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I am designing a soft start circuit for my bench power supply.

The voltage input of the IC that control the mosfets in the circuit can not be under 4.3 V even when the output of the power supply is under 5 V.

I've been doing some research on the web and I found a circuit so I decide to simplify it for simulation in order to understand it better.

This is the circuit:

enter image description here

enter image description here

V1 refers to a 5 V voltage regulator and V2 refers to the output of the power supply that can be between 0.8-20 V.

What I can not understand for example is why when V2 is under 5 V, voltage output is 4.36 V and not 8.6 V (4.36 + 3.3 V). It's like depending what voltage has V2, the voltage output depends of V1 or V2.

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3 Answers 3

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Everyone is saying what it's not, but I'll throw my two cents in to explain what this circuit is:

The technique you're using here is called OR diodes.

The voltage at the load is not the combined total of supplies. It is the voltage of the higher voltage supply (minus any forward voltage drop). This technique is typically used when you have several power sources to a board.

For example, if you have a battery backup you want to fall back on if your main supply fails. If the main supply is, say, 30V. You could make design the backup supply for 25V. That way, when the main supply is working, the diode leading to it is reverse biased and the battery is effectively taken out of the circuit, but if the main supply fails, you can be sure that the power to the board will never go under 25V (again, minus the forward voltage of the diode, typically ~0.7V for napkin analysis purposes.

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  • \$\begingroup\$ Also very useful for running a normally battery-powered device while it's connected to the "charger" i.e. a PSU that powers both the load and the device's built in charger. \$\endgroup\$ Commented Feb 3, 2022 at 9:07
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The voltages do not add together. Instead, each diode forward biases when the difference across it is about 0.7V. The diode with the bigger voltage wins.

The output will be the higher of V1 or V2, less one diode drop. Considering your two cases:

  • V2 at 4V, V1 at 5V: only D2 conducts, setting the output to be 5V - (forward drop) = 4.36V. Winner is V1.

  • V2 at 20V, V1 still at 5V: only D1 conducts, so the voltage will be 20V - (forward drop) = 19.3V. Winner is V2

If you want less forward drop, consider a Schottky diode like a BAT54 or similar. Schottky diodes have a forward drop of only 0.3V.

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The output voltage is set by the power supply with higher voltage. It can be expressed as,

$$ V_L = max(V_1, V_2) - V_{f} $$

The 1N4148 is a normal silicon diode, which has 0.6-0.7V forward voltage drop. You can see the detailed forward voltage drop characteristics on 1N4148 datasheet.

For less power consumption on diodes, it's better to use schottky-barrier diods, which has lower forward voltage drop. For example, if you use 1N5819, the output voltage will be greater than 4.5V in your second schematics.

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