hi,
here v0.8 : https://gsm.tsaitgaist.info/SIMtrace/v0.8/ Compared to v0.7+, I added 2 1n4148 after the parallel voltage regulator outputs. I wanted to avoid the regulators to fuzz each other, but I don't know if it's right. Maybe an additional zener diode is also required.
Now I'm looking for the parts and footprints.
Kevin
Hi Kevin,
On Mon, May 09, 2011 at 07:26:58PM +0200, Kevin Redon wrote:
here v0.8 : https://gsm.tsaitgaist.info/SIMtrace/v0.8/ Compared to v0.7+, I added 2 1n4148 after the parallel voltage regulator outputs. I wanted to avoid the regulators to fuzz each other, but I don't know if it's right. Maybe an additional zener diode is also required.
I understand the intention, but I doubt it's a good idea. First, the 1N4148 is quite low-power/low-current, i.e. it will be the weakest element in the chain. Second, it's a silicium diode and thus causes a voltage drop of 0.7V, i.e. we will have 3.3-0.7= 2.6V only, which is really bad.
I think you should just leave them out. You could simplify the design by removing the DC/DC converter (SC120) and simply attach the battery connector to the +5V input. The TPS73633 is a ultra-low-drop regulator, i.e. from voltages >= 3.375V on it will provide a stable 3.3V output.
This means a battery pack with 3 or 4 1.2V cells should do just fine.
What do you think? We can save the step-up regulator and loose the design risk of leaking power from the USB into the step-up regulator at the same time...
Regards, Harald
Hallo Kevin.
here v0.8 : https://gsm.tsaitgaist.info/SIMtrace/v0.8/ Compared to v0.7+, I added 2 1n4148 after the parallel voltage regulator outputs. I wanted to avoid the regulators to fuzz each other, but I don't know if it's right. Maybe an additional zener diode is also required.
Hmm. The diodes will cause a voltage loss of about 0,7V ;-/ Maybe you will need something more tricky or just add a little switch. I think nobody will change the power supply method every minute.
regards. Philipp
Hi!
Or maybe you could do this:
3xMigon ----->|------+ +--------+ | | | +----------+ 3,3V +-------- | | Reg. | USB 5V ----->|------+ +---+----+ | _|_
Simple, but effective ;-) And you need only one regulator.
regards. Philipp
then it would be 4xAAA/AA (3xAA will only work if fully charged)
the other idea would be :
0.7-4.5 --- VREG 5V --->|----+ | +--- VREG 3.3V | USB 5V ---------------->|----+
or another nice idea (like the arduino does for the 5v) :
BATT --- VREG 3.3V ---+ | FDN304 -- 3.3V | VUSB --- VREG 3.3V ---+
as you wish, Kevin
On 09.05.2011 21:18, Philipp Fabian Benedikt Maier wrote:
Hi!
Or maybe you could do this:
3xMigon ----->|------+ +--------+ | | | +----------+ 3,3V +-------- | | Reg. | USB 5V ----->|------+ +---+----+ | _|_
Simple, but effective ;-) And you need only one regulator.
regards. Philipp
Hi Kevin,
On Tue, May 10, 2011 at 09:57:29AM +0200, Kevin Redon wrote:
then it would be 4xAAA/AA (3xAA will only work if fully charged)
fine with me. I honestly don't care about the number of batteries... and the PCB will not be smaller than 4x AAA/AA anyway.
the other idea would be :
0.7-4.5 --- VREG 5V --->|----+ | +--- VREG 3.3V | USB 5V ---------------->|----+
that should work, but you should use something like a 1N4001 as the diode, to cope with more current. However, I would want to have that tested on a breadboard before sending the gerber off for PCB manufacturing, just to make sure the power supply side is working. We still have a problem if somebody decides USB _and_ the battery power supply :/
In fact, now that I'm thinking off it, we should use one of those jacks which disconnnect one contact when plugging in the power supply. This way we can actually interrupt the USB 5V once a power supply plug has been plugged in. This should be done independent of the decision on the voltage regulator.
or another nice idea (like the arduino does for the 5v) :
looks not that trivil to me, as you don't only need the FDN304 but also the LM358 op-amp and external circuitry. Probably a bit overkill.
So I think we should route the 5V USB via the power plug, and simply use a low-drop-out 3.3V regulator, like in the current schematics. This way the LDO input will be USB 5V _until_ somebody plugs in an external power supply or battery pack, which will have to provide at least 3.4V for stable operation.
Regards, Harald
I'm back. I was looking around for a solution for such a trivial task (power switching), and there is no easy answer.
the other idea would be :
0.7-4.5 --- VREG 5V --->|----+ | +--- VREG 3.3V | USB 5V ---------------->|----+
not effective.
In fact, now that I'm thinking off it, we should use one of those jacks which disconnnect one contact when plugging in the power supply. This way we can actually interrupt the USB 5V once a power supply plug has been plugged in. This should be done independent of the decision on the voltage regulator.
most dc jack can provide such a solution, but I think the USB 5v should be used if the two sources are connected. USBVCC is more stable, moreover if the user leaves it connected the battery will be drained. Another point is that the normally closed circuit uses the barrel (normally the jack is used to plug wall power, cutting the board battery line). Having +5v on the barrel is not common (if the user has an of-the-self power adaptor).
or another nice idea (like the arduino does for the 5v) :
looks not that trivil to me, as you don't only need the FDN304 but also the LM358 op-amp and external circuitry. Probably a bit overkill.
agree
So I think we should route the 5V USB via the power plug, and simply use a low-drop-out 3.3V regulator, like in the current schematics. This way the LDO input will be USB 5V _until_ somebody plugs in an external power supply or battery pack, which will have to provide at least 3.4V for stable operation.
yes, I also like this solution, but has drawbacks (see above)
I found two other solutions : - using a automatic power switching mux (TPS2115). too expensive and take to much space. - using a manual switch to select the power source
then it would be 4xAAA/AA (3xAA will only work if fully charged)
fine with me. I honestly don't care about the number of batteries... and the PCB will not be smaller than 4x AAA/AA anyway.
that is the decision I took. Simple, cost effective, saves place.
that should work, but you should use something like a 1N4001 as the diode, to cope with more current
1N5819 allows 1A, it should be enough. 1N4001 has a forward voltage of 1.0V, 1N5819 only 0.6-0.7V.
here the schema : https://gsm.tsaitgaist.info/SIMtrace/v0.9/ I now an doing the BOM. I'm half-way through, but it takes time.
Kevin
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Harald Welte -
Kevin Redon -
Philipp Fabian Benedikt Maier