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PNP transistor switching voltage
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Audioguru:

--- Quote from: imo on August 13, 2018, 03:06:42 pm ---There is nothing wrong with that circuit now. A typo has been fixed.
If you have a better circuit you may post
--- End quote ---
Your R3 is in the wrong place and draws way too much current, more than the transistor Q2b that it turns off. My fix changed the value of R3 and placed it correctly parallel to the base-emitter of Q2.
I also increased the value of R1 so that it dopes not overload the MCU.
David Hess:
Below are some other variations.

The upper left is the variation Hero999 suggested with Q1 operating as a constant current source.  R1 and the resistor in series with the collector are optional.  The collector series resistor may be desirable to limit power dissipation in Q1; it could be replaced with a zener diode.  I like this one for general use.

The upper right circuit has a non-inverting level shifter (a digital low turns the PNP on) allowing an optional series RC feedforward network to speed up switching of Q2.  This configuration might be used in a switching regulator.  The resistor in series with the collector of Q1 (not marked) is again optional and reduces power dissipation; it could be replaced with a zener diode.  The resistor in the series RC circuit is often zero ohms.

The lower circuit uses a zener diode for the level shift but depends on the 12 volt supply being stable.  The resistor in series with the zener diode (12V - 0.6V - 2.5V = 8.9 volt zener diode) limits the drive current.  The optional series RC circuit speeds up operation.  The resistor in the series RC circuit is often zero ohms.

Note that the last two circuits require the logic output to provide the entire drive current plus the charge through the feedforward network if used.  If this is a problem, then a buffer may need to be added or a more powerful logic output used.  The advantage of these circuits is much faster switching because non-inverting operation allows the feedforward network to drive charge directly into and out of the base of the output transistor.  In practice, a Baker clamp on the output transistor might also be added to further speed up operation.
Zero999:

--- Quote from: imo on August 13, 2018, 03:06:42 pm ---There is nothing wrong with that circuit now. A typo has been fixed.
--- End quote ---
It was not a typo, but a missing/incorrectly placed part, which would cause smoke!

Don't worry about it, this kind of mistake is easy to make and is one of the reasons why forums like these exist.



--- Quote ---If you have a better circuit you may post, there could be several variants, sure.  :clap:
--- End quote ---
I was thinking of this, which has a faster switching speed and higher input impedance.

Q1 could have a pull-down resistor, if there's a possibility the base could be left floating, otherwise it's not needed.


--- Quote ---PS: As I wrote above we need more info on his requirements (LOAD, speed) as the currents/transistor_types have to be designed based on the requirements. The above circuit is "for example" only, the resistor values are "an example" only.

In case he wants to switch an inductor, or an electromotor, or a welding machine he needs much more components to add, indeed.

--- End quote ---
If the load is inductive, all that's needed is a diode in reverse parallel with it.


--- Quote from: Audioguru on August 13, 2018, 04:14:55 pm ---
--- Quote from: imo on August 13, 2018, 03:06:42 pm ---There is nothing wrong with that circuit now. A typo has been fixed.
If you have a better circuit you may post
--- End quote ---
Your R3 is in the wrong place and draws way too much current, more than the transistor Q2b that it turns off. My fix changed the value of R3 and placed it correctly parallel to the base-emitter of Q2.
I also increased the value of R1 so that it dopes not overload the MCU.

--- End quote ---
Yes, the base discharge resistor should be across the base-emitter connections, not the base resistor and emitter nodes.
IanMacdonald:
BC337 has a minimum hfe of 60 at 300mA, so assuming no more collector current than that, you need at least 300/60 mA of base current. 5mA.
 
A 1k resistor will pass just under 5mA if used in the later circuit, just under 12mA in the first circuit. Therefore either would be acceptable. (60 gain is a guaranteed minimum, it will be more like 100)

The thing you don't want  is insufficient base current because that will cause the transistor to be in linear mode instead of switching, in which case it will get hot. Whilst it can switch 300mA no bother (actually up to 800mA with a higher base drive) it can only dissipate 625mW in free air. Which means that at 300ma, the emitter-collector volts drop needs to be kept under 2 volts whilst on.

Since the base-emitter junction is a diode it will drop about 0.7 volts in the forward direction. Therefore you MUST have a base current limiting resistor of some kind. Otherwise the B-E junction will be a near dead-short across the 12v supply. Likewise for the interfacing NPN transistor.

Hope this explains how to work these things out.

https://www.onsemi.com/pub/Collateral/BC337-D.PDF
danners430:
Hey guys, sorry to have neglected this - very, very busy afternoon all of a sudden!!

I've already had a look in terms of transistor specs, but here are the requirements since you asked so nicely [emoji13]

The system is to switch a model railway signal, which can only be done on the high side, as they use a common negative. Each transistor will be switching a maximum of 2 LEDs, usually just 1, from a 12V supply, and switched using a microcontroller.

VCC = 12V
Vmcu = 5V
Il = 20mA
Rl = 600r
Ib = 5mA

Hope this helps :-)

Sent from my ONEPLUS A3003 using Tapatalk

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