High-side vs. low-side NPN transistor - Page 1
Forum section: electronics / beginners

author: permal
author_rank: Regular Contributor
timestamp: January 12, 2019, 03:15:48 pm
content: 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 thistutorial.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.

author: spec
author_rank: Frequent Contributor
timestamp: January 12, 2019, 03:24:56 pm
content: 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.

author: permal
author_rank: Regular Contributor
timestamp: January 12, 2019, 03:33:31 pm
content: CTRL is driven by an output of an MCP23017 (I2C IO expander).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 12, 2019, 03:43:04 pm
content: 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.pdfThis is probably much more effort and complication than you need so don't worry about it.Tim

author: spec
author_rank: Frequent Contributor
timestamp: January 12, 2019, 03:48:31 pm
content: Quote from: permal on January 12, 2019, 03:33:31 pmWhat supply lines are the MCP23017 using?

author: permal
author_rank: Regular Contributor
timestamp: January 12, 2019, 03:55:00 pm
content: Quote from: spec on January 12, 2019, 03:48:31 pmIts on 3.3V

author: spec
author_rank: Frequent Contributor
timestamp: January 12, 2019, 04:09:35 pm
content: Quote from: permal on January 12, 2019, 03:55:00 pmOK thanksWhat 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?

author: permal
author_rank: Regular Contributor
timestamp: January 12, 2019, 04:17:02 pm
content: Quote from: T3sl4co1l on January 12, 2019, 03:43:04 pmUhm, can you repeat that in English please?Are you saying my measuring is causing side effects?Quote from: spec on January 12, 2019, 04:09:35 pmI 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.

author: rstofer
author_rank: Super Contributor
timestamp: January 12, 2019, 04:29:03 pm
content: Quote from: permal on January 12, 2019, 04:17:02 pmIn 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

author: spec
author_rank: Frequent Contributor
timestamp: January 12, 2019, 04:39:36 pm
content: Quote from: permal on January 12, 2019, 04:17:02 pmOK will doQuote from: permal on January 12, 2019, 04:17:02 pmThe 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?

author: spec
author_rank: Frequent Contributor
timestamp: January 12, 2019, 04:57:26 pm
content: + permalCan 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.

author: permal
author_rank: Regular Contributor
timestamp: January 12, 2019, 05:00:40 pm
content: Quote from: spec on January 12, 2019, 04:57:26 pmAbsolutely, as soon as I finish dinner. Sorry for the delay and thanks so much for your time.

author: spec
author_rank: Frequent Contributor
timestamp: January 12, 2019, 05:02:46 pm
content: Quote from: permal on January 12, 2019, 05:00:40 pmNo sweat. Take your time and enjoy dinner.

author: iMo
author_rank: Super Contributor
timestamp: January 12, 2019, 05:12:30 pm
content: 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.

author: Zero999
author_rank: Super Contributor
timestamp: January 12, 2019, 05:16:19 pm
content: 2.7V what you should expect with a base voltage of 3.3V.The base voltage is 3.3VThe base-emitter voltage is 0.6VSo the emitter voltage will be 0.6V less than the base.See the simple formulae below:VBE= 0.6VVB= 3.3VVE= VB= VBE= 3.3 - 0.6 = 2.7VWorking back:VB= 2.7V + 0.6 = 3.3V

author: iMo
author_rank: Super Contributor
timestamp: January 12, 2019, 05:22:58 pm
content: @Zero999: his base is not 3.3V and the emitter resistor is not 1k..

author: rstofer
author_rank: Super Contributor
timestamp: January 12, 2019, 05:37:19 pm
content: Quote from: imo on January 12, 2019, 05:22:58 pmThat 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.

author: iMo
author_rank: Super Contributor
timestamp: January 12, 2019, 05:40:02 pm
content: 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

author: rstofer
author_rank: Super Contributor
timestamp: January 12, 2019, 05:57:57 pm
content: Quote from: imo on January 12, 2019, 05:40:02 pmThe OP shows an output voltage and no load resistor, hence infinite output resistance and zero output current. Therefore, Vbeis 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.

author: permal
author_rank: Regular Contributor
timestamp: January 12, 2019, 06:03:08 pm
content: 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): 2.50VVb = 3.07VQuote from: spec on January 12, 2019, 04:39:36 pmI 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.

author: Zero999
author_rank: Super Contributor
timestamp: January 12, 2019, 06:16:44 pm
content: 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= 3.15VVBE= 0.34VVE= VB= VBE= 3.15 - 0.34 = 2.81V

author: permal
author_rank: Regular Contributor
timestamp: January 12, 2019, 06:46:42 pm
content: Ok, so my lesson here is that Vbehas a larger impact than what I've previously understood and thus that Vbehas 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 Veis 0V, then you can make use of entire Vbeto drive the transistor open.Likewise, for a PNP transistor, you want to use that on the high-side because you pull Vblow to drive it open, and thus get the full Vcbto do that.Did I get that right?

author: spec
author_rank: Frequent Contributor
timestamp: January 12, 2019, 07:38:51 pm
content: Quote from: permal on January 12, 2019, 06:03:08 pmExcellent that is what I expected and fits the theory. I will post an explanation of what is going on soon.

author: spec
author_rank: Frequent Contributor
timestamp: January 12, 2019, 07:48:43 pm
content: Quote from: permal on January 12, 2019, 06:03:08 pmMOSFETS 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.pdfhttps://www.onsemi.com/pub/Collateral/BC546-D.PDFhttps://www.onsemi.com/pub/Collateral/TIP42C-D.PDFhttps://www.diodes.com/assets/Datasheets/ZXTN25050DFH.pdfhttp://www.vishay.com/docs/65900/SI2312CD.pdfhttps://www.vishay.com/docs/64004/si2323dds.pdf

author: IanB
author_rank: Super Contributor
timestamp: January 12, 2019, 07:56:35 pm
content: Quote from: permal on January 12, 2019, 06:46:42 pmAnother 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 VBmust 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.

author: rstofer
author_rank: Super Contributor
timestamp: January 12, 2019, 08:03:21 pm
content: Logic Level MOSFETS might be better than BJTs but it sometimes takes 10V between the gate and source (equivalent to base and emitter) to turn a MOSFET fully on. You want to drive it to saturation to reduce heating.There can be enormous gate capacitance and this slows the transition to whatever the final gate voltage is going to be. During the transition period, the MOSFET is heating. Add two transitions (rising and falling edge) and that is the reason that PWM is typically at a low frequency. The designers try to make the transition time a small percentage of the total period. Take 1 kHz = 1 mS. If the rise time and fall time are 250 ns, then half the time is spent in the transition region where heating occurs. I just grabbed those numbers, they aren't real, but it is important to understand that uCs can't always deal with conventional MOSFETs and may have a problem with Logic Level MOSTFETs as well. You just have to work the numbers!Which leads us to why they invented MOSFET driver chips. These puppies can dump AMPS into the gate to get through the transition in a great big hurry.MOSFETs are great devices but the user better read the datasheet and understand what is going on. Vgs is a big deal, so are all the capacitances.

author: spec
author_rank: Frequent Contributor
timestamp: January 12, 2019, 08:49:44 pm
content: + permalHere is a mini teach-in covering Bipolar Junction Transistors (BJTs) (normal transistors) and Metal Oxide Silicon Field Effect Transistors (MOSFETS). To keep it simple, this teach-in only covers the two transistor types in their linear mode (amplifiers) and omits the switching mode. But have no fear, once you have got the linear mode, the switching mode is pretty straight forward.BJTs are available in complimentary types, NPN and PNP. Likewise MOSFETs are available in complimentary types, N type and P type. The only difference between the two types is that all voltages and currents are the opposite polarity. To simplify matters only the N versions of BJTs and MOSFETs will be covered further. So there will be NBJTs and NMOSFETs.In most fields of engineering you can form simple models of components that describe their fundamental operation, and it is very important not to let any of the more esoteric aspects cloud your view.NBJTsAn NBJT has three terminals, emitter, base, collector.In a linear circuit the collector is always more positive than the emitter, 5V to 50V say.The base is like a diode connected to the emitter.To make a diode conduct current you put 600mV across it, with the base being the more positive, If you put less than 600mV across the base emitter no current will flow, but if you put more than 600mV across the base emitter a large current will flow.The current (Ib) flowing from the base to the emitter will be amplified by the NBJT and will cause a higher current to flow from the collector to the emitter.The ratio of the collector current(Ic) and the base current (Ib) is the current gain of the NBJT, called HFE in data sheets.And that completes the basic model for an NBJT.But there is one very important characteristic to remember about NBJTs: the collector current will always be HFE * Ib, regardless of what the collector/emitter voltage is. This characteristic has some very important consequences, and is fundamental to many linear NBJT circuits.NMOSFETsAn NMOSFET has three terminals, source, gate, drain. These are analogous to the emitter, base, collector of an NBJTIn a linear circuit the drain is always kept more positive than the source, say 5V to 50V.The gate is completely insulated from the rest of the NMOSFET and cannot pass any current. This is radically different to the BJT base.If the gate voltage is the same as the drain voltage no current will flow from the drain to the source.As you make the gate more positive than the source a voltage will be reached when a small current starts to flow from the drain to the source, say 100uA. This gate/source voltage is the threshold voltage (Vth) of the NMOSFET.As you increase the gate/source voltage so the drain/source current will increase, exponentially.And that is it for NMOSFTsBut there is one very important characteristic to remember about NMOSfets: provided that VGS is kept constant the drain current will always remain constant regardless of what the collector voltage is. This characteristic has some very important consequences and is fundamental to many NMOSFET linear circuits.Here endeth the first leson.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 12, 2019, 09:16:48 pm
content: Quote from: permal on January 12, 2019, 04:17:02 pmI can't explain it much better than that; I know it's written in jargon, so do take the time to find out (or ask) about what the terms mean. Bias, forward and reverse, current flow, voltage difference...I have no idea what your level is, aside from having posted in "beginners". Unless I know better, I'm just going to go with what comes to mind. Help me help you, as they say.Tim

author: permal
author_rank: Regular Contributor
timestamp: January 12, 2019, 09:53:02 pm
content: Thank you all so much for these explanations and extra kudos to @Spec for his/her summary. I will be reading these thoroughly and will surely be back with more questions.Again, thank you!

author: spec
author_rank: Frequent Contributor
timestamp: January 12, 2019, 10:39:24 pm
content: Quote from: permal on January 12, 2019, 09:53:02 pmLast time I looked I was a his

author: spec
author_rank: Frequent Contributor
timestamp: January 13, 2019, 08:05:01 am
content: Quote from: rstofer on January 12, 2019, 08:03:21 pmOf course you have to look at the data sheets. And it is obvious that you have to take note of all the device characteristics. That is what design is all about.

author: spec
author_rank: Frequent Contributor
timestamp: January 13, 2019, 08:07:55 am
content: Quote from: rstofer on January 12, 2019, 04:29:03 pm2N3055

author: permal
author_rank: Regular Contributor
timestamp: January 13, 2019, 11:27:11 am
content: I'm starting to like MOSFETsQuestion re. them: In the SOA-diagrams a pulse length is mentioned, but it is assumed you know of *what*. I understand it as the pulse length of where Vgsis larger than its threshold voltage, i.e. Vgs> Vgs(th), is that correct?What if you have a high duty cycle, would that effectively be considered as a longer pulse? Am I correct in that the the rise time of Vgsfrom 0 to lowest voltage where RDS(on)is specified [N-channel] matters, i.e. if it is short enough the MOSFET would pass through the linear mode very quickly, effectively reducing the power dissipation and thus allowing for a longer pulse?@Spec: In your example schematics (thanks) you put a R120 in series with the gate - why is that needed if the gate of an NMOSFET already is so high (1014Ohms?) Is it purely as a protection for the MCP23017?

author: rstofer
author_rank: Super Contributor
timestamp: January 13, 2019, 04:13:32 pm
content: Quote from: spec on January 13, 2019, 08:07:55 amJust a common pass transistor. Not the one in the OP, just an example I am familiar with. Nevertheless, I would expect the datasheets to have the same information. Probably not in the same order but essentially the same. The numbers will be slightly different, of course, but the ideas are identical.