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September 12, 2009
about 6 years ago
Hadn’t noticed another RF layout issue regarding L1. See that trace that joins L1 to the RF trace perpendicularly? Well, it shoudn’t exist. The corresponding pad of L1 should be “embedded” into the RF trace, with no trace length in between. Trace lengths emerging from RF traces act like stubs, and those effectively behave as RF shorts to ground for some frequencies. So it’s better not to have them…
Are you sure that the RF trace that goes from the IC to the SMA connector is the correct width? Looks to me like if you had used the microstrip model for calculating that trace width, when what you have is mostly behaving as a coplanary waveguide with bottom ground plane (CPWG model). Using 1,5mm FR4 PCB material the trace should be about 0,8mm wide and be separated by a 0,16mm gap to the upper ground plane to get a 50ohm line, using CPWG. What you have looks more like a ~3mm width, which is the 50ohm width for a microstrip model, which doesn’t have an upper ground plane.
Anyway, the trace impedance here doesn’t really matter that much, because the length of that minute trace is much shorter than the wavelength of the 1.57GHz GPS signal, but I think you should correct this mistake for future versions of the board.
about 7 years ago
This is not about max. voltage rating, this is all about CURRENT!!!!!!!!!
The 200k resistor biases a pass transistor that can’t tolerate much more than 15mA. That’s the driving current available for the bus. If you limit it too much, then you can’t drive heavily loaded buses, but if you don’t limit it properly, you basically blow the pass transistor.
That resistor is suggested to have a value of 200k for a high side voltage of 3.3V and MAX CURRENT!!!. If you apply more than 3.3V to the high side with a 200k resistor you’re overloading the pass transistor and if you go to 5V you’ll blow it within minutes.
This board should either have a higher value bias resistor to be keep pass transistor currents in the safe side at the maximum high side voltage, or have selectable resistor values to be able to optimize the current setting for some high side voltage values.
Hope you’ve now understood what’s the deal here. Surely the IC can tolerate 5+ volts, but under certain conditions that are not met in this desing. This design’s absolute max high side voltage ISN’T limited by the absolute max voltage spec of the IC, but by the ABSOLUTE MAX CURRENT spec for the pass transistor.
That design will eventually blow up! BEWARE! The 200k bias resistor is incorrectly sized for the stated voltage range. If you apply 5.5V with a 200k resistor then the pass transistor will basically blow for overcurrent and I’m not kidding. I recently made a design with a PCA9306 to which I was applying 5V at the high side. I didn’t completely read the datasheet so I just used a 200k resistor. Lasted for about 2 hours and lots of time was spent on debugging because I wasn’t expecting the PCA to be the broken component. I even changed it and the new one lasted seconds. Then I read the datasheet and changed the resistor to a more proper value (in order to be safe and considering that the I2C bus wasn’t heavily loaded I chose 470k) and the circuit is working without any single problem since then. So please review the desing and either add multiple resistors so the users can select the appropriate one for their application or choose a higher value one that does a trade off.
about 7 years ago
Duct tape doesn’t protect against high powered lasers, they just burn it in less than a second if properly focused. The next barrier is your retina. Eclipse goggles doesn’t protect either, as they just burn too. Those are made to handle a broad spectrum emission with an overall low power density. With a laser, you’re talking about power densities that are some orders of magnitude away from what you could receive from the sun, at a very specific wavelength. Same happens to welding glass or most other household items. So please, do take this seriously.
For the pot-modding dude: you won’t get 100mW’s out of this, it just aint gonna happen. Considering an optical efficiency of 15% being generous for such a cheap module, you’d need a 50mW pump diode to get 5mW’s, and a 700mW pump diode to get 100mW. A 50mW rated diode just won’t emit 700mW’s. 100mW (15mW of total output) could be done, 200mW (30mW of total output) attainable for some hours, but above that, it would just burn.
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