5V to 24V switch with transistors

Hey,
so I had a lot of problems with my recent transistor circuit and I think that I have a big missunderstanding in how transistors actually work but I dont get it ( breakdown of the circuit is attached ).
My idea was to get a BC337 triggering a BC557 with 5V and the BC557 outputting 24V. Between the Arduino and the transistor circuit is a shift register which I didnt refer to in the picture because I dont think that the problem is there.
And ofc my GNDs are put together.
Best regards

You've connected the transistors in reverse and blown them up.

For an NPN switching transistor you always wire:

emitter -> ground
base -> resistor -> arduino pin
collector -> load-
+ve supply -> load+

The way to blow up a transistor reliably is reverse-bias the base-emitter junction by more than 5V or so, which you have done.
For switching the base-emitter junction should either have 0V across it, or be forward biased via a current-limiting resistor.

For switching the base current should be 5 to 10% of the load current. The transistor gain parameter is not relevant.

With a PNP you swap all the voltages around, so its emitter to +ve supply, collector to load+, ground to load-

MarkT:
You've connected the transistors in reverse and blown them up.

For an NPN switching transistor you always wire:

emitter -> ground
base -> resistor -> arduino pin
collector -> load-
+ve supply -> load+

The way to blow up a transistor reliably is reverse-bias the base-emitter junction by more than 5V or so, which you have done.
For switching the base-emitter junction should either have 0V across it, or be forward biased via a current-limiting resistor.

For switching the base current should be 5 to 10% of the load current. The transistor gain parameter is not relevant.

With a PNP you swap all the voltages around, so its emitter to +ve supply, collector to load+, ground to load-

Thank you very much for your detailed answer!

I dont want you to make my work but I still have my problems with it, would it be possible to get a circuit for better understanding?

Best regards

OP's diagram:
Unbenannt.PNG

Hi,
This may help.
Note the BC557 is only good to 100mA current.
What is the application?


Tom... :slight_smile:

TomGeorge:
Hi,
This may help.
Note the BC557 is only good to 100mA current.
What is the application?


Tom... :slight_smile:

Thx it works, could you explain why I need those capacitors?
I have a vacuum fluorescent display which I want to controll with an arduino and some shift registers via multiplexing (picture attached).
Best regards, helped me a lot!

Hi,
Good to see it worked.
The caps are there for bypassing any digital noise that might get on the 24V supply.
Just a case of habit when doing basic power supply work with digital circuits.

Tom… :slight_smile:

If you use TPIC6B595 shift registers & low side switching you don't need those external transistors. They can switch loads of up to 50V and 150 mA (500 mA total per package).

Actually if you must do high side switching, it seems you can use those TPIC shift registers also as replacement for your existing shift register & BC337 - add a pull-up resistor (10k should do) to keep the BC557 firmly off when the output is off.

Azoni's circuit looks good except you need to add an Emitter to Base resistor on the output device, otherwise expect some artifacts on the display. C2 could also cause this type of problem, if it were me I would save it for another project. Check the gain of your output transistor, it is probably in the 10 to 20 range, not Hfe. A avalanche rated MOSFET would work nicely in this circuit. This response is to help you get started in solving your problem, not solve it for you.
Good Luck & Have Fun!
Gil

With regard to an avalanche rated MOSFET: Does that mean that the
body diode is a zener diode?
Herb

herbschwarz:
With regard to an avalanche rated MOSFET: Does that mean that the
body diode is a zener diode?

I have asked @gilshultz what it means before, as s/he's always suggesting this, even when inappropriate (like here). I never got a response, and I seriously doubt whether @gilshultz him/herself even knows what it is.

For lack of response and out of curiousity I've read up on it myself a bit. It has to do with the MOSFET being able to absorb some reverse current - this is helpful with an inductive load, probably not relevant to your lights. I don't know exactly how a vacuum fluorescent light works but doubt it's an inductive load.

This "avalanche rated" is just that: a rating, which is not measured for all MOSFETs. By definition, only MOSFETs where this property has been measured are "avalanche rated". Without fully understanding what it is and how much energy the inductive load is releasing when switched off it's probably best ignored. The fact that it's not even measured for most MOSFETs by the manufacturer tells me it's a rather unimportant parameter for most situations.

Adding a flyback diode to an inductive load does the job just fine, and seeing the size of the MOSFET (grain of rice) and some values of this avalanche rating vs. the coils (size of my fist) I'm routinely switching with them I doubt it's going to do much for any but the smallest of inductive loads.

Thank-you, wvmarle. FYI: Vacuum fluorescent lights are like neon lights.
When they first draw current and light up, they exhibit negative resistance!
Herb

There is no such thing as negative resistance… You mean resistance drops as they switch on?

Just read up on those displays. Interesting. Less current than LEDs? That’s not much! Main disadvantage is that they’re all common cathode, so high side switching.

The circuit as in #4 will do great. OTOH there just have to be driver ICs out there for just this application, doing the multiplexing and maybe even brightness control for you.

I only know the facts of the lowly NE-2 neon
glow tube that was used as a pilot light in
many AC powered units for many years. When
powered up by a DC power supply with a 100K
current limit resistor, it takes 90 volts to get one
to draw current and glow. While it is glowing, its
terminal voltage is only 60 volts! So, 90V when
off and 60 V when on! Once the neon gas ionizes
it takes much less voltage to remain ionized.
That is the reason for the resistor, otherwise the
NE-2 would destroy itself.
Herb

wvmarle please read Vishay’s Application Note AN-1005, http://www.vishay.com/docs/90160/an1005.pdf, It will partially explain what I am saying. “Some power semiconductor devices are designed to withstand a certain amount of avalanche current for a limited time and can, therefore, be avalanche rated. Others will fail very quickly after the onset of avalanche. The difference in performance stems from particular device physics, design, and manufacturing.” Also read APT9402 by Kenneth Dierberger UNDERSTANDING THE DIFFERENCES BETWEEN STANDARD MOSFETs AND AVALANCHE ENERGY RATED MOSFETs. You can get it at: http://www.ohm.com.tr/doc/Microsemi---Understanding-the-Differences-Between-Standard-MOSFETs-And-Avalanche-Energy-Rated-MOSFETs.pdf. Then when you understand what you just read and the testing (I have done it many times in characterizing MOSFETs) you might understand what I am saying. In that time frame the usage of a MOSFET as a high performance diode evolved.
Gil

FYI: In electronics, negative resistance is a property of some electrical circuits and devices in which an increase in voltage across the device’s terminals results in a decrease in electric current through it. You can simply check this out with a google search. Gil

RE: NE-2 negative resistance: Please see Neon NE-2 Circuits You Can Build
for some more explanation and circuits to enjoy.
Herb

Some additional information. You may want to goggle "MOSFET UIS", you might find my name. When doing this I got well over 100,000 hits. When you use an avalanche rated MOSFET many data sheets but not all will give you a Unclamped Inductive Switching (UIS) rating. This is how much energy the MOSFET will dissipate from a single pulse. This is a term we used in the early 90's to indicate how much energy the "body diode" will conduct before failure. The major failure mode is the FET junction gets hot and fails. There are repetitive tests for this and TTF tests as well. The MOSFET is connected with the source connected to ground, the gate is controlled and fully enhance for these tests. The drain is connected to one end of an inductor the other end is connected to the positive supply. Look at the test circuits on the data sheets, this is the typical configuration. For testing the inductor is charged, the MOSFET switched off and the return energy from the inductor is measured. I did these tests with a scope with both voltage and current probes, triggering from the gate. When the MOSFET turns off, the field in the inductor collapses, reverses polarity and the voltage will rise indefinitely until something dissipates the energy. You will find that MOSFETs do conduct in reverse and can even conduct in the third quadrant, this mode is used in SMPS and reverse battery protection is many circuits. This response is to help you get started in solving your problem, not solve it for you.
Good Luck & Have Fun!