I'm fixing an electronic device (a Sony Discman D-50) but there is a part of the circuit that I don't fully understand.

This is the part of the circuit I'm talking about. I've highlighted the part I'm confused about.

enter image description here

There is a 9V DC input from the power supply that goes to a 5V voltage regulator.

There is a diode (D802, the diode component part is 1S2835) with the cathode pin connected to this 9V rail. The anode is connected to the positive pin of a 2.2uF 20V tantalum capacitor.

Now, there is a connection that goes from the junction of the diode and the capacitor to pin 8 of the main IC of this device, which is the Reset pin and expects an input voltage (3.8V).

The problem I'm having is that there is almost no voltage (~0.2V) on this pin so I started troubleshooting the issue, but my problem is that I don't really understand how this diode-capacitor circuit works - and I would like to know.

I've built this small circuit on a breadboard to try to understand it but I still get the same voltage as on the circuit board of this device.

You can find the service manual for this device here in case it is of any help. The schematic I show in the screenshot is on page 26.

What am I missing?


2 Answers 2


It looks like the diode is there to discharge the capacitor when the power is removed.

And the capacitor looks like a part of RC-delay circuit intended to reset the main IC when power is applied. Is there a resistor between Reset (pin 8 of the main IC) and its Vcc somewhere? Datasheet for MB88541 chip describes that reset pin as having an internal pullup resistor, see e.g. page 15 here.

  • 2
    \$\begingroup\$ There may be a pullup or a current source inside the IC. \$\endgroup\$
    – Dave Tweed
    Apr 9, 2021 at 10:47
  • \$\begingroup\$ There is no other connection to the Reset pin (not externally at least) I've edited my question adding a link to the service manual in case you want to take a look. Thank you. \$\endgroup\$
    – JV Lobo
    Apr 9, 2021 at 10:52
  • \$\begingroup\$ I think Dave Tweed and Bravale are right: the charging current likely comes from the IC itself. \$\endgroup\$
    – Igor G
    Apr 9, 2021 at 10:54
  • \$\begingroup\$ Page 17 of the service manual shows if the signals are input or output ones and according to that, Pin 8 of the main IC is an input signal. \$\endgroup\$
    – JV Lobo
    Apr 9, 2021 at 10:59
  • \$\begingroup\$ See the datasheet for MB88541 (that IC). Pin description section (page 15) says: "This pin is a hysteresis input with an internal pullup resistor". \$\endgroup\$
    – Igor G
    Apr 9, 2021 at 11:09

If it is a reset pin, probably some current will flow from chip, and load capacitor up to 3.8 V. When 9 V power voltage is removed, the capacitor will be discharged by the diode, and will be prepared for next start. Check if capacitor is short-circuit and is draining current to ground.

  • \$\begingroup\$ I don't understand. Do you mean that the current will flow out from the Reset pin? According to schematics, this pin will get current and not outputting it? \$\endgroup\$
    – JV Lobo
    Apr 9, 2021 at 10:53
  • \$\begingroup\$ It is usual in digital circuits that some inputs have internally a resistor or current path to +VCC. \$\endgroup\$
    – Bravale
    Apr 9, 2021 at 11:05
  • 1
    \$\begingroup\$ It may be then that the IC pull resistor is not working. I actually got the device to work. I had to replace some things like the voltage regulator, the DC-DC converter, and some capacitors. But I got it working by connecting a jumper wire from the 5V output of the voltage regulator to the reset pin. But I wanted to understand this last issue and do a proper fix on it. I will look into the resistor from VCC to the reset pin option you are proposing. Thank you! \$\endgroup\$
    – JV Lobo
    Apr 9, 2021 at 12:04
  • 1
    \$\begingroup\$ Thanks. I don't know what was the original issue of this device but it could be probably that. As it had some burnt components... the DC input is center negative which is not usual nowadays. I have a Schottky diode ready to install too :) \$\endgroup\$
    – JV Lobo
    Apr 9, 2021 at 12:22
  • 1
    \$\begingroup\$ Yikes, center-negative and no reverse polarity protection! No wonder there were burnt components.... cool that you were able to fix it! :-) \$\endgroup\$ Apr 9, 2021 at 21:45

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