18:36
  • 7
    \$\begingroup\$ That circuit is not isolated, it is live with mains voltage. Poking a live circuit randomly with conductive metal objects is not safe and an electric shock can be lethal. \$\endgroup\$
    – Justme
    Aug 31, 2022 at 18:58
  • \$\begingroup\$ wolyure - Hi, Your last edit removed the schematic, but that is an important part of the question. Also it has been used in the answers. It is not appropriate to remove it now. Therefore I have rolled-back (reversed) your last edit. You can make edits to your question to clarify & improve it, but not to remove important information. Thanks. \$\endgroup\$
    – SamGibson
    Sep 1, 2022 at 7:17

2 Answers 2

4
\$\begingroup\$

The FB (feedback) pin is a high impedance input. You don't say what value your resistor divider resistors are (R5 and R6) but they are likely relatively high value resistors as well.

Most likely you are injecting noise into the feedback pin, raising the level above the reference point and causing switching to stop or stop periodically. Your wire is acting as an antenna, coupling whatever it's picking up into your circuit.

Another possibility is that parasitic capacitance added to the FB node could cause the control loop to go unstable, which might cause oscillations that would prevent the output from reaching the correct voltage.

\$\endgroup\$
0
\$\begingroup\$

@JohnD is correct that the wire acts as an antenna which broadcasts as well as receives the nearby radiated PFM noise which has spectrum well into the AM band with PFM rates up 150 kHz.

I presume you wanted to measure the sensed voltage externally on a DSO. The better way would have been to use a twisted pair magnet wire with FB and GND and keep it short to permit connection to two null-balanced 10:1 probes in A-B Math mode since DSO ground is not 0V on the regulator which is a diode drop from Neutral. This attenuates the antenna effect with twisted pair at 1MHz. Connect both probes to the same signal to get a flat line then test in A-B mode. Often I would twist both probe cables together for even better common mode rejection.

Any feedback signal beyond the control range might be detected by the OVP circuit to shut down. Even though the Aux transformer will be a low voltage winding ratio and thus lower impedance at low frequencies, the antenna or crosstalk effects are common on this sensitive port with PFM pulses non-isolated.

The IC has many protections for the shutdown which are a good feature to have.

Internal Protections
 Under Voltage Lock Out (UVLO)
 Output Over Voltage Protection (OVP)
 Output Short Protection (OSP)
 Over Current Protection (OCP)
 Thermal Fold-back Protection (TFP)
 Over Temperature Protection (OTP)
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.