# How bootstrap capacitor voltage is added to source voltage?

Circuit diagram below is a high side n-channel mosfet with bootstrap(while C1 is discharging)

It says that after charging of bootstrap capacitor:

The upper optocoupler opt1 gets ON for 1s and lower optocoupler Opt2 remains OFF for 1s. The capacitor C1 now tries to maintain the 12V across it and this raises the source voltage to 12V. This makes the diode D1 reverse biased as its cathode voltage is now 24V for maintaining the 12V across the capacitor. The capacitor C1 now starts discharging through upper optocoupler Opt1 and the gate of the MOSFET Q1 develops 24V.

BUT I cannot understand how the 12 volts from the source is added to the 12 volts of the charged capacitor. I mean, at the time of discharging, the capacitor is not in series with the 12 volts source. (it seems to me that the capacitor and source is in parallel to gate of mosfet)

I think the current does added up but not the voltage. How so the source voltage is added to capacitor voltage to become 24V?

• The charged ("big") capacitor behaves just like a voltage source. If VC =12V and at the beginning when MOSFET turns-on the source voltage for example Vs = 1V, thus the voltage at D1 cathode is 12V +1V = 13V so now the source voltage is 13V - Vgs = 10V and the D1 cathode voltage is 10V +12V = 22V
– G36
May 26, 2020 at 12:27
• Look at the example here electronics.stackexchange.com/questions/338838/… and here electronics.stackexchange.com/questions/111831/…
– G36
May 26, 2020 at 12:30
• Where does 10V come from? May 26, 2020 at 12:47
• Which opto is OPT1? If C1 was a battery, would you understand the circuit? May 26, 2020 at 12:49
• It's the upper one May 26, 2020 at 13:39

The optocouplers are just your turn-on and turn-off switches.

You require a bootstrap capacitor because you are trying to drive an N-channel Mosfet on the high side, and therefore you will need a voltage between Gate and Source of the mosfet which is greater than its threshold value.

When the mosfet is on you will be sending 12V to the load and therefore the node between the source and the load is at ~12V so to keep the mosfet on you need a voltage above 12V.

Notice how the capacitor (negative terminal) is at the node between the load and the source of the mosfet.

1. When the mosfet is off, lower optocoupler on, the capacitor will charge to 12V approx through the load and partially through the optocoupler.
2. When you turn on the mosfet, upper optocoupler on.

a. the capacitor will provide a voltage to the gate with respect to the source above the threshold value.

b. The mosfet will turn on and the node at the source of the mosfet will climb to ~12 volts

c. The voltage between the gate and the source of the mosfet will not change because it is provided by the capacitor, but if you were to observe the voltage referenced to gnd, the voltage will climb whatever the voltage drop on the load provides as "offset".If you measure the gate voltage referenced to ground you are measuring two things : Vgs + Vload. For the mosfet to stay on, it only requires that Vgs be above its threshold value, the capacitor provides this "bootstaped"/"floating" voltage source.

d. So with respect to gnd, you will have initially a value above 12V, idealy 12V + 12V, which will discharge slowly through the resistances R3 and R4 and the reverse leakage current of the diode. If switching is fast, you will not even notice any discharge.

3. You turn off the mosfet and the mosfet source node voltage goes back down to Gnd and the bootstrap capacitor is charged again whatever charge it lost.

Cycle repeats.

• I'm not sure if I understood it right: The 12 volts from capacitor will go all to the Vgs while the 12 volts from the source will all go to the load resistance, so all in all it is like there is a voltage of 24V at the gate of MOSFET. May 26, 2020 at 13:41
• The key is where you reference the voltage to. Mosfet Vgs (voltage difference between the gate and the source of the mosfet) does not go above 12V. Voltage at the gate of the mosfet referenced to Ground will be the voltage drop on your load + the voltage of the capacitor. Voltage at the source of the Mosfet referenced to ground will be ~0 when the mosfet is off and ~12V when the mosfet is on. May 26, 2020 at 17:29

This is a little bit of a simplification. When the switch is open (ignoring your load resistor) the voltage from drain to source on Q1 is 12V, the voltage at the source is more or less 0. Now say you close that opto coupler. Q1 requires very little current to turn on (only enough to charge gate capacitance) so that cap is going to stay pretty full, AKA it has 12V on it still. When the switch is closed though, that source node is now at 12Vish. Since the cap is still charged, and there is 12V at it's "low" side, the cap is now sitting at 24V wrt ground.

• I'm not sure if I understood it right: The 12 volts from capacitor will go all to the Vgs while the 12 volts from the source will all go to the load resistance, so all in all it is like there is a voltage of 24V at the gate of MOSFET. May 26, 2020 at 13:42
• Not quite. Redraw your circuit without the FET. You see that the cap will fill up to 12V, it is now charged. When you turn on that FET, the 12V from your source is now at the source of the FET (right at the load resistor). Since the cap had nowhere to discharge, it is still at 12V. The cap voltage doesn't change! The only thing that changes is you are now "adding" the 12V from the source again. I like @Andy aka's answer though. Think of the cap as a battery and it will make more sense. May 26, 2020 at 14:47