I'm using a AP2141:
to disconnect a circuit from another. It is connected as following:
IN 4V EN 5V OUT 10k pull-down (for test purpose)
On the OUT pin I read the same voltage applied to the IN pin, 4V. I checked it really is an AP2141 (active low).
I can't understand why I read 4V instead of 0V. There are several boards on my desk, all with the same behavior.
Perhaps I misunderstood something in the ds?
Thanks! Marco
Unless your "load" draws some current greater than the switch leakage it will show input voltage.
You DO have the 2141, Enable=LOW, not the 2151, Enable=HIGH ? ...Jim Thompson
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| James E.Thompson | mens |
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I love to cook with wine. Sometimes I even put it in the food.
Il 06/04/2013 19:40, Jim Thompson ha scritto:
I connected a 180 ohm resistor across output terminals...
Unless the label on the package is wrong, I do have a 2141 ;)
Marco
Is it legal for EN > IN?
I also note "All IN pins must be tied together externally" and "All OUT pins must be tied together externally"
Maybe you have the "frog" problem.... croak >:-} ...Jim Thompson
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| James E.Thompson | mens |
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| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
Il 06/04/2013 21:59, Jim Thompson ha scritto:
There is no mention of the contrary on the ds.
I noticed that too ;) I tied them toghther.
Two "frog" problems on a single board! Double croak :(
Marco
Il 06/04/2013 21:01, Marco Trapanese ha scritto:
Very strange. Some other tests:
1) I disconnected the outputs and the enable pins from the circuit. Loaded the output ones with a 1k resistor.Powered the board and after that I forced the voltage on the enable pin with a piece of wire. It works opposite: applying a high level leads to conduct (output voltage = input voltage). Bring the enable down to zero volt disconnect the load (output voltage = 0V).
With 4V at input I can rise the enable voltage up to 5V without any problem.
2) I changed my board adding an n-mos to invert the enable signal. But again it doesn't switch off even with 0V on the enable pin. I removed the mos and did again the test with the wire *after* power up and it works (anyway like 2151).Bottom line: I'm afraid the datasheet doesn't tell the truth also this time. I can't understand at all the polarity of the enable signal. The figure 1 at page 6 is very clear though. Perhaps there are some timings to respect at power up. I suspect because the enable voltage will rise before the input, it may likely latch in some undesirable condition.
Marco
Il 07/04/2013 17:01, Jamie ha scritto:
Do you mean the enable pin? Or the voltage input pin? If you're talking about the enable pin at first I had a pull-down resistor, then I used a mos to tie to gnd the enable pin.
Marco
Maybe you need a pull down R on the input pin?
Jamie
Verify your parts. You said in another post that you had tied the inputs together so you must be using a SOP or MSOP package. The part number is on the top. In this case dont trust the package label.
Measure your EN voltage, make sure it's below 0.8V or above 2V
-- Chisolm Republic of Texas
Il 07/04/2013 19:52, Joe Chisolm ha scritto:
I'm talking about the marks on the case. Shall I trust them?
I measured 4 mV and about 5V. But I tried also from 3V with an external power supply.
Marco
Looking at the data sheet, which I just did, your device P.N. is designated to be active in the low state. This means you should have a pull UP not DOWN, R on the EN input. And pulling the EN pin low via the processor or a open collector/Drain device with pull up R still attached should work.
Jamie
Il 07/04/2013 22:46, Jamie ha scritto:
Perhaps I didn't explain well. I have a pull-down because I want the device is ON by default. When I plug an external power supply the enable pin will be forced to 5V.
I'd expect the device will disconnect its load. But it doesn't happen, the load is still connected.
Inverting the logic, that is pull up the enable pin to turn on the device and tie to 0V to disconnect the load will work only if I power the board first and *after* I toggle the pin.
Marco
Oooops! That can be tricky... did you make yourself a latch? ...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
Il 08/04/2013 16:49, Jim Thompson ha scritto:
I've just ripped off those switches... and I make the boards work using standard mosfet :)
Marco
Probably the best bet... you now know what the real schematic is ;-)
Making hot swap that's automatic based on application of an alternate supply can be really tricky... all kinds of possible latch-up scenarios. ...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
I think you picked the wrong part to use, it looks like a one shot switch. Which is common in power management start up systems because you only need to do this once. In other words, it latched in. To turn off such a device, you need to kill the front end, which is doable via a pmos being reversed biased from the uC for example.
Jamie
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