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In the figure below I have two power supplies (+ VS, -VS) of 100V to power a PA107DP amplifier.

The 2 power supplies must be supplied by an external voltage of 24V (the connector of the external power supply is just above the 100V power supply (R24-100B)).

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At the connector output of the external power supply there is a Schottky diode (D5.)

  • What is the role of the Schottky diode in this case?
  • How do I choose the Schottky diode? This diode is 50V / 10A. Why was a diode with these characteristics chosen?
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  • \$\begingroup\$ One common problem: there's a MCU somewhere and it gets supplied through the programming interface when programmed. Such designs are often made so that the prog interface supplies go straight to the 3V3 plane. If a voltage regulator has no input power and then receives this 3V3 on its output pin, they usually say 'poof'. The schottky will even out the input and output voltages in such cases, preventing damage to the regulator. \$\endgroup\$ – Lundin Jan 20 at 14:20
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What is the role of the Schottky diode in this case?

If you connected the incoming supply to J5 the wrong way round, the diode (D5) would block that voltage and thus prevent U3 and U4 from burning or breaking.

How do I choose the Schottky diode? This diode is 50V / 10A. Why was a diode with these characteristics chosen?

It has to be able to handle the full current that flows and, given that this may be several amps, a 10 amp rating may be appropriate for reliability of the device. It also has to have a reverse voltage rating greater than 24 volts and, for Schottky type diodes, a 50 volts rating is appropriate (given that they can be quite leaky).

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The Schottky diode is there to protect the circuit in case some one accidentally connects the power to the socket backwards (wrong polarity.)

The designer chose a Schottky diode because the voltage drop across a Schottky diode is lower than for most other diodes. A Schottky diode has a voltage drop of around 0.3 volts when conducting. Plain silicon diodes will have a voltage drop of 0.7 volts or more.

The current rating of the diode was chosen from the current requirements of the circuit. The diode must be able to pass the required current without burning out. The PA107DP can pass upto +-5A amperes peak to peak. 10A is too small if you actually need high current from the amplifier - more so because you are using boost converters to get 100V from 24V. 100V at 2 amperes would be over 8 amperes at 24V.

The voltage rating was probably not really considered. It needed to be higher than the expected input voltage, so I expect the designer looked at the reverse voltage on the 10A diode, saw that it was high enough, and went on to other things.

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The Schottky diode placed just after the incoming power connector J5 is there to protect the circuit from reverse polarity, ie applying the positive and negative the wrong way around.

If the incoming power to the anode of the diode is positive, the diode is forward biased and will allow current to flow. If this is reversed by mistake, and the incoming power to anode of the diode is negative, the diode is reverse biased and will block the flow of current, thus protecting the rest of your circuit.

A Schottky diode in particular has a lower forward voltage than other types of diode - typically a Schottky Vf is 0.4V, compared to 0.7V of a silicon diode.

The diode is chosen to allow enough current to flow (including peak to peak) with some safety margin. The 10A rating should be more than the total current consumption of the rest of the circuit.

The same is true of the voltage - if the power is plugged in backwards, the Schottky needs to be rated to block the reverse voltage.

Another factor to consider is the diode's power dissipation:

PD (Watts) = Forward Voltage across the diode * Forward Current flowing through the diode. This might be too high, depending on the application, and an alternative would be to use a P-channel MOSFET instead. With the P-FET, the Power Dissipation can be calculated using (I^2)R from the RDS[on] resistance of the FET and the forward current, which is normally drastically less than that of a diode, Schottky or otherwise.

There are other disadvantages of course, such as cost and more components, but this is for the designer to consider. I certainly find myself using P-FETs almost exclusivley for this purpose.

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