Start-up Aux Circuit in SMPS - Page 1
Forum section: electronics / beginners

author: sean0118
author_rank: Frequent Contributor
timestamp: August 04, 2014, 03:33:20 am
content: Hi everyone,I'm currently attempting to design a circuit to power theFAN7631on start-up. I'm currently planning on using the method outlined in the third post in thisthread. (I didn't want to post in that thread because it's in General Chat.)The general idea seems pretty straight forward, the capacitor is charged slowly through Rstartup. Once the voltage over the capacitor (Vcc) reaches the minimum threshold voltage for switching the FAN7631 will start conducting. Then the auxiliary winding of the transformer will start supplying current before the Vcc drops below the minimum voltage required for switching.Now, it seems the minimum starting voltage (HVcc) for switching the high-side MOSFET is 9.2V, while that of the low-side (LVcc) is 12.5V. Now assuming I'm looking at the correct datasheet values it looks like the high-side MOSFET will start switching first and Vcc will drop before the low-side starts switching.Assuming I am using the correct values would a solution be to add a resistor to the HVcc line so both HVcc and LVcc start switching at 12.5V? I have noticed in the datasheet they add a series resistor and diode to the HVcc line, is this why? Let me know if my questions are unclear or if you require more information.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 04, 2014, 05:03:24 am
content: So what's the problem..?Doubtful the high side will start switching first; when the bridge is initially off, no voltage charges the bootstrap capacitor, and the high side remains off. Once the low side turns on, it dumps some charge in there, hopefully charging it in less than a half cycle, so the high side can immediately start up and run.Tim

author: sean0118
author_rank: Frequent Contributor
timestamp: August 04, 2014, 11:55:30 am
content: Ah, right that makes it easier.But what's this so called 'bootstrap capacitor' you speak of? Is that the capacitor between the HVcc and CTR pins?I can sort of see that it would start to charge when the low side turns on. But what is its purpose? Also, what is the purpose of the series resistor and diode that feed it? Is it related to having HO float at HVcc + CTR?

author: David Hess
author_rank: Super Contributor
timestamp: August 04, 2014, 02:21:54 pm
content: Quote from: sean0118 on August 04, 2014, 11:55:30 amThat is the bootstrap capacitor. When the low side MOSFET is on which pulls CTR low, the high side bootstrap capacitor charges from the low voltage supply which goes to LVCC.QuoteYes. When the high side MOSFET turns on, CTR rises and HVcc is held above CTR by the capacitor to power the high side gate driver. This allows an N-channel MOSFET to be used for the hide side switch instead of a P-channel MOSFET.The diode and resistor charge the HVcc bootstrap capacitor. The resistor just limits the current and makes the switching softer.Pin LS is *only* the under-voltage lockout so a resistor divider is used to divide down the high voltage supply to the proper threshold. The regulator is powered through LVCC which is where the low voltage bootstrap supply should be applied but that is not shown in the application example.

author: sean0118
author_rank: Frequent Contributor
timestamp: August 30, 2014, 02:02:42 am
content: Sorry for the late reply, but I wanted to thank you because your posts did help me understand the driving and Vcc circuits much, much better. I'm going to add a terminal so I can supply Vcc externally, then if that works, I'll connect the trickle charge resistor and auxiliary winding.

author: salbayeng
author_rank: Frequent Contributor
timestamp: August 30, 2014, 05:06:40 am
content: Yeh the bootstrap circuit can take a while to get your head around.You can tack in some extra circuitry to precharge the bootstrap capacitor, (e.g. if you are doing something a bit unusual with a FAN73832, which has the same driver arrangement) , but you generally need to put a zener across the bootstrap capacitor and other diode steering in place.The resonant converter uses 50% duty cycle , so the bootstrap circuit will work flawlessy, (other applications with 95% duty and/or large deadtimes and/or low frequencies, can have bootstrap issues)