I need to use IR2302 or IR2104 gate driver for my synchro buck application, but I could not find a good explanation about the capacitor between pins VB and VS, I guess it is called bootstrap capacitor but I could not understand what is the purpose and what should be the value or even a type for this capacitor. Is a regular film capacitor will work or I need an electrolytic capacitor or Should the capacitor be a polarized or not? I couldnt find any information about this in the datasheets of both.
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1\$\begingroup\$ First hit on google: ti.com/lit/an/slua887a/slua887a.pdf. Plastic would be my first choice. \$\endgroup\$winny– winny2025-04-23 07:47:40 +00:00Commented Apr 23 at 7:47
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\$\begingroup\$ @winny Why wouldn't you entrust the next best MLCC in your BoM with this role, if I may ask? \$\endgroup\$tobalt– tobalt2025-04-23 08:08:20 +00:00Commented Apr 23 at 8:08
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\$\begingroup\$ @tobalt I would, but I rarely have large value MLCCs in my BOMs with sufficient voltage rating. \$\endgroup\$winny– winny2025-04-23 08:27:26 +00:00Commented Apr 23 at 8:27
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\$\begingroup\$ Why do you prefer MLCC instead of SLC? \$\endgroup\$Hubert Jo.– Hubert Jo.2025-04-23 08:49:38 +00:00Commented Apr 23 at 8:49
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2 Answers
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The purpose is to be able to fully turn ON the top N-channel FET. You properly know that an N-channel FET needs to have a positive Vgs potential. So imagine without the bootstrap capacitor, what will happen. The driver is supplied a voltage don't need to be very larger, enough to turn ON/OFF the external MOSFETs, so lets say Vcc=5 V.
When the top FET is supposed to conduct, HO goes high with 5 V at the gate. It will at first turn ON the top FET, but as it turns ON the voltage at the source will also try to increase and lets say the switching voltage (voltage at the drain) is larger than Vcc=5 V, some 12 V. You can now see that it will be difficult to have a Vgs that is going to be higher than 5V and it will properly find an intermediate level with high loss in the FET and this is not good.
So what happens if we apply the bootstrap capacitor.
Source: Datasheet of IR2302
Consider a load is connected and as soon as the bottom FET conducts, the bootstrap capacitor is charged through the diode, to Vcc-Vd through the load and the bottom FET. But now when the top FET is conducts, the voltage at the source of the top FET will again increase, but this time the capacitor which is charged up, will make the gate voltage rise up, so effectively you will have the capacitor voltage as Vgs.
As an example with a switching voltage of 12V. That would give as the N-FET source 12V. But the gate voltage will now be 12V + V_capacitor or 12V + 5V. You now see you have effectively a Vgs of Vcc=5V or whatever you chose for Vcc.
Green path: Bootstrap charge
Orange path: Bootstrap discharge
A standard choice is 100 nF, the size is also dependent on the frequency the capacitor has to charge.
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\$\begingroup\$ Also, is this 100 nF capacitor independent of the circuit's power rating? I'm working on a synchronous buck converter rated at about 50V and 10A. \$\endgroup\$Hubert Jo.– Hubert Jo.2025-04-23 21:00:43 +00:00Commented Apr 23 at 21:00
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\$\begingroup\$ @HubertJo. The capacitor is charged up to Vcc. Supply voltage of the driver. \$\endgroup\$Tyassin– Tyassin2025-04-24 09:14:53 +00:00Commented Apr 24 at 9:14
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\$\begingroup\$ I dont think its up to Vcc it should be more than the middle voltage to provide bootstrap \$\endgroup\$Hubert Jo.– Hubert Jo.2025-04-28 07:54:22 +00:00Commented Apr 28 at 7:54
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\$\begingroup\$ @HubertJo. I am not sure what you mean. The bootstrap capacitor can only be charge to the voltage you supply to the driver IC: Vcc \$\endgroup\$Tyassin– Tyassin2025-04-28 08:41:35 +00:00Commented Apr 28 at 8:41
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\$\begingroup\$ doesnt the gate driver use the voltage of the bootstrap capacitor to increase the voltage for Vgs above the output voltage? Like Vgs should be 5V but since s is at 30V it uses this capacitor voltage to provide 35 V to the gate. Isnt it the process ? \$\endgroup\$Hubert Jo.– Hubert Jo.2025-04-28 12:12:46 +00:00Commented Apr 28 at 12:12
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This capacitor presents the power supply for the high-side n-type MOSFET's gate driver. It will be roughly charged to the same voltage you use to drive the low-side FET's gate, e.g. 10-15 VDC typically. Capacitance wise, it should be substantially larger than the input capacitance of the high-side n-type MOSFET.
A very usual choice is a 100 nF MLCC with 25-50 VDC rating. But you can use whatever you already have in the BoM.
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\$\begingroup\$ Why do you prefer MLCC instead of SLC? \$\endgroup\$Hubert Jo.– Hubert Jo.2025-04-23 08:49:23 +00:00Commented Apr 23 at 8:49
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1\$\begingroup\$ @HubertJo. What do you mean by SLC? Single layer capacitor? I think single layer caps are only used for some niche applications in the pF range. Almost all capacitors are multilayer. MLCC has become a colloquial synonym for "SMD ceramic capacitor", at least I tend to use it as such \$\endgroup\$tobalt– tobalt2025-04-23 09:17:28 +00:00Commented Apr 23 at 9:17
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\$\begingroup\$ Also, is this 100 nF capacitor independent of the circuit's power rating? I'm working on a synchronous buck converter rated at about 50V and 10A. \$\endgroup\$Hubert Jo.– Hubert Jo.2025-04-23 21:00:36 +00:00Commented Apr 23 at 21:00
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\$\begingroup\$ @HubertJo. it doesn't depend on the power rating directly, but on the FET input capacitance as I wrote. Still 100 nF is plenty. \$\endgroup\$tobalt– tobalt2025-04-24 03:18:17 +00:00Commented Apr 24 at 3:18