Hi,
Can you recommend a schematic of dual power supply (inputs) switch with one load output based on SLG59M1606V?
Regards,
Bartosz
Device:
Device Number:
SLG59M1606V
Hi Bartosz. Thank you for interesting in our prodcuts. Can you please give me some information regarding your application:
1. What is your input voltages for both power rails?
2. Do you need automatic turn off when reverse voltage is detected?
3. What is your max switching current?
Hi,
1. First: 4.3V @ 3A, Second: 5V @ 1A
2. Yes (the first is the primary power source)
3. As above.
Hi Bartosz,
谢谢你的回应。电压水平和铜rrents are acceptable for this SLG59M1606 but unfortunately reverse current blocking works only when device is off, so you will need to switch power rail manually. For your application we can suggest you to use dual channel SLG59M1639 for 4.3V @3A and SLG59M1748 for 5V @1A. Please see attached schematic for your reference. With this load switches 1639 will automatically detect reverse voltage from 5V. I would also recommend you to read this AN of how to use our LS's in such applications as yours//www.xmece.com/AN-1096. Please let me know if you have any questions.
Hi,
I'm confused, "unfortunately reverse current blocking works only when device is off, so you will need to switch power rail manually" - can you explain it? I've sent a schematic block diagram with two pwr switches. I expected it's doable like it's illustrated in AN-1096 Fig. 1A
Regards,
Bartosz
Hi Bartosz,
SLG59M1606 has reverse current blocking when device is off, that means that if you turn ON one power rail, lets say 4.3V and then apply 5V to VOUT, so this 5V will pass to input of 4.3V power rail, however if you set ON to low, then reverse current will not pass from VOUT to VIN. In case of using SLG59M1639, so besides reverse current blocking, this device has additionally reverse voltage detection, that means that if you apply voltage to VOUT that is higher by at least 50mV than VIN, so device will automatically detect it and shutdown, so you don't need to set ON to low manually. As a conclusion, your application is duable in case of using SLG59M1639 and SLG59M1748 as I described in a previous response. If you will switch 4.3V and 5V power rails manually and the behavior brake before make is ok, so you can use SLG59M1606V as well. Please let me know if you have any questions.
Hi
Just to make sure, in case of using SLG59M1639 and SLG59M1748 as you described in a previous message, keeping EN pin of SLG59M1639 high and SLG59M1748 EN pins low, will allow me to create switch that will work automatically, won't it? What about quiescent current?
Regards,
Bartosz
Hi Bartosz. Yes, you will need to switch only EN in SLG59M1748 to automatically switch between 5V or 4.3V but please make sure you meet requirements for reverse voltage detection for 1639 that is described in AN that I sent you earlier ( I mean that there should be enough current on VIN side of 1639 to detect reverse voltage). Regarding quiescent current, so once 1639 shutdown it will consume the same current as when it is on because ON signal will be still high. Please let me know if you have any questions.
Hi Petro,
Can we use SLG59M1649 in that case instead of SLG59M1639?
This part of datasheet is unclear for us:
"When enabled, internal reverse-current protection will quickly open the switch in the event of a reverse-voltage condition is detected (a VOUT + 50 mV > VIN condition opens the switch)."
" I mean that there should be enough current on VIN side of 1639 to detect reverse voltage" - there will be 60V battery connected, what do you mean by enough current on VIN?
So, I need to control EN signal of SLG59M1748C, don't I? or just connect to logic high all time.
The second switch I've sent in power block draft at the beginning - can be replaced by two SLG59M1639 and switch automatically?
Hi Bartosz.
SLG59M1649 has internal discharge when it is off (but only when off by enable pin), so if this is not a problem for you, then you can use it as well.
regarding the DS statement, so yes, when VOUT is > than VIN by 50mV, then 1639 will shutdown but to detect this voltage, there should be enough current on VIN side. When you have 4.3V on VIN and apply 5V to VOUT, so this 5V will pass to VIN. The RDS(ON) of 1639 when its VIN1, VIN2 and VOUT1, VOUT2 are paralleled will be around 23mohm, so to generate 50mV difference between VIN and VOUT, you will need to have around 2A current consumption on VIN side. In AN you can see that we use resistor on VIN side for this condition. Based on you schematic it will mean, you need to continuously burn 2A current directly on the output of your DC/DC if you don't use this power rail for any other blocks that may totally consume 2A (please see attached image for better understanding).
Yes, you need to control EN signal if you will always have 5V on VIN and just need to switch between 5V or 4.3V, otherwise you will always have 5V on common VOUT. If you plan that this 5V will be completely disconnected from USB and it it needs to be turned on once USB is connected, so you can set EN high all the time.
For secon load switch it may not work because you described that input should accept 4.3V only and in case you will have 5V there 1639 will automatically switch to that power rail, that is not good for you. You will probably need to check which voltage is present and if it is 5V, then manually turn off that load switch. Or you will need to have some Overvoltage protection. BTW: What is the reason of second power switch if your IC charger supports both 4.3V and 5V, and anyway you make 4.3V at the end?
Please let me know if you have any questions.
Can we have a quick call (hangouts/zoom/teams)?
My email isbziolek@taoglas.com
Regards,
Bartosz
Sure, I just wrote you on hangouts.
Hi,
Pin ON of SLG59M1649V is active high or low? I think there is a mistake in datasheet (1.01) in block diagram. Should be SW Closed ->SW Open. The description is correct: "A low-to-high transition on this pin initiates the operation of the power switch. ON is an asserted-HIGH, level-sensitive CMOS input with ON_VIL < 0.3 V and ON_VIH > 1 V. As the ON input circuitry does not have an internal pull-down resistor, connect the ON pin directly to a GPIO controller – do not allow this pin to be open circuited." Could you confirm?
Regards,
Bartosz
Hi Bartosz.
It is active high and the description in block diagram is correct. When we talk about load switches, MOSFETs as well as about contacts in general, closed means turned ON and open means Turned off.https://thegrid.rexel.com/en-us/knowledge/product-faqs/w/wiki/928/contac....