Start Up resistor  - Page 1
Forum section: electronics / projects

author: whatty
author_rank: Contributor
timestamp: March 20, 2023, 04:46:40 am
content: Hi i had designed a smps using uc3845 it works fine but after switch on of smps 10-15 seconds the start-up resistor starts to heat up why....?I am using a start-up resistor of 18k 2W and 100uf 25v capacitor for starting the IC upon starting the resistor heats up how to tackle it ... i understand there can be high power loss even i adjusted the resistor to 470k 2w but the ic didn't start-up .

author: srb1954
author_rank: Super Contributor
timestamp: March 20, 2023, 07:09:06 am
content: Quote from: whatty on March 20, 2023, 04:46:40 amWhat is your incoming supply voltage?Please post a full schematic so we can comment.

author: Psi
author_rank: Super Contributor
timestamp: March 20, 2023, 07:19:05 am
content: The boostrap resistor is a common failure mode in SMPS that use them, they tend to run hot and not be spec'ed large enough to handle the heat.

author: Zero999
author_rank: Super Contributor
timestamp: March 20, 2023, 08:48:54 am
content: Did you try 100k, as recommended on the data sheet?There are various ways to avoid high power dissipation.https://www.industrial-electronics.com/switching-power-supply_1-8.html

author: whatty
author_rank: Contributor
timestamp: March 20, 2023, 09:09:20 am
content: Yep used 100k the controller is not kick starting if i use 100K i am getting after bridge rectifier 330v dc. If i use 470k, 100k the ic is not kick starting but if i use 18k its starting but the the start-up resistor then gets heat up. The capacitor of 100uf is used along the vcc

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 20, 2023, 09:19:24 am
content: How many watts were you expecting it to dissipate? Surely not 2W. Show your calculation?Tim

author: whatty
author_rank: Contributor
timestamp: March 20, 2023, 10:10:56 am
content: Yep i calculate Rstart=(vin-vcc)/iout; R=100k calculated by the above equation the controller ic needs atleast 1ma on starting so by power dissipation is expected 0.1w as by I2R. Even upon on the vcc voltage comes only up to 8.4v (which is turn on voltage) There is output in secondary. The auxiliary is rated for 15v the vcc is not raising beyond 8.4v but it still heats up the start-up resistor.

author: jmelson
author_rank: Super Contributor
timestamp: March 20, 2023, 04:21:22 pm
content: What you need is a circuit that detects the converter has started, and then disconnects the start-up resistor.Jon

author: Zero999
author_rank: Super Contributor
timestamp: March 20, 2023, 07:52:14 pm
content: There's something wrong with the circuit. My guess is the auxiliary winding, which powers the IC doesn't generate a high enough voltage to power the IC.Please post a schematic and picture of the physical build of the circuit.

author: whatty
author_rank: Contributor
timestamp: March 21, 2023, 09:10:34 am
content: Hi i have attached the schematic as shown in schematic i have connected all components in bread board. Please right now avoid the leakage inductance of primary side.What issue i have faced: When R5 start-up resistor was 18k the controller starts up and there was output at the secondary but the resistor R5 heats up after some time it was burned. The measured voltage across vcc of IC is 8.4v above which the voltage doesn't goes up even tried measuring the transformer auxiliary pins (on keeping the multi-meter on AC mode the reading shows 0.16v)Then i changed the resistor R5 to 100K but there was no output at the secondary and measured the voltage across vcc of ic it was 4v above that it was not going. Also measured the transformer auxiliary voltage showed same 0.16 vDoes this means the auxiliary was not excited and problem is with the dot notation...?

author: Zero999
author_rank: Super Contributor
timestamp: March 21, 2023, 10:57:50 am
content: I can't see the schematic at the moment as I'm at work and the firewall has decided to block it for some reason.A breadboard is a bad idea, especially for mains voltages.The power dissipation in an 18k resistor will be roughtly 6W, so it will burn, if it's not suitably rated.No, a multimeter is not suitable for measuring high frequency voltages, such as the output of the auxiliary winding.The UC2845 is not specified to work below 12V. The data sheet says VVCCshould be between 12V and 28V. The output from the auxiliary winding is too low to keep the IC poweredhttps://www.ti.com/lit/ds/symlink/uc2845.pdf

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 21, 2023, 11:14:49 am
content: Coupling factors are not shown, but assuming k --> 1, I get about 7.5V aux at C3 in operation. UVLO falling max is 8.2V so it's not guaranteed to remain operational there.Throw on a few more turns to the aux winding. Running at 10-15V is fine.Quote from: Zero999 on March 21, 2023, 10:57:50 amOops, take a closer look at VCC_ON/VCC_OFF (page 8 ). -2/-4 are 10 to 16V hysteresis range, -3/-5 are 7.6 to 8.4V (typ.).Not sure why -5 was chosen here, it's a simple flyback. Normally that is used for one- or two-switch forward converters. The -3/-5 are low voltage parts suitable for say a fixed 12V supply; the narrow hysteresis band makes them difficult to use in an offline application. Recommend changing to UC3842. Adjust R1-C1 as needed (there's no frequency divider).The feedback circuit is also poor and no meaningful compensation is shown, and no EMI filtering is shown, but this will address the immediate issue at least.Tim

author: whatty
author_rank: Contributor
timestamp: March 21, 2023, 03:11:32 pm
content: increasing turns leads to only 8.4v at auxiliary i am doubting that does if coupling co-efficient is poor does that causes problem as i had measured coupling co-efficient by shorting both auxiliary and secondary both together and got a K=0.92 leakage inductance=130uh but i haven't measured the coupling co-efficient between primary and auxiliary. will check it and uploadAlso yes the feedback circuit poor I will correct it

author: Zero999
author_rank: Super Contributor
timestamp: March 22, 2023, 10:16:56 pm
content: Quote from: whatty on March 21, 2023, 09:10:34 amWhat the heck? The PDF says you're using the LT1246, not the UC3845.What's the output voltage supposed to be? The LT1009 is not an adjustable reference like the TL431. It has a tiny trim range. This circuit will give about 4V out.

author: whatty
author_rank: Contributor
timestamp: March 23, 2023, 04:13:42 am
content: Yep i wanted to create a schematic to show you the pin-out of that ic is same as uc3845.