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High-side vs. low-side NPN transistor
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Zero999:
I got a bit confused when I read some of the replies which mentioned a 3.3V supply and 1k resistor.

Yes, with a potential divider after the 3.3V, making the base voltage 3.1V and a 10M mulitmeter loading the collector, the base-emitter drop will be lower, giving about 3.8V.

The lower the base current, the lower the base emitter voltage, so in this case it drops just over 0.3V, rather than the usual 0.6V. I've added some arrows which shows the reading you'd get on multimeter.

The key to understanding this is all voltages are relative to one another, with ground or 0V, the common point in the circuit from where all voltages are measured from.

VB = 3.15V
VBE = 0.34V
VE = VB = VBE = 3.15 - 0.34 =  2.81V
permal:
Ok, so my lesson here is that Vbe has a larger impact than what I've previously understood and thus that Vbe has to be equal to Vbe(sat/on) as specified by the spec for the transistor to fully open. This in turn is why an NPN shall be used for low-side switching; if Ve is 0V, then you can make use of entire Vbe to drive the transistor open.

Likewise, for a PNP transistor, you want to use that on the high-side because you pull Vb low to drive it open, and thus get the full Vcb to do that.

Did I get that right?

spec:

--- Quote from: permal on January 12, 2019, 06:03:08 pm ---With a 1k resitor what I get is this:
V across resistor (so to ground): 2.50V
Vb = 3.07V
--- End quote ---
Excellent that is what I expected and fits the theory. I will post an explanation of what is going on soon.


spec:

--- Quote from: permal on January 12, 2019, 06:03:08 pm ---
--- Quote from: spec on January 12, 2019, 04:39:36 pm ---It would be much better to use MOSFETs for the switching rather than BPJTs like the ZXTN25050FHTA, Are you OK with MOSFETs?

--- End quote ---
I don't know. I went with "regular" transistors because I though I understood them (which I've now learned that I didn't fully). All I want is to be able to turn on/off 5/12V (250-500 mA) on eight different outputs. I've already overcome lots of other obstacles in this project so MOSFETs would just be one more thing to learn. What's the benefit, lower gate(?) voltages?

--- End quote ---
MOSFETS are better in every way for your switch application, but their three main advantages are:

* They only require a voltage to turn them on, no current.
* When they are fully turned on they can have a much lower resistance than a BJT (ordinary transistor)
* They can handle higher currents more easilyMOSFETS are easy to learn, perhaps easier than BJTs. I will explain how both types work in another post.

But in the meantime, attached is a schematic showing circuits for low side and high side switches using BJTs or MOSFETs.

The low side BJT switch will conduct around 150mA with a voltage drop of around 100mV. The high side BJT switch will conduct around 1A5 with a drop of around 500mV.

On the other hand, both of the MOSFET switches will conduct 4A with a voltage drop of 250mV, or less.

http://ww1.microchip.com/downloads/en/devicedoc/20001952c.pdf

https://www.onsemi.com/pub/Collateral/BC546-D.PDF

https://www.onsemi.com/pub/Collateral/TIP42C-D.PDF

https://www.diodes.com/assets/Datasheets/ZXTN25050DFH.pdf

http://www.vishay.com/docs/65900/SI2312CD.pdf

https://www.vishay.com/docs/64004/si2323dds.pdf
IanB:

--- Quote from: permal on January 12, 2019, 06:46:42 pm ---Ok, so my lesson here is that Vbe has a larger impact than what I've previously understood and thus that Vbe has to be equal to Vbe(sat/on) as specified by the spec for the transistor to fully open. This in turn is why an NPN shall be used for low-side switching; if Ve is 0V, then you can make use of entire Vbe to drive the transistor open.

Likewise, for a PNP transistor, you want to use that on the high-side because you pull Vb low to drive it open, and thus get the full Vcb to do that.

Did I get that right?

--- End quote ---

Another way to look at it is that to turn on an NPN transistor you have to make current flow from B to E, which in turn means that VB must be greater than VE (or current wouldn't flow in that direction). This of course means that VE < VB, which is what you are measuring.
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