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RG1 1.8v regulator

Former Member

Ok, so in a different thread I threatened to remove RG1 and do some current measurements on it's output after seeing those thermal images that show it's not generating any heat...

 

Well, I did it tonight. Some photos here: https://picasaweb.google.com/selsinork/RPi18v

 

The jumper pins in the output let me either just put a jumper on and verify the Pi boots ok, or wire a multimeter in series to get some current readings.

 

The results were interesting to say the least. I had to go back and check I was reading the multimeter correctly, that it wasn't broken etc.

 

On initial power up I see a negative current for a second or so which then reverses to about 0.5mA (yes half a milliamp, that's not a typo) for a few seconds while we get the first sd-card accesses. Once we're booted and sitting at the login prompt the current reading fluctuates from around 0.001mA to maybe 0.04mA. 

 

I'm using the 40mA range on a decent Fluke multimeter, so I've no reason to doubt the results. There's obviously going to be some inaccuracy down at that level due to length of meter leads etc, but the result is fairly clear.  You'll understand why I was checking the meter was working and I was reading it correctly though image

 

 

So from there onto the next test, lets try completely disconnecting RG1 and see if the Pi boots while using the LAN9512 1.8v 'output'.  Yes it does! 

 

I think that's reasonably good indication that jamodio got it spot on, the lan9512 shouldn't be connected to the 1.8v plane and it's heat problems are going to be largely due to supplying current on it's 1.8v filter pin that it was never designed to do.

 

So anyone willing to pull RG1 off a Pi and verify my results ?

Parents

  • Hi

     

    I am in the process on testing how little power the RPI can use and what heat there is, so i will just share some small limit results with you.

     

    In the test setup i have:

    Disconnect the LAN9512 from the main 1V8 power rail on the board.

    Add caps to te LAN9512 as the SMSC pdf says

    Removed the RG1

    Removed the RG2

     

    I have then connected external power supplys for the 1V8, 3V3 and 5V0 Rail and on each rail there is an volt and amp meter so i can see how many V and mA there is used on each rail.

     

    Here is when the RPI is just power up and is idle

    Rail 1V8          1,8000V * 0,06338A = 0,114084W

    Rail 3V3          3,3400V * 0,02460A = 0,082164W

    Rail 5V0          4,9900V * 0,04605A = 0,022979W

    Idle mode total = 0,426038W

     

    Here is when the RPI calc PI

    Rail 1V8          1,8000V * 0,06351A = 0,114318W

    Rail 3V3          3,3300V * 0,02710A = 0,090243W

    Rail 5V0          4,9200V * 0,07973A = 0,392272W

    RPI calc PI total = 0,596833W

     

    Note: the reason it is the 5V0 rail there is having higher mA when calc PI is because i have not removed the RG3 from the board, the RG3 is feed the CPU with 2V5 from the 5V0 rail.

     

     

    wow this is much lower then i was thinking it was using and it seems there is alot to save with an very good switch mode regulator..

     

     

    So this was just to share some test details with your guys.

     

     

    Tooms

Reply

  • Hi

     

    I am in the process on testing how little power the RPI can use and what heat there is, so i will just share some small limit results with you.

     

    In the test setup i have:

    Disconnect the LAN9512 from the main 1V8 power rail on the board.

    Add caps to te LAN9512 as the SMSC pdf says

    Removed the RG1

    Removed the RG2

     

    I have then connected external power supplys for the 1V8, 3V3 and 5V0 Rail and on each rail there is an volt and amp meter so i can see how many V and mA there is used on each rail.

     

    Here is when the RPI is just power up and is idle

    Rail 1V8          1,8000V * 0,06338A = 0,114084W

    Rail 3V3          3,3400V * 0,02460A = 0,082164W

    Rail 5V0          4,9900V * 0,04605A = 0,022979W

    Idle mode total = 0,426038W

     

    Here is when the RPI calc PI

    Rail 1V8          1,8000V * 0,06351A = 0,114318W

    Rail 3V3          3,3300V * 0,02710A = 0,090243W

    Rail 5V0          4,9200V * 0,07973A = 0,392272W

    RPI calc PI total = 0,596833W

     

    Note: the reason it is the 5V0 rail there is having higher mA when calc PI is because i have not removed the RG3 from the board, the RG3 is feed the CPU with 2V5 from the 5V0 rail.

     

     

    wow this is much lower then i was thinking it was using and it seems there is alot to save with an very good switch mode regulator..

     

     

    So this was just to share some test details with your guys.

     

     

    Tooms

Children
  • in reply to Former Member

    Tooms wrote:

     

    wow this is much lower then i was thinking it was using and it seems there is a lot to save with a very good switch mode regulator..

    Thank you for sharing your measurements.  Those numbers do seem low, but calculating pi probably doesn't use much RAM.  I believe the power consumption goes up a lot with:

     

    1.  Large computations that don't fit in cache, so there's lots of DRAM activity.  Large compiles are good candidates.

     

    2.  Heavy GPU use such as 3-D graphics and video decode.  Those would also have a lot of DRAM traffic.  I don't know how much HDMI adds in when active.  I know my RasPi draws more current when X Windows comes up.

     

    3.  Ethernet, which has 50 Ohm pull-up connected to +3.3V.  USB probably uses a lot when it's driving.

     

    I remember my board being quite a bit warmer with Ethernet hooked up, so I typically just connect it for a few seconds to get the correct time and then leave it disconnected.

  • in reply to Former Member

    Tooms,

     

    When you're not feeding the 'pi externally with those powersupplies, all these are derived with a linear regulator from the 5V. So while your measurement says

     

    Rail 1V8          1,8000V * 0,06351A = 0,114318W

     

    That would be 1.8V rail: 5V * 63mA = 315 mW if you hadn't modified your 'pi.

     

    So in fact, we can calculate the power-use of you'r pi by just adding all the currents:

     

    Itot = 63 mA + 27 mA + 79 mA = 169 mA total.

     

    My measurements said that the 'pi was using about 400mA (+/- 40mA depending on system load and Video). Others have done similar measurements and also measured values in that range. So your measurements differ by more than a factor of two from previous measurements.

     

    Can you double check?: It seems there  is a measurement error.

     

    Things to try: Restore the 'pi to the original state and measure 5V consumption. I expect on the order of 400mA. Not 170mA as you're seeing now.

     

    P.S. Those linear regulators are very "efficient"in that they don't "waste" any current. They are powered from the voltage difference between output and input. Almost all the current is fed into the load. An 1.8V regulator delivering 63mA, will draw very close to 63mA from its powersupply. Even an old one like the LM1117.

  • in reply to johnbeetem

    John Beetem wrote:

     

    Tooms wrote:

     

    wow this is much lower then i was thinking it was using and it seems there is a lot to save with a very good switch mode regulator..

    Thank you for sharing your measurements.  Those numbers do seem low, but calculating pi probably doesn't use much RAM.  I believe the power consumption goes up a lot with:

     

    1.  Large computations that don't fit in cache, so there's lots of DRAM activity.  Large compiles are good candidates.

     

    2.  Heavy GPU use such as 3-D graphics and video decode.  Those would also have a lot of DRAM traffic.  I don't know how much HDMI adds in when active.  I know my RasPi draws more current when X Windows comes up.

     

    3.  Ethernet, which has 50 Ohm pull-up connected to +3.3V.  USB probably uses a lot when it's driving.

     

    I remember my board being quite a bit warmer with Ethernet hooked up, so I typically just connect it for a few seconds to get the correct time and then leave it disconnected.

     

    Hi

     

    yes this was just an small test, my plan is todo this things when load testing

     

    * copy files via network

    * take images via webcam (my not do this because then the webcam load is part the setup)

    * calc PI

    * use omxplayer to play an mpeg4 movie

     

    so by this it will do more load on it when testing.

  • in reply to rew

    Roger Wolff wrote:

     

    Tooms,

     

    When you're not feeding the 'pi externally with those powersupplies, all these are derived with a linear regulator from the 5V. So while your measurement says

     

    Rail 1V8          1,8000V * 0,06351A = 0,114318W

     

    That would be 1.8V rail: 5V * 63mA = 315 mW if you hadn't modified your 'pi.

     

    So in fact, we can calculate the power-use of you'r pi by just adding all the currents:

     

    Itot = 63 mA + 27 mA + 79 mA = 169 mA total.

     

    My measurements said that the 'pi was using about 400mA (+/- 40mA depending on system load and Video). Others have done similar measurements and also measured values in that range. So your measurements differ by more than a factor of two from previous measurements.

     

    Can you double check?: It seems there  is a measurement error.

     

    Things to try: Restore the 'pi to the original state and measure 5V consumption. I expect on the order of 400mA. Not 170mA as you're seeing now.

     

    P.S. Those linear regulators are very "efficient"in that they don't "waste" any current. They are powered from the voltage difference between output and input. Almost all the current is fed into the load. An 1.8V regulator delivering 63mA, will draw very close to 63mA from its powersupply. Even an old one like the LM1117.

     

    yes it is looking very low but i have looked and recheck the setup and it all seem to be correct.

     

    My RPI is also using around 400 +/- mA so yes it is low the 170mA

     

     

    But for more testing i have to make an plan of testes an follow, but this post was just to share with your guys that it seems to alot to gain bu replacing the regulators.

     

     

    Thomas

  • in reply to Former Member

    Well, IMHO, the first thing to do is to measure where that missing 230mA goes to....

     

    P.S.

    Rail 5V0          4,9900V * 0,04605A = 0,022979W

    is wrong. That's 0.23W.

  • in reply to rew

    Hi

     

    your correct, it is an calc error from my side

     

     

    Here is an new calc of the idel mode

     

    Rail 5V0          5.01V * 0,04595A     =     0,2302095W

    Rail 3V3          3,35V * 0,02830A     =     0,0948050W

    Rail 1V8          1,83V * 0,06317A     =     0,1156011W

    Total watt                                         =  0,4406156W

     

     

    Tooms

  • in reply to rew

    Roger Wolff wrote:

     

    P.S. Those linear regulators are very "efficient"in that they don't "waste" any current. They are powered from the voltage difference between output and input. Almost all the current is fed into the load. An 1.8V regulator delivering 63mA, will draw very close to 63mA from its powersupply. Even an old one like the LM1117.

    However, linear regulators are inefficient in that they waste power by converting the voltage drop into heat.  A switching regulator uses less average input current to produce the same average output current.