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I designed a PCB with the following schematics.

schematic

simulate this circuit – Schematic created using CircuitLab

The circuit uses a MAX6369 external watchdog timer which has an open-drain output(WDO) which is pulled low at a fixed interval. When the output is pulled low, the reset pin will be pulled low as well and light up an LED through a PNP transistor. However, this circuit does not work well as it lights up the LED at the fixed interval but does not pull the reset pin low. After debugging I noticed that when the transistor is removed the circuit works fine. I concluded that I forgot to put a base resistor which causes the transistor to be saturated and draws too much current from the watchdog. This prevents the reset pin to be pulled low properly.

So my questions are, Is my conclusion right? Is there anyway to save this circuit without adding additional components and just replacing existing components such as replacing the transistor with a low base current transistor?

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  • \$\begingroup\$ Can you share a photo of the physical circuit? \$\endgroup\$ – The Photon Jan 18 at 5:37
  • \$\begingroup\$ looks like D1 is backward \$\endgroup\$ – jsotola Jan 18 at 5:40
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Yes, you can replace the transistor with a "pre-biased transistor" which has the resistor built in or you can replace it with a P-channel MOSFET. The diode is flipped and you'll have to pull it low for reset.

The pinouts, luckily, arecompatible, check the case, they are available in several similar packages, such as SOT-23 and SC-70.

enter image description here

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  • \$\begingroup\$ Thanks, I will see if I can source this part. As I commented in the previous answer, I don't think there is a problem with diode orientation. \$\endgroup\$ – Max Jan 18 at 6:08
  • \$\begingroup\$ For wired-or the diode is the wrong way. \$\endgroup\$ – Spehro Pefhany Jan 18 at 6:28
  • \$\begingroup\$ Diode direction depends on where the actual /RESET pin is connected, it could be okay, hard to tell from the snippet. \$\endgroup\$ – Spehro Pefhany Jan 18 at 7:00
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The obvious solution is to insert Rb in series with base = 20xRe = 1k+/- 50% and reverse D1 to enable active low reset.

The present circuit prevents the output going low due to Vbe attached to Vcc.

Rb can easily go to 5k and R2 go to 100 Ohms and drive the RED LED with 15mA instead of 27mA while still working at Vcc=-10% with 11 mA to the LED.

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  • \$\begingroup\$ The reset pin needs to be able to pass current to ground to reset. In this configuration current can flow from the reset pin to ground. I have tested this and it works and I have even tried flipping it and it stopped working. I can't add the Rb because I have already made the PCBs for this circuit and can't change it. \$\endgroup\$ – Max Jan 18 at 6:07
  • \$\begingroup\$ You failed to mention this In your question. This works. yours does not. \$\endgroup\$ – Sunnyskyguy EE75 Jan 18 at 6:58
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Your circuit is like this

schematic

simulate this circuit – Schematic created using CircuitLab

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