EEVblog Electronics Community Forum
Electronics => Projects, Designs, and Technical Stuff => Topic started by: MobileWill on December 08, 2016, 07:30:14 am
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I am working on my first switched mode PS design and just assembled the first PCB. The problem that I can't figure out is the output voltage is just about the same as the input. I am inputting 5V and should be getting 3.3V out. I am using the Intersil isl85415. Also on the phase and boot pins I am getting a weird readings. Also the I can't detect anything on the feedback pin. The board itself seems to be okay as far as assembly and such.
Here is a link to the datasheet
http://www.intersil.com/content/dam/Intersil/documents/isl8/isl85415.pdf (http://www.intersil.com/content/dam/Intersil/documents/isl8/isl85415.pdf)
Attached are the images of the schematic, board, and the scope results of boot/phase.
I am at a loss of what to do next. I am used to doing digital designs.
Thanks for any help.
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What does it do if you raise and lower the 5v between 4v-6v? You're not getting anything on the feedback pin, according to the circuit, that voltage divider between the output and ground should have something on it.
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I tried 9v in and the output is about the same.
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I see that the device has a ground pad with some vias in it. Are you sure that it has properly soldered? I have had issues with the vias wicking solder away from the joints in cases like this which caused all sorts of issues.
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From the sides it looks good. I don't have a way to tell if the ground pad is soldered. The ground pins I know are connected to ground. I thought about making another board and see if I get the same problem. I don't see anything wrong with the design.
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If you read nothing on the FB pin it's normal the output goes to the input voltage?
"Are we there yet"?
*checks FB voltage, still zero*
"Let's give it some more"
"Are we there yet"?
*checks FB voltage, still zero*
"Let's give it some more"
"Are we there yet"?
*checks FB voltage, still zero*
"Let's give it some more"
....
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If you measure zero voltage on the feedback pin and the output voltage is equal to the input, chances are you're bypassing and the internal FET is in "OFF" state. Did you check if the feedback resistors are properly soldered and whether they have the correct values? It happens a lot soldering the incorrect resistor value or missing a solder pad on the resistors.
Hope it helps
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if you have zero fb then either the feedback path is broken (resistor or pcb trace), there is a short to gnd around fb pin, or fb pin is not soldered. eithet case you can check with a multimeter.
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Thank you everyone for pointing me in the right direction. After some probing I realized I used a 22.1Ohm instead of 22.1K for the resistor divider, oops. So that is why I didn't see a voltage, I was dividing down too low. Which is good and means my design is probably correct!
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So it worked adding the correct value. But I found another problem. The diodes are acting strange. If I apply 5v usb power I am measuring 5v on the DC jack and vice versa. Doesn't make sense to me. What am I not seeing?
Sent from my SAMSUNG-SM-N920A using Tapatalk
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What type of diode are you using?
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What type of diode are you using?
PMEG4005EJ
http://www.nxp.com/products/discretes-and-logic/diodes/medium-power-schottky-diodes-200-ma/0.5-a-very-low-vf-mega-schottky-barrier-rectifiers:PMEG4005EJ
Sent from my SAMSUNG-SM-N920A using Tapatalk
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Check the polarity of the diodes. You should also check the continuity (or resistance) with your multimeter between the dc jack pin and the usb power pin, there could be a short somewhere
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Check the polarity of the diodes. You should also check the continuity (or resistance) with your multimeter between the dc jack pin and the usb power pin, there could be a short somewhere
Well the PS works so the polarity is correct, both diodes are facing the same way. I checked and no shorts. I did the diode test and get OL on one direction and 0.124v the other. So everything seems to check out.
I can see the voltage drop across the diode and a slight on in reverse across the 2nd one, which is 0.006 of a volt drop.
Edit:
I guess it is fine. If I put a load then it works like a diode. I guess I didn't know with no load you can measure a voltage on the other side. Once there is a load it stops current and voltage drops to like 0.1. So I guess its okay then when both USB and the DC Jack are supplying power.
So my question the diode is rated at 0.5A forward current. Does that mean I can't pass more than than that in forward bias? What happens when you draw 0.5A or more?
Thanks.
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I can see the voltage drop across the diode and a slight on in reverse across the 2nd one, which is 0.006 of a volt drop.
That is the diode's leakage current in combination with multimeter's high input impedance.
So my question the diode is rated at 0.5A forward current. Does that mean I can't pass more than than that in forward bias? What happens when you draw 0.5A or more?
That current is thermally limited. You can push more through it, but only intermittedly. That can be pushed a bit by good thermal coupling to the PCB (planes around the pads, or via to planes), or forced air flow.
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What happens when you draw 0.5A or more?
It will blow up, the most probable failure mode being a short circuit, so it acts like a piece of wire after that.
If the 0.5A rating is so called "absolute maximum rating", maybe specified at 25degC package temperature, this destruction will happen even before 0.5A, depending on how you handle the heat. Normally you would do the thermal analysis and then add some safety leeway. For 0.5A continuous current draw, you would most likely pick a diode rated to 1A or so.
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So if the buck is 500mA at 3.3v the input current will be less, correct?
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Iin = Vout * Iout / {efficiency} / Vin. the higher Vin the lower Iin.
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Thank you very much. I found the 1A version of the same diode.
http://www.digikey.com/scripts/DkSearch/dksus.dll?Detail&itemSeq=214096183&uq=636171873707431964 (http://www.digikey.com/scripts/DkSearch/dksus.dll?Detail&itemSeq=214096183&uq=636171873707431964)
It seems in some situation the input current can be higher than the output. Especially with lower efficiency and input voltage.