LT1374 Adjustable Dual Output SEPIC - Page 1
Forum section: electronics / general

author: jarez95
author_rank: Newbie
timestamp: October 21, 2020, 01:11:40 pm
content: Hi,I am designing a high efficiency power supply which requires a clean +-5V rail for a matrix of Op-amps providing a target +60dB gain on an input signal. However I am running into a few issues with getting the dual output SEPIC to work on the LT1374-SYNC as per the application note 'DN100' and the bottom of the datasheet of the 1374. I have also made a small modification by adding a voltage divider between the positive output and the FB pin, this should allow the output voltage to be adjustable (I have tested this with the standard LT1374 eval board), I actually need the SEPIC to have an output voltage around +-6.3 V as I will pass this into dual linear regulators to get the final +-5V.I have implemented the circuit on perfboard and I am testing it with 12V (cannot be changed) and it seems to be working quite well, developing the +ve and -ve voltages, however I am not able to get below -5.5 V i.e to my target at -6.3 V and the negative voltage stops changing linearly after -5V. For example when I adjust the potentiometer further the positive voltage continues to adjust up to 10 V however the negative voltage remains stuck in the region of -5 V. When loading the -ve rail with say 10 to 20 Ohms, which should result in 300mA to 600mA being drawn, the voltage on the rail drops drastically to -2 or -3V, however the app note states that the max load should be 1.5A for the 1376 model, so I should expect it to be able more for the 1374 model rated at 3A.Initially I thought it was because of the 'C4' capacitor in the app-note being rated at 10V and the output being +-5V that I quickly thought it might be the capacitor limiting the voltage, I installed a 47uF rated at 16V but that didn't make a difference and it would seem that the full voltage range is not imposed on the capacitor anyway. I have used the exact same component values as indicated by the schematic at the bottom of the LT1374 datasheet, though having more and bigger capacitors at the output.Changing the input voltage from 12 V upwards, adjusts the magnitude of the negative voltage but doesn't change the positive voltage, if I go to 15V for example I am able to achieve the +-6.3V, it effectively changes this non-linear breakdown point indicated by the black arrow. The fact that I can obtain these voltages tells me that I do not have an issue with any of the voltage ratings on my components. I am still having issues with the negative rail being able to sustain load and I am still not able to get -6.3V with an input voltage of 12V only. Please see the below plot:I can also observe that putting the +ve rail under load allows the negative voltage to sustain load a bit better, this is because the +ve rail is connected to the sense line through a variable resistor, however there is no such feedback for the -ve rail. In my application the load will be balanced 22R load on both +ve and -ve rails, however at powerup this load will not be present, whilst a microcontroller verifies the voltage and then activates a relay once it has passed checks. If anyone has any suggestions regarding voltage stability it will be a big help.Any help or guidance will be appreciated as it is first time designing/working with a system that generates -ve voltages and working with switching power supplies.Best Regards, Jarez

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: October 21, 2020, 01:44:38 pm
content: Hi,Here is a link to the design note DN100:https://www.analog.com/media/en/reference-design-documentation/design-notes/dn100f.pdfFirst the design note was written by a legend Carl Nelson. You can read about Carl Nelson here:https://www.electronicdesign.com/technologies/analog/article/21805566/an-interview-with-analog-guru-carl-nelsonThe circuit is not a dual SEPIC. The positive output is a buck regulator. The negative output is a SEPIC.The DN100 explains that the maximum negative current is limited:Total output current of this circuit is limited by the maximum switch current of the LT1376. The following formula gives peak switch current, which cannot exceed 1.5A.This formula, in the spirit of simplicity, issimplified, so caution must be used if it indicates close to 1.5A peak current.IPEAK = 0.25A + (I+) + (2)(I–) (Must be less than 1.5A)Maximum negative load current is limited by the +5V load. A typical limit is one half of 5V current, but a more exact number can be found from:Max Negative Load = (I+)(0.07)(VIN – 2)Try increasing the load on the positive side.Regards,Jay_Diddy_B

author: julian1
author_rank: Frequent Contributor
timestamp: October 21, 2020, 09:21:49 pm
content: QuoteDoes the circuit only regulate on the positive output? If so load that up first, then check if the voltage on the negative output still sags under load.If the current requirements are not symmetrical - eg. DC rather than AC, then you probably want separate power supplies for each rail that can regulate load independently.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: October 21, 2020, 09:56:47 pm
content: Hi,Only the positive output is regulated by the feedback loop. The negative output is semi-regulated because L1A and L1B are coupled.In this application, this is fine, because the rails are going to regulated with LDOs.Jay_Diddy_B

author: julian1
author_rank: Frequent Contributor
timestamp: October 21, 2020, 10:05:35 pm
content: QuoteAhh, that's clever.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 22, 2020, 09:57:32 am
content: The buck diode D1 says it only works while there's more load on the positive output than negative.If this is unreasonable (can't tolerate the efficiency hit of a bleeder resistor?), consider a synchronous switching regulator, which can sink as well as source current, thus keeping the negative rail going, even when the positive doesn't need as much current. (You may have to strap it for continuous operation, which will again cost some efficiency, at light load.)If efficiency is paramount, and the difference in load currents is expected to be fairly large, just use two independent converters. Also consider a relatively high switching frequency (400kHz+?) and LC filtering, rather than LDOs.Shielding may also be desirable to keep noise down. In my experience, board-level filtering alone, can get in the range of 60dB, which may be enough. If you need more like 100dB of filtering, you'll probably need shielding and feed-thrus. (A low-EMI converter may also be desirable; LT has a family for that.)Tim

author: jarez95
author_rank: Newbie
timestamp: October 22, 2020, 12:48:44 pm
content: Hi,Thanks all for the replies, I will try to answer everyone's questions and queries in this post.Yes you are correct Jay_Diddy, the positive rail has to be loaded for the negative rail to work properly I did indeed observe this. From that rule of thumb statement 'A typical limit is one half of 5V current', I would then load the positive rail with 11R to get the negative rail to work at 22R. I could come up with a load resistor and place that on the positive rail to try and compensate in some way, but it would not be my preferred solution. As I also mentioned I am looking for a way that both rails will produce the correct voltage under basically no load so that it can pass the check from the uC (I have preloads of 1k resistors on all rails). I also don't think I can rely on my load being fixed since the current draw of the matrix will depend on the input signal and if it is present.Do you think I can make some clever closed loop control circuit with op-amps to regulate both the positive and negative rails feeding into the single feedback input of the IC? I have drafted a few circuits in my head involving a few op-amps and voltage inverters but I have doubts as to whether it would be stable.Tim, efficiency is not so important with this portion of the design but of course the SMPS is critical in getting the +ve and -ve rails as this system will be purely DC powered at 12V. To be honest I might have to look into the pure positive to negative converter, though the circuit makes me a little uneasy with the VSW connected to ground. (I know it would work but it is still a bit frightening). I would like to try and get this dual setup working with the single piece of magnetics since I also accidentally brought 125 of these transformers and have gone through the trouble of prototyping this circuit. I will probably take the switching frequency up to 960 KHz since this is the closest multiple to the sample rate on my FPGA. FYI there will be 11 other LT1374 on the same circuit board, generating various other voltage rails, and the whole system will be powering up some very expensive bespoke hardware.Thoughts welcome!Best Regards, Jarez

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: October 22, 2020, 02:59:14 pm
content: Hi,Here is an example of the kind of circuit that Tim is talking about:It uses the LT8641-1 in the forced continuous mode. The LT8641-1 is from the silent switcher family of parts. These are designed to be low noise.Link:https://www.analog.com/en/products/lt8640.htmlThe negative output is one Schottky diode drop less than the positive rail.I have attached the LTspice model.Regards,Jay_Diddy_B8640-1 dual.asc(3.42 kB - downloaded 56 times.)