Transistor wierdnes (?) - Page 1
Forum section: electronics / beginners

author: permal
author_rank: Regular Contributor
timestamp: January 07, 2018, 04:47:28 pm
content: Given the attached schematic and assuming that I2C_PWR_CTRL is 0V, what is the expected output at pin 3 of Q309 (I2C_PWR)? I2C_PWR goes off to the Vdd pin of an IC.I expected it to be 0V, since the transistor isn't conducting, but I'm getting ~2.7V, regardless of I2C_PWR_CTRL.Is there an error in this circuit or is the transistor broken (I've not yet desoldered it)? Measuring Vce and Vbe, both gives ~0.7V.

author: rstofer
author_rank: Super Contributor
timestamp: January 07, 2018, 05:03:01 pm
content: Pull the transistor and see what you have at pin 3. I'll bet there is some voltage coming back from the I2C device.I don't think the circuit is going to work very well. At a minimum, you're going to give up some voltage. Your base voltage won't quite go to 3.3V but regardless, the emitter voltage will be 0.7V less or 2.6V. If the base only goes to, say, 3.0V, the emitter will be just 2.3V.You might be better off with a PNP transistor where you can pull the base down, relative to the emitter, and only have about Vce drop (probably around 0.2V in saturation). You might need to add a pull-up resistor on the base to keep it from floating around. You will still need a series resistor in the base to limit the base current.

author: permal
author_rank: Regular Contributor
timestamp: January 07, 2018, 05:56:56 pm
content: Quote from: rstofer on January 07, 2018, 05:03:01 pmYour're correct. I measure 3.8V at pin three when the transistor is desoldered. Very unexpected. I wonder where it gets its power from? Must be some back feed through some other pin. The I2C devices are working though so I don't really understand thisQuote from: rstofer on January 07, 2018, 05:03:01 pmAny chance you could draw a simple schematic showing what you mean with that? Do PNP generally give less of a drop across C-E?

author: rstofer
author_rank: Super Contributor
timestamp: January 07, 2018, 07:45:18 pm
content: Quote from: permal on January 07, 2018, 05:56:56 pmThere are voltages on the I2C device pins. There are transistors and diodes inside the device. Sooner or later, something conducts back to the Vcc rail. This will almost always happen. Just think about the Schottky clamp diodes on each pin. They conduct to the Vcc or Gnd rail any time the pin voltage is above or below the rail voltage. That's pretty easy when the Vcc rail doesn't have a source!QuoteI don't have a simple way to draw schematics. You just use the circuit you have but instead of the NPN, you use a PNP. The emitter goes to Vcc, the collector goes to the load and the base goes to your control circuit. You should have some kind of base resistor because you will be pulling it nearly to Gnd. If your control circuit can't pull the base up to Vcc - 0.6V or so, you will also need a pull-up resistor to guarantee the transistor shuts off.Rather than do a lot of calculating, I would put 10k between the base and Vcc and 1k between the base and the control circuit. Then I would check the voltages around the transistor to make sure Vce is about 0.25V when the transistor is on. Vcol should be 0V when the transistor is off.Note that your logic circuit it now inverted. A low level on the base turns the transistor on, a high level turns it off.A common 2N3905 or 3906 will probably work:https://www.centralsemi.com/get_document.php?cmp=1&mergetype=pd&mergepath=pd&pdf_id=2n3905.PDFThe voltage drop between collector and emitter should be the same for both NPN and PNP. The question is whether you can get the base high enough with your NPN configuration to force the transistor into saturation.You can see VceSat is about 0.25V for the 2N3905 at 10 mA of load. It varies with load and gets as high as 0.95V just before the magic smoke escapes.You didn't specify the I2C device you are controlling so I have no idea how much of a load it presents. The supply current for the Microchip MCP23017 IO Expander is only 1 mA.http://ww1.microchip.com/downloads/en/DeviceDoc/20001952C.pdfMaybe those resistors should be calculated. For the case of the expander, we have 3x as much current in the base as in the collector. Kind of a waste.

author: danadak
author_rank: Super Contributor
timestamp: January 07, 2018, 11:47:36 pm
content: Does the I2C device have pullups on the board connected to the on, or its owninternal pullups ? Some devices do.That could account for the voltage you see.The other possibility is power supply sequencing issues on two connected partsthat see different ramp rates on power up.Regards, Dana.

author: bson
author_rank: Unknown
timestamp: January 07, 2018, 11:47:51 pm
content: Quote from: permal on January 07, 2018, 04:47:28 pmAre the I2C bus pullups tied to I2C_PWR? If not, the bus inputs on some I2C device will leak across the TVS input protection and onto the I2C_PWR supply. This would manifest itself as the emitter on your BJT seeing approximately the pullup voltage (5 or 3.3V?) minus one diode drop (~0.65V).

author: permal
author_rank: Regular Contributor
timestamp: January 08, 2018, 05:46:11 pm
content: Quote from: bson on January 07, 2018, 11:47:51 pmOh, good thinking. They are indeed connected to the regular 3.3V rail.I'm going to drop this idea about cycling the power for I2C in the next revision of the H/W. Thanks all for your input, helpful as always.