# Voltage Drop across NPN BJT C&E?

Discussion in 'General Electronics Discussion' started by chopnhack, Oct 5, 2016.

1. ### chopnhack

1,573
354
Apr 28, 2014
Scrapping the MOSFET idea due to static sensitivity, I would like to use a N channel BJT to turn on/off a circuit. The base will be taken high by a pic output. With some testing tonight using a S9018 transistor, I found that by biasing the base with various resistances, I was able to get different voltages across the collector and emitter.

With Vcc at 2.97VDC using a base resistor of:
3k0 = 2.350V was measured between the collector and emitter
4k7 = 2.517V
6k8 = 2.620V
7k5 = 2.660V
8k2 = 2.680V ~320μA

So very little current was required to saturate the base and diminishing returns are seen past 8k2 resistance. Actually using a trimpot at 10k resistance, my circuit ceased functioning... not sure why (perhaps not enough current?).

Questions:
1. What governs the voltage drop seen between collector and emitter? What values can I compare on the datasheet to find a more efficient BJT? I am reading a loss of about 0.29VDC.
2. What values on the datasheet are responsible for determining the current necessary to saturate the base? If I wanted to saturate the base using a smaller current, how could I find that transistor?
The model below reflects what I used to 'see' the immediate results during rapid testing (brightness of LED). I removed the LED and the 100Ω resistor and used a jumper when getting the values listed above.

2. ### hevans1944Hop - AC8NS

4,631
2,159
Jun 21, 2012
It isn't the base that saturates. It is the collector-emitter circuit. Saturation occurs when the depletion layer at the base-collector junction has been reduced to zero by forward biasing the base-emitter junction. The collector current increases and the voltage between the collector and emitter decreases with increasing base-emitter forward bias voltage, reaching a minimum of 0.1 to about 0.2 V from collector to emitter when collector saturation occurs. This is for common small-signal transistors such as the 2N2222 or 2N3904. Note that for this particular transistor, saturation is specified at 0.5 V with 10 mA of collector current and 1 ma of base current, which works out to hfe = 10.

You probably don't want a detailed discussion of how a BJT "really" works, but base current has nothing to do with it. Base current is a result of forward biasing a real base-emitter junction with a finite base thickness. In an "ideal" BJT the base is infinitesimally thin and forward biasing the base-emitter junction just changes the thickness of the depletion layer at the base-collector junction, allowing majority carriers from the emitter to drift across the base into the collector region without combining with minority carrier holes in the P-doped base region. If the base were zero thickness, clearly there would be no base current, and all the electrons that drifted from the emitter region into the collector region would become the collector current. None would combine with holes in the infinitesimally thin base to create a base current in an external circuit. Well, I am sure someone of a (more) pedantic inclination will step in here and make a more sophisticated explanation, but that's the best I can do without math.

You are lucky that, when the LED and 100 Ω resistor were replaced with a jumper wire, the transistor didn't fail. This no doubt occurred because the large value of base resistor, 3000 Ω minimum, did not allow much base-emitter forward voltage drop to occur as a result of the small base current. Most of the voltage applied to the base resistor was dropped across the base resistor, leaving very little voltage to forward bias the base-emitter junction.

If this circuit were being driven by a coin-cell, not much power was available to crispy-fry the transistor. In any event, at no time was the transistor driven into saturation, which would result in a collector-emitter voltage on the order of 0.5 volts or less.

Repeat this experiment with a 1 kΩ resistor between the collector and Vcc and raise Vcc to 10 V so you can get 10 mA of collector current at saturation. Depending on the "gain" of the transistor, you may be surprised at how low a value of base resistor is needed to drive the collector into saturation and close to common potential. Do the experiment for different values of collector resistor and plot the results. Pay attention to the power dissipation in the transistor (collector voltage x collector current) to avoid bricking it.

The datasheet for this NPN (not N channel... there is no channel) epitaxial transistor was published in November 2002, so the technology is very old. Back in the day it was probably considered to be a high-frequency capable device... current gain-bandwidth product around 1 GHz. Epitaxial construction allows for a very thin base layer and higher frequency response, but it was apparently difficult to control because this transistor has a huge range of hfe specifications: 28 to 198 sorted into six bins.

And don't give up on the MOSFET. The 2N7000 is not particularly sensitive to static.

#### Attached Files:

• ###### Fairchild NPN Epitaxial Transistor SS9018 datasheet.pdf
File size:
41.7 KB
Views:
73
Tha fios agaibh likes this.
3. ### BobK

7,682
1,688
Jan 5, 2010
Static sensitivity of MOSFETs is easily handled by having a resistor (even 1M) from gate to source. The gate should never be floating.

Bob

4. ### chopnhack

1,573
354
Apr 28, 2014
Hop, I will repeat with 5v and appropriate resistor to get 10mA and see what I get, hopefully tonight.

If the fet version is doable I will order a sample of bss108 or bsh135's. I thought that in circuit they are less vulnerable, but hadn't researched it.

5. ### hevans1944Hop - AC8NS

4,631
2,159
Jun 21, 2012
MOSFETs can be vulnerable to gate oxide punch-through because that insulating layer between gate and body channel is very thin. That is why @BobK said to always have some high-valued resistor between the gate and source pins to drain off static build-up. You can, of course, use less resistance if the circuit calls for it. Driving the gate directly from a CMOS totem-pole output is generally safe, although with PICs you might want to add that 1 MΩ resistor between gate and source because the PIC outputs can be high-impedance inputs (both output transistors turned off) until programmed as outputs. Still, if the PIC output is connected directly to the MOSFET gate, there should still be enough impedance to ground to bleed off any static charge no matter whether the PIC "output" is programmed as an output or as an input.

The problem you might have, placing the MOSFET in series with the coin-cell 3 V supply, is providing sufficient gate-to-source voltage from the PIC output to turn the MOSFET on. I've never tried this with only a 3 V supply. You are basically trying to create a source-follower but may lack the drive voltage from the PIC output to turn the MOSFET on hard enough to power up your comparator circuit.. As others have stated, there are MOSFETs with lower threshold voltages you might try. And you can try using the BJT approach you mentioned in this thread, although the same problem with turning the base-emitter voltage on hard enough to drive the collector into saturation still occurs.

Try this: connect the drain of the MOSFET to Vcc and connect the source to common through a 1 kΩ resistor to simulate the comparator circuit load. Connect the gate to the PIC output pin. You can also connect the 1 MΩ resistor from gate to source to guarantee there will be no static build-up problems. See if you can get most of the coin-cell voltage dropped across the 1 kΩ load from source to common by setting the PIC output to logic "1" state. Try the same experiment with a 5 V supply.

6. ### LvW

604
146
Apr 12, 2014
For my opinion, your explanation is logical and in accordance with your motto (Einsteins rule).
It is the changing thickness of the depeletion layer which matters only - and the thickness depends on the bias and the superimposed signal voltage only.

7. ### Colin Mitchell

1,416
314
Aug 31, 2014
I don't agree with a lot of the things you have said in #1 and in some of the other posts. Some of it is quite incorrect. I am surprised you are saying things like this.

8. ### LvW

604
146
Apr 12, 2014
Colin - of course, it is your legal right not to agree with some contributions.
However, don`t you think that it would be helpful if you could list all the points of disagreement and explain (justify !) your own view ?

9. ### Ratch

1,094
334
Mar 10, 2013
For your information, BJTs do not operate using "channels". The operate using diffusion. Also they are bipolar (charge carriers are both holes and electrons), so the designation of N-channel or P-channel does not apply or make sense for BJTs.

Ratch

10. ### Colin Mitchell

1,416
314
Aug 31, 2014
"Repeat this experiment with a 1 kΩ resistor between the collector and Vcc and raise Vcc to 10 V so you can get 10 mA of collector current at saturation. Depending on the "gain" of the transistor, you may be surprised at how low a value of base resistor is needed to drive the collector into saturation"

Suppose the gain is 100.
To get 10mA collector current we use 1k load.
The base requires one hundredth of the load current.
The base resistor will be about 100k. I don't call 100k base resistor a "low value"

"Note that for this particular transistor, saturation is specified at 0.5 V with 10 mA of collector current and 1 ma of base current, which works out to hfe = 10. "
Nowhere does it give these figures.

"Most of the voltage applied to the base resistor was dropped across the base resistor, leaving very little voltage to forward bias the base-emitter junction."

You seem to be denying or omitting the fact that the base current controls the collector-emitter current.
For a beginner it is easiest to say the transistor is amplifying the base current by the gain of the transistor.
When the base resistor is 3k, 0.3mA base current flows and the transistor will deliver 300mA collector current.
The transistor is capable of delivering this without being damaged.

"reaching a minimum of 0.1 to about 0.2 V from collector to emitter when collector saturation occurs. This is for common small-signal transistors such as the 2N2222 or 2N3904. "

How is it, the sheet specifies 1v saturation for 50mA base and 500mA collector current.

11. ### Ratch

1,094
334
Mar 10, 2013
No, Vbe controls the Ic of a BJT. The base current is an indicator, not a control of Ic. A BJT by itself is a transconductance amplifier (voltage Vbe controls current Ic). If you load the base with lots of resistance to make it a current generator, then you have a current amplifier circuit, not a single BJT.

Ratch

Last edited: Oct 6, 2016
12. ### LvW

604
146
Apr 12, 2014
This is just a claim - are you able to give any proof or explanation/justification?

13. ### hevans1944Hop - AC8NS

4,631
2,159
Jun 21, 2012
We are talking about a specific, ancient (November 2002), epitaxial-constructed transistor, the Fairchild SS9018, as linked to by the OP. This particular transistor has an hfe range of 28 to 198, sorted into six bins. It doesn't even make it up to 200 on its best day, Colin.

Try reading the datasheet, which I have kindly attached. On the first page, Under Electrical Characteristics, it specifies Vce (sat) is 0.5 V maximum for Ic = 10 mA and Ib = 1 mA. My rough calculation says that implies hfe = 10 under those conditions. Note that hfe is a measurement not a cause. The base current didn't cause the collector current to equal 10 mA. The base-emitter forward bias voltage did.

Yes, I absolutely do deny that the base current controls the collector-emitter current. It is the base-emitter forward biased junction voltage that controls the collector current. The fact that generations of technicians and electrical engineers think base current controls collector current is right up there with the 18th Century belief that phlogiston is responsible for combustion. It is bad physics and laziness on the part of teachers who promulgate it. Believe whatever you like, but base current is an effect, not a cause.

How is it, the datasheet (2N2222) specifies 0.3 V saturation for 15 mA base and 150 mA collector current?

Is there no resistance in the emitter-collector circuit that accounts for the increased voltage drop when the collector current increases from 150 mA to 500 mA? What do you reckon the saturation voltage would be, collector to emitter, if the collector current were raised to its maximum value of 800 mA? What do you reckon the saturation voltage would be, collector to emitter, if the collector current were lowered to, say 100 mA?

Oh, wait! Datasheet Figure 3, Collector-Emitter Saturation Voltage, says 0.1 V at 100 mA collector current. Looks like the "saturation voltage" is somewhat dependent on collector current, especially at collector currents greater than 100 mA. From 0.1 mA to 100 mA (three orders of magnitude change in collector current) the saturation voltage, collector to emitter, remains under 0.1 V, even decreasing slightly until 100 mA is approached and it thereafter increases with collector current.

Okay, I will now admit that sometimes I use an assumed hfe value to "guesstimate" collector current as a function of base current, and plug in 0.7 V for Vbe for "back of the napkin" figurin'... hell, my instructors were lazy too. That doesn't make it right, though. And it may account for my movement into digital electronics instead of continuing to "dabble" in analog circuit design with discrete transistors back around 1970 or thereabouts. Besides, all the heavy lifting had been done by then. Some really sharp analog engineers figured out how to package transistor amplifiers with all the details hidden from the outside world. No need to worry about how to bias a transistor, how to accommodate parameter variations, how to compensate for temperature changes... it's all done for you in this one little package. Just add power, connect inputs, use the outputs. Reminded me of component hi-fi stereo systems from the 1950s and 1960s. Plug and play.

#### Attached Files:

• ###### Fairchild NPN Epitaxial Transistor SS9018 datasheet.pdf
File size:
41.7 KB
Views:
71
Last edited by a moderator: Oct 7, 2016
LvW likes this.
14. ### Colin Mitchell

1,416
314
Aug 31, 2014
I think he probably means C9018.

15. ### chopnhack

1,573
354
Apr 28, 2014
No, it is an S9018 that I received from a Chinese E-bay seller as part of a sample of various transistors, so I don't believe its necessarily a Fairchild SS9018 as I linked, but it seemed to have similar characteristics as the Fairchild model. That datasheet was not only in English, but seemed more complete so that is why I included it.

Taking a rain day down here... if I have power tomorrow, perhaps I will have some time to work on this. Luckily before the storm I etched a few sot23-3 to dip conversion boards like I did for my sot23-5 comparator (@hevans1944 ) and ordered some BSS138's. If the order is not delayed due to the storm, perhaps I can cobble that up by sunday, if they get here!

From what I have read above it seems that the FET will be safe to use if not left floating, easy enough to do.

16. ### hevans1944Hop - AC8NS

4,631
2,159
Jun 21, 2012
Following Colin's remark, I Googled several variants of the "9018" part number. Everything I found that was a transistor was an NPN epitaxial transistor, and they were all probably derivatives of the original Fairchild part.

Epitaxy, the growth of thin layers one on top of the other, is how my government contractor customer built their gallium arsenide hetero-junction bi-polar transistor (HBT) integrated circuits, but they sub-contracted the epitaxial layering and doping of the original un-doped wafer to a third party who specializes in that sort of thing, growing aligned crystal layers with precise doping profiles. So what if a ten dollar GaAs wafer becomes a hundred dollar wafer after six or seven precisely doped layers are grown on it? Then add another thousand bux per wafer for adding the integrated circuitry. All my customer did was mask the wafer and add some circuits to it. And all I did was isolate those hundreds of thousands of tiny vertical transistors with oxygen ion beams, sending them through apertures in a thick photolithographic resist that my customer applied and developed to expose the isolation "trenches" between individual HBTs. It had to be a pretty thick resist because it had to stop and absorb up to 3 MeV oxygen ions.

This is pretty routine stuff today, but it was bleeding-edge state-of-the-art when Fairchild did the SS9018 in 2002. Although epitaxial transistors had been around since the early 1960s here and late 1950s in the USSR, their main goal initially was not high frequency response but to increase the collector-base breakdown voltage.

It is difficult to accurately control all the process parameters that determine how fast a crystal layer, aligned with the layer below it, grows inside the epitaxial coating chamber. The reactive gasses that are used are very dangerous, being toxic and explosively hypergolic if exposed to atmosphere. It took many years to work out all the kinks to where a company can now make money by epitaxy alone. My employer got out of the business of trying to make epitaxial thin films before I got there in 1996, but the former employees who did do that left behind several very dangerous gas bottles that I had to pay big bux to remove and safely dispose. You are playing with a bit of history, @chopnhack! Go fer it!

17. ### LvW

604
146
Apr 12, 2014
Yes - to me this is really a phenomenon: About 60 years after the BJT was invented we still can find in textbook and other knowledge sources two fundamentally different descriptions how the BJT works - in spite of the fact that there are several observable effects and circuit properties which clearly indicate voltage control (Re-feedback, Early-effect, tempco -2mV/K, diff. amp with tanh characteristic, log amplifiers, current mirrors,...).

hevans1944 likes this.
18. ### BobK

7,682
1,688
Jan 5, 2010
LvW,

We all know that it is actually the voltage that controls the collector current. We also know that the voltage controls the base current. Both of these functions are exponential. When you compose them, you get a linear relationship between the base current and the collector current. Given a black box with these functions, you could not prove that the base current is not controlling the collector current. So, given that it is a linear function instead of an exponential function, we choose to use it in calculations. This is not wrong since it provides the same answer.

Bob

hevans1944 likes this.
19. ### hevans1944Hop - AC8NS

4,631
2,159
Jun 21, 2012
Given a black box, some texts I have read (for beginners) posit a little guy inside the box who turns the wiper arm on a rheostat in response to whatever current is supplied to the input. I am guessing he uses a D'Arsonval current meter, but since the box is sealed and opaque, there is no way to find out.

Good point though, Bob. if you can't see what the man behind the curtain is doing, you might as well go with the (current) flow until you get out of the Land of Oz and find yourself in Kansas again... or Missouri, which is the "show me" state. I bet even Missouri engineers use base current to control collector current... because, as you pointed out, the calculations work. And, as previously "confessed," I use that paradigm too.

And while I am at it, many thanks to the moderator who removed my pithy observations about rainwater. My father used to throw that line at me when he didn't believe whatever it was I was telling him. It was his way of accusing me of trying to pull the wool over his eyes. Which may have been partly (or entirely) true, since I was a post-pubescent teenager at the time. Anyway, the comment was unnecessary and I apologize for offending anyone who happened to read it.

Hop

LvW likes this.
20. ### Ratch

1,094
334
Mar 10, 2013
Everyone here knows that I have been playing Whack-a-Mole with every BJT base current control freak that pops up here. They just can't seem to understand that base current is an indicator of the emitter/collector current, not the control of it.

The physics of the BJT determine what is in control. And, Vbe controls the diffusion, which in turn controls the emitter/collector current. Also, because the BJT is a diffusion device, it has a nonlinear curve for Vbe vs Ie/Ic.

Ratch

Last edited: Oct 8, 2016
hevans1944 likes this.