Author Topic: High-side vs. low-side NPN transistor  (Read 8812 times)

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Offline permalTopic starter

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High-side vs. low-side NPN transistor
« on: January 12, 2019, 03:15:48 pm »
Hi,

In another recent thread I learned that NPN transistors must be placed on the low-side of the load to fully open, as also demonstrated in this tutorial.

Before I redo my design I'd just like to have a behavior I'm seeing in my (non-compliant) circuit explained, see the attached image.

My question is why can I only measure 2.7V at the output to ground using a DMM? I'm thinking that there is such a low current that the voltage drop, Vce, should be negligible and thus read near 5V? Is this due to the transistor not yet having saturated because the emitter is offset from ground through the DMM?

Thanks in advance.

« Last Edit: January 12, 2019, 03:18:11 pm by permal »
 

Offline spec

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Re: High-side vs. low-side NPN transistor
« Reply #1 on: January 12, 2019, 03:24:56 pm »
Hi permal,

That circuit can never output a voltage higher than 4V4: there will always be the NPN transistor's 0.6V VBE drop from the 5V line.

But in order to find out what is going on with your circuit we would need to know what is driving CTRL.

« Last Edit: January 12, 2019, 03:28:21 pm by spec »
 

Offline permalTopic starter

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Re: High-side vs. low-side NPN transistor
« Reply #2 on: January 12, 2019, 03:33:31 pm »
CTRL is driven by an output of an MCP23017 (I2C IO expander).
 

Offline T3sl4co1l

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Re: High-side vs. low-side NPN transistor
« Reply #3 on: January 12, 2019, 03:43:04 pm »
For a 3.1V input, emitter voltage rising above 3.1V would imply not only that the base-emitter junction is reverse biased, but also that leakage current is flowing in the same direction, i.e., from emitter to base.  This can only happen if something is supplying current into the emitter!

The C-E path won't do this, because it's not actually a path, the B-C junction has leakage to the base.  In this region the transistor only acts like two reverse biased diodes smooshed together.

You can still get 5V output for some base voltages -- but you need to apply more than 5V to do it.  This is impractical in most circuits, as spec is referring to.

If the output is switched frequently (as is the case for switching supplies), a bootstrap supply can generate this extra voltage.  A number of LT converter chips come to mind, which internally use a bipolar high-side switch with an external bootstrap capacitor that supplies the transistor while it's on.  Example: https://www.analog.com/media/en/technical-documentation/data-sheets/lt1616fs.pdf  This is probably much more effort and complication than you need so don't worry about it. :)

Tim
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Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Offline spec

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Re: High-side vs. low-side NPN transistor
« Reply #4 on: January 12, 2019, 03:48:31 pm »
CTRL is driven by an output of an MCP23017 (I2C IO expander).
What supply lines are the MCP23017 using?
 

Offline permalTopic starter

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Re: High-side vs. low-side NPN transistor
« Reply #5 on: January 12, 2019, 03:55:00 pm »
What supply lines are the MCP23017 using?
Its on 3.3V
 

Offline spec

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Re: High-side vs. low-side NPN transistor
« Reply #6 on: January 12, 2019, 04:09:35 pm »
What supply lines are the MCP23017 using?
Its on 3.3V
OK thanks

What do you want?

Do you just want an explanation of what the circuit is doing (which is correct)

Or do you want a circuit for a low side switch, driven by the MC23017,  for the 5V line.

Or do you want a circuit for a high side switch, driven by the MC23017,  for the 5V line.

Or do you want all three?
 

Offline permalTopic starter

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Re: High-side vs. low-side NPN transistor
« Reply #7 on: January 12, 2019, 04:17:02 pm »
For a 3.1V input, emitter voltage rising above 3.1V would imply not only that the base-emitter junction is reverse biased, but also that leakage current is flowing in the same direction, i.e., from emitter to base.  This can only happen if something is supplying current into the emitter!

The C-E path won't do this, because it's not actually a path, the B-C junction has leakage to the base.  In this region the transistor only acts like two reverse biased diodes smooshed together.

You can still get 5V output for some base voltages -- but you need to apply more than 5V to do it.  This is impractical in most circuits, as spec is referring to.

If the output is switched frequently (as is the case for switching supplies), a bootstrap supply can generate this extra voltage.  A number of LT converter chips come to mind, which internally use a bipolar high-side switch with an external bootstrap capacitor that supplies the transistor while it's on.  Example: https://www.analog.com/media/en/technical-documentation/data-sheets/lt1616fs.pdf  This is probably much more effort and complication than you need so don't worry about it. :)

Tim
Uhm, can you repeat that in English please? :) Are you saying my measuring is causing side effects?

What do you want?
I want to understand why I get such a low voltage reading in this particular scenario - 2.7 < 4.4. So yes, an explanation is what I'm asking for.

Re the low/high side switches, as long as I put the transistor on the right side, I should be good if I've understood things correctly.
 

Offline rstofer

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Re: High-side vs. low-side NPN transistor
« Reply #8 on: January 12, 2019, 04:29:03 pm »

I want to understand why I get such a low voltage reading in this particular scenario - 2.7 < 4.4. So yes, an explanation is what I'm asking for.

Re the low/high side switches, as long as I put the transistor on the right side, I should be good if I've understood things correctly.

In normal operation, an NPN transistor base will be about one diode junction drop higher than the emitter.  It's 0.5V in the case of the 2N3055.

So, if the base is 3.1V, I wouldn't expect the emitter to be higher than 2.6V.


See Vbe(on) - second page here:
https://www.onsemi.com/pub/Collateral/2N3055-D.PDF



 

Offline spec

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Re: High-side vs. low-side NPN transistor
« Reply #9 on: January 12, 2019, 04:39:36 pm »
I want to understand why I get such a low voltage reading in this particular scenario - 2.7 < 4.4. So yes, an explanation is what I'm asking for.
OK will do

Re the low/high side switches, as long as I put the transistor on the right side, I should be good if I've understood things correctly.
The NPN transistor, ZXTN25050FHTA, is only really suitable for a low side switch (you have it in the high side), which is the ideal way to have a 3V3 logic chip switch a load connected to 5V. I will post a schematic to illustrate.

It would be much better to use MOSFETs for the switching rather than BPJTs like the ZXTN25050FHTA, Are you OK with MOSFETs?
« Last Edit: January 12, 2019, 04:50:50 pm by spec »
 

Offline spec

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Re: High-side vs. low-side NPN transistor
« Reply #10 on: January 12, 2019, 04:57:26 pm »
+ permal

Can you connect a 1k resistor from the emitter of the ZXTN25050x transistor to 0V and measure the voltages between the transistor's emitter and 0V and between the transistor's base and 0V. If you post the readings that will help with the explanation.
 

Offline permalTopic starter

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Re: High-side vs. low-side NPN transistor
« Reply #11 on: January 12, 2019, 05:00:40 pm »
+ permal

Can you connect a 1k resistor from the emitter of the ZXTN25050x transistor to 0V and measure the voltages between the transistor's emitter and 0V and between the transistor's base and 0V. If you post the readings that will help with the explanation.
Absolutely, as soon as I finish dinner. Sorry for the delay and thanks so much for your time.
 

Offline spec

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Re: High-side vs. low-side NPN transistor
« Reply #12 on: January 12, 2019, 05:02:46 pm »
+ permal

Can you connect a 1k resistor from the emitter of the ZXTN25050x transistor to 0V and measure the voltages between the transistor's emitter and 0V and between the transistor's base and 0V. If you post the readings that will help with the explanation.
Absolutely, as soon as I finish dinner. Sorry for the delay and thanks so much for your time.
No sweat. Take your time and enjoy dinner.  :)
 

Offline iMo

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Re: High-side vs. low-side NPN transistor
« Reply #13 on: January 12, 2019, 05:12:30 pm »
I think there has been an answer already - the base of an NPN is always by Vbe (0.5-0.65V) higher than the emitter. With pretty low emitter currents the Vbe is a bit lower.
« Last Edit: January 12, 2019, 05:15:17 pm by imo »
 

Online Zero999

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Re: High-side vs. low-side NPN transistor
« Reply #14 on: January 12, 2019, 05:16:19 pm »
2.7V what you should expect with a base voltage of 3.3V.

The base voltage is 3.3V

The base-emitter voltage is 0.6V

So the emitter voltage will be 0.6V less than the base.

See the simple formulae below:
VBE = 0.6V
VB = 3.3V
VE = VB = VBE = 3.3  - 0.6 = 2.7V

Working back:
VB = 2.7V + 0.6 = 3.3V


« Last Edit: January 12, 2019, 05:19:03 pm by Zero999 »
 

Offline iMo

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Re: High-side vs. low-side NPN transistor
« Reply #15 on: January 12, 2019, 05:22:58 pm »
@Zero999: his base is not 3.3V and the emitter resistor is not 1k..
 

Offline rstofer

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Re: High-side vs. low-side NPN transistor
« Reply #16 on: January 12, 2019, 05:37:19 pm »
@Zero999: his base is not 3.3V and the emitter resistor is not 1k..

That doesn't matter.

His base is 3.1 and the emitter is about 0.5V less which is as expected.  Within reason, the value of the load resistance doesn't change things very much.  There will always be a one diode junction drop between the base and emitter because there actually is a diode in there.

You simply don't use NPN transistors for high side switching unless you can get the base one diode junction drop higher than the required output voltage and this is usually 0.2V less than Vcc (the Vbe(sat) voltage)  In other words, the base needs to be about 0.3V higher than Vcc to get the maximum output voltage.  This is often very difficult to do.

Short answer, don't use NPN for high side switching.  That's why they invented the PNP variant.
 

Offline iMo

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Re: High-side vs. low-side NPN transistor
« Reply #17 on: January 12, 2019, 05:40:02 pm »
It does not matter from EE perspective, sure. It does matter when the OP is showing a specific schematics and asking on specific voltage at specific node :)
 

Offline rstofer

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Re: High-side vs. low-side NPN transistor
« Reply #18 on: January 12, 2019, 05:57:57 pm »
It does not matter from EE perspective, sure. It does matter when the OP is showing a specific schematics and asking on specific voltage at specific node :)

The OP shows an output voltage and no load resistor, hence infinite output resistance and zero output current.  Therefore, Vbe is going to be as low as it will ever be - about 0.5V for the 2N3055.  This value will increase as the load increases as shown in Figure 8 of the datasheet I linked above.

At 10A, Vbe(sat) could be as high as 1.8V but the OP circuit isn't delivering 10A, it's delivering zero current.  The reason the voltage has to rise is that the transistor is going to require a LOT of base current - probably on the order of 1A, and to push 1A through the base-emitter junction is going to require higher base voltage.


 

Offline permalTopic starter

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Re: High-side vs. low-side NPN transistor
« Reply #19 on: January 12, 2019, 06:03:08 pm »
Lots of people chiming in now. I appreciate the attention.  :-+

@Spec: With a 1k resitor what I get is this:
V across resistor (so to ground): 2.50V
Vb = 3.07V

It would be much better to use MOSFETs for the switching rather than BPJTs like the ZXTN25050FHTA, Are you OK with MOSFETs?

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?

@imo / @Zero999: I see your discussion, but opt not to comment on it at this time as I'm trying to understans what you're saying. It has become apparent to me that my understanding of transistors have been slightly off.

 

Online Zero999

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Re: High-side vs. low-side NPN transistor
« Reply #20 on: January 12, 2019, 06:16:44 pm »
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
 

Offline permalTopic starter

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Re: High-side vs. low-side NPN transistor
« Reply #21 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?

 

Offline spec

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Re: High-side vs. low-side NPN transistor
« Reply #22 on: January 12, 2019, 07:38:51 pm »
With a 1k resitor what I get is this:
V across resistor (so to ground): 2.50V
Vb = 3.07V
Excellent that is what I expected and fits the theory. I will post an explanation of what is going on soon.


 

Offline spec

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Re: High-side vs. low-side NPN transistor
« Reply #23 on: January 12, 2019, 07:48:43 pm »
It would be much better to use MOSFETs for the switching rather than BPJTs like the ZXTN25050FHTA, Are you OK with MOSFETs?
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?
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 easily
MOSFETS 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
« Last Edit: January 12, 2019, 10:12:03 pm by spec »
 

Offline IanB

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Re: High-side vs. low-side NPN transistor
« Reply #24 on: January 12, 2019, 07:56:35 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?

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.
« Last Edit: January 12, 2019, 09:00:55 pm by IanB »
 


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