Simple Single Transistor Driver - Page 1
Forum section: electronics / beginners

author: Gizmo
author_rank: Regular Contributor
timestamp: July 11, 2017, 10:45:13 am
content: I have a situation where I just need to be able to drive a signal low which is pulled high normally by a weak pull-up. The output signal is 5v TTL logic and the input is 3.3v LVTTL logic -- normally I'd use a level shifter, but in this case I just need to be able to pull the signal low, and there's just the one of them, so I'm looking for an alternative.Would using an NPN transistor here with the emitter tied to ground and the collector connected to the signal work ok? As far as I can tell, with a suitable current limiting resistor (which may or may not be required as its coming from a CPLD output) on the base this should pull down the signal to ground when the current is high enough on the base.The VC will naturally drop, but will never be below VE, and a dont think VB>VC is problem?

author: Benta
author_rank: Super Contributor
timestamp: July 11, 2017, 12:04:09 pm
content: A transistor with a base resistor in series will work fine. Remember that your output logic will be inverted by the transistor.

author: KL27x
author_rank: Super Contributor
timestamp: July 11, 2017, 12:26:10 pm
content: How do you think the level shifter IC does it? There are transistors driving the outputs. The low output is done with an N transistor. Now you know.A signal FET might work, too, and save you a base resistor. You might even get by with just a shottky diode between the output and input, if the resulting voltage is low enough. That would take the littlest amount of routing/wiring.

author: Gizmo
author_rank: Regular Contributor
timestamp: July 11, 2017, 01:07:37 pm
content: Quote from: KL27x on July 11, 2017, 12:26:10 pmWith the voltage difference / emitter being a higher voltage, this would not work with a FET?

author: tszaboo
author_rank: Super Contributor
timestamp: July 11, 2017, 01:17:13 pm
content: Congratulations, you described the open collector output. Now, that you know, how it is called, you can read up on websites, likewww.google.com

author: KL27x
author_rank: Super Contributor
timestamp: July 11, 2017, 01:31:34 pm
content: QuoteI don't follow. Emitter/source is gonna be on the ground. What is it 'posed to be higher than, and why do you care?

author: Gizmo
author_rank: Regular Contributor
timestamp: July 11, 2017, 01:37:22 pm
content: Quote from: KL27x on July 11, 2017, 01:31:34 pmYes, I'm being an idiot. It's gate to source voltage, not gate to emitter. Its a problem with a PNP MOSFET when the higher voltage is on the source, not for an NPN with a grounded source. Sorry.

author: KL27x
author_rank: Super Contributor
timestamp: July 11, 2017, 03:42:49 pm
content: QuoteI'm not sure what you mean, here, either. The PFET only switches on when the source voltage is higher than the gate. You couldn't do a whole lot with a PFET if this "problem" never happened.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: July 11, 2017, 03:57:46 pm
content: Quote from: Gizmo on July 11, 2017, 01:37:22 pmYou sound pretty confused, your terminology is all over the place.BJT = base, collector, emitter, comes in NPN or PNP varietyMOSFET = gate, drain, source, comes in Nch or Pch varietyNo such things as "gate to emitter" voltage on either a BJT or MOSFET, since you don't have a gate on a BJT or an emitter on a MOSFET. An IGBT has a gate, collector, and emitter, but no need to go there with this circuit. No such thing as a PNP or NPN MOSFET either.A simple logic-level Nch MOSFET would do fine. I'm a fan of the BSS138 personally, but there are a lot of options. You just want the Vgs(th) to be less than half of your drive voltage. Since your drive is 3.3v, that means a Vgs(th) of <=1.6V.

author: Zero999
author_rank: Super Contributor
timestamp: July 11, 2017, 06:11:34 pm
content: How fast does it need to be?Here's an example showing 3V to 12V level shifting. It will work with 5V too. If higher speeds are required, R1 can be bypassed by a small capacitor: 47pf to 1nF and your input may need a pull-up if the weak one isn't strong enough to discharge the base-emitter fast enough.

author: KL27x
author_rank: Super Contributor
timestamp: July 12, 2017, 10:55:35 am
content: Ah, ok.I think mebbe OP was concerned that [Vgs/Vbe] is going to be greater than [Vds/Vce], since he wants to drive the gate/base with +5V and have only 3.3V on the collector/drain. No, this is not a problem. The BJT is current driven. It doesn't care what the voltage is, as long as it's high enough to overcome the voltage drop between B and E. The FET is voltage driven, but it also doesn't care, as long as you observe maximum Vgs. In fact, the majority of FETs are used in a way where  is smaller than  once the FET switches fully on.

author: Mechatrommer
author_rank: Super Contributor
timestamp: July 12, 2017, 11:44:33 am
content: Quote from: Hero999 on July 11, 2017, 06:11:34 pmnice circuit (non inverting 1 transistor level shifter).. just to be aware of the mcu/fpga/chip will sink Q1's collector current during output low (1.2mA @ 12V shift in the example), and also the output low is 2 transistor's Vce(sat), Q1's Vce(sat) and mcu/fpga/chip's transistor's Vce(sat)... if that matters, ymmv..

author: bson
author_rank: Unknown
timestamp: July 13, 2017, 02:32:32 am
content: You can also use a 1G version of a 74LVC04; the inputs are 5V tolerant with a 3.3V supply. This way you won't have any potential issues with slow transition times.https://assets.nexperia.com/documents/data-sheet/74LVC1G04.pdf:5 V tolerant inputs for interfacing with 5 V logic.

author: Zero999
author_rank: Super Contributor
timestamp: July 14, 2017, 11:31:17 am
content: Quote from: Mechatrommer on July 12, 2017, 11:44:33 amYes, it's common base (kind of, not true common base because of the resistor) so has no current gain, just voltage gain but is the only way to do, non-inverting, low voltage to high voltage level translation, with only one transistor.