0
\$\begingroup\$

I have 4 of these 2n3904 transistor setups on my board:

enter image description here

A1 -- requires 5v on these 2 connected pins to be enabled. I control this by driving r2 high at 3.3v which should allow 5v to flow through the collector.

I am considering this alternative approach of using a sn74lv1t34dbvr .

It appears I could drive this as follows where the ESP32 is the microcontroller I want to control power to the FSTD chip, via the SN74 regulator chip I have added to the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab Does this make sense? Is this correct? Would you recommend this approach rather than my original idea of using the transistor?

\$\endgroup\$
  • 1
    \$\begingroup\$ The sn74lv1t34dbvr is NOT a voltage regulator. It is a buffer for digital signals. It accepts a signal at any typical logic voltage, and outputs a signal using the supplied voltage. You can use it to translate logic levels from low voltage (like 1.8V) to higher (like 5V.) \$\endgroup\$ – JRE Jan 11 at 20:49
  • \$\begingroup\$ If P1 and P2 require any substantial amount of current (more than 25mA continuously,) then this chip is not a good choice. If P1 and P2 are only enable signals, then that'd be OK. If P1 and P2 are the power pins, then that's not OK. \$\endgroup\$ – JRE Jan 11 at 20:53
  • \$\begingroup\$ It would be more useful to post the schematic diagram of the 2N3904 circuit instead of the layout. \$\endgroup\$ – JRE Jan 11 at 20:59
  • \$\begingroup\$ @jre According to it's datasheet here: mouser.com/ds/2/149/FSTD16211-1010341.pdf It looks to me as if VCC goes into pin 17, so the OE1 and OE2 pins should be signal only, making this all ok? \$\endgroup\$ – RenegadeAndy Jan 11 at 21:01
  • \$\begingroup\$ OE1 and OE2 are digital signals. Yes, switching those signals is an appropriate task for the sn74lv1t34dbvr. \$\endgroup\$ – JRE Jan 11 at 21:09
1
\$\begingroup\$

The sn74lv1t34dbvr is not a voltage regulator. It is a digital signal buffer that can also translate logic levels.

As such it is predestined for the task you want to perform: drive a 5V logic level input from a 3.3V logic level output.

You haven't posted the circuit around the 2N3904, so I'm not sure if you have it set up to invert the signal or not.

If you need an inverter, consider using an inverting buffer instead. Like the SN74LV1T04. It will also translate logic levels.

Or, just invert the program logic driving the signal.


Whether you should do it or not depends on other things.

The 2N3904 costs about half what the chip costs. The transistor and a couple of resistors is cheaper than the chip.

The chip is in an SOT23-5 housing. It is a small part. Some people aren't comfortable hand soldering them.

Maybe you need board space more than you need to save money. That'd make the chip a better choice.

\$\endgroup\$
  • \$\begingroup\$ It turns out I need to sink 5v, and pull those 2 pins low to enable it...which was my oversight. Would this be able to perform that job also in my current arrangement? \$\endgroup\$ – RenegadeAndy Jan 11 at 21:21
  • \$\begingroup\$ See additions to my answer. \$\endgroup\$ – JRE Jan 11 at 21:29
  • \$\begingroup\$ I am looking at that chip you posted - I don't really see where in the datasheet that it says it can handle taking a 5v signal to low? I am a bit confused, its probably saying it in language I don't quite understand correctly, can you highlight it for me? \$\endgroup\$ – RenegadeAndy Jan 11 at 21:36
  • \$\begingroup\$ It says it can translate between low voltage logic and high voltage logic. If you put in a low on the input, it will put out a high. The voltage level of the high is given by the voltage you are powering the chip from. Power it from 5V, and you get a 5V high. Put in a high on the input (3.3V) and it will put out a low - pulled down to zero volts. That is how logic circuits work. It isn't explicitly spelled out because everyone knows how logic circuits work. \$\endgroup\$ – JRE Jan 11 at 21:41
  • \$\begingroup\$ Hmm ok - I suppose for people in the industry this is all ok, for others its making it hard to embrace, i find that datasheet to be really really bad, doesn't even call out what the pins are for. So - if I put a vcc of 5v, then put a 5v on the input, it will allow my output to be pulled low from 5v. Which I want in this case right? If I had vcc of 3v, then presumably it wouldn't be too happy with 3v on input pulling the 5v signal low? \$\endgroup\$ – RenegadeAndy Jan 11 at 21:45
-1
\$\begingroup\$

enter image description here

THis chip is active low not high.

All transistors Base to Collector are inverting.

Therefore any open drain, open collector or 5 to 3V level shift or transistor will do. In high volume, designers choose parts from ROhm and TI that combine the 2 resistors for each transistor at a cost of 1 penny or less.

\$\endgroup\$
  • \$\begingroup\$ So are you saying that my current design would not work. Please answer the question specifically. If it not please show what would work . \$\endgroup\$ – RenegadeAndy Jan 11 at 21:16
  • \$\begingroup\$ Sorry I dont support people who vote -1 \$\endgroup\$ – Sunnyskyguy EE75 Jan 11 at 22:01
  • \$\begingroup\$ Especially when they ask the wrong question with errors. \$\endgroup\$ – Sunnyskyguy EE75 Jan 11 at 22:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.