5
\$\begingroup\$

I have recently started studying BJTs with an hands-on approach.

I am trying to make a simple switch by working in the cut-off and saturation regions and, while the equations seem quite clear to me now, I still can't wrap my head around WHICH values of the datasheet should be actually used in said equations. My circuit, I've replicated it on a breadboard and I'm using a DMM to measure the values

For example, let's look at this datasheet by Onsemi, I want to switch ON a fake load (a resistor) to feed it 90mA, so I went straight to this graph: The infamous graph, named FIGURE 4 on the datasheet and estimated a VCEsat of 105mV, a VBEsat of 900mV and an hfe of 10 (as shown in the graph!!).

So what I did then, I've calculated my resistances:

Rc = (Vcc-VfLED-VCEsat)/Ic= (5-0.105)/0.09 = 54ohm
Ib = Ic/hfe = 0.09/10 = 9mA
Rb = (Vcc-VBE)/Ib = (5-0.9)/0.009 = 457ohm

So I was expecting to measure around 90mA going into my collector. But that didn't happen.

Here are the actual measured values:

VCE=0.126V (in my expected range)
VBE=0.88V (in my expected range)
Ic=76.4mA (what??)
Ib=7.6mA (ok it's 10% of Ic but it's still not right)

As you can see, I'm getting capped at 7.6mA, using even smaller resistances at the Base doesn't change much... The voltages and the hfe on the other hand are normal...

A person on another forum told me to make the same circuit but they gave me values of the resistors: 1kohm on the base and 44ohm on the collector, and it actually worked: I've got Ib=3.6mA and Ic=94.4mA. This means that the actual hfe needed to switch my load should've been 26, not 10, while at the same time the measured VCE was 0.22V so I'm well within the saturation region.

Other people also recommended to do the same thing, but everyone basically refused to explain to me why stuff suddenly started working, or why it doesn't work in my (textbook) way.

Honestly, I can see by myself that, on another table, the datasheet states that I can switch 100mA with an Ib of 5mA.

The table

However, I do not understand then WHY is there also a graph (FIGURE 4) that explicitly shows an hfe of 10 (so it's a graph about the saturation region) and VCEsat/VBEsat values that aren't the ones I'm supposed to use in this case.

Of course datasheets are made like this for a reason, I want to understand which reason is it though, and I also want to understand in which case I'm supposed to use the graph and in which the table.

It's not about not knowing the equations, but about understanding what to put inside.

P.s. Before you ask:

  • Yes my PSU is stable, and has a max output of 2A.
  • Yes I'm taking the measurements in the right positions.
  • Yes I have an high quality DMM that isn't playing tricks on me.
  • No I don' think temperature is involved that much, I might be wrong tho.
  • Yes the pins are oriented correctly.
\$\endgroup\$
36
  • 1
    \$\begingroup\$ @PeteW it's 73mA, this is weird... 5V divided by 54ohm should give at least 90mA... \$\endgroup\$ Commented yesterday
  • 2
    \$\begingroup\$ Something I do not understand if for RB = 470Ω and RC = 56Ω, and you measured Vce = 0.126V and Ic =76.4mA. This means that your load resistance is RL = (5V - 0.126V)/76.4mA = 64 Ω. This means that your ammeter series resistance is around 10Ω. And everything is ok. You simply forgot to include the ammeter series resistance into consideration. \$\endgroup\$ Commented yesterday
  • 3
    \$\begingroup\$ Be aware that any meter you use to measure current will have its own resistance. And so will reduce the measured current. A typical DVM has 200-500mV burden when in the circuit. Even the test leads will have significant resistance that will affect the measured current when you place it in the circuit. You may measure the current more accurately by measuring the voltage across the resistors and calculating the currents using Ohms law. \$\endgroup\$ Commented yesterday
  • 1
    \$\begingroup\$ @PeteW I only have one multimeter, but tomorrow I can take it to work and test it with the Fluke \$\endgroup\$ Commented yesterday
  • 1
    \$\begingroup\$ @all - Hi all, Please remember that if/when someone works out an answer in comments, it should be actually posted by them as an answer per SE policy (that allows it to be upvoted / áccepted & allows the site to record the post as answered, rather than remaining unanswered). So when things reach that point, please avoid leaving the answer only in comments as that harms the site. TY \$\endgroup\$ Commented yesterday

2 Answers 2

Reset to default
8
\$\begingroup\$

If we believe your measurements:

VCE=0.126V  
VBE=0.88V 
Ic=76.4mA

The transistor is in saturation Vbc = 0.88V - 0.126V = 0.754V.

So from the transistor point of view, everything is alright. The transistor is saturated.

Therefore, the collector current is only determined by the load resistance and supply voltage.

Load resistance = Rc + ammeter series resistance (3.3Ω) + wire resistance. And this is where you should look for the reason why the current is less than the expected value. Simply do not use ampmeter. Just use the voltmeater only, and measure the voltage drop across the load resistance, Vcc and Vce, next use Ohm's law to calculate the current.

The value of the ammeter series resistance can be found in the user manual here: https://brymen.eu/wp-content/uploads/biall/102092/102092.INSTRUKCJA_EN..2014-11-04.1.pdf

\$\endgroup\$
1
  • \$\begingroup\$ Hi, I haven't managed to test the internal resistance of my DMM. However today evening I've tried to measure the voltage across the Rc of my first circuit (54ohm), which is 4.2V. # This means that Ic=4.2/54=0.077A which is still well under my expected ~90mA. # So I don't think it's my ammmeter who's causing trouble # Considering that my VCE is still 0.125V, and 4.2+0.125 = 4.325V, I have no idea where the other ~0.7V went, could VBC be interfering? It's 0.75V after all, however nobody ever said that it had to be considered in the calculation to find Rc. \$\endgroup\$ Commented 1 hour ago
4
\$\begingroup\$

Your interpolation of the log graph is a bit suspect (it's about halfway between 100mV and 200mV and you guessed 105mV but we'll ignore that... )

As G36 says you need to take the ammeter burden into account if you are attempting to measure current directly, and it's usually more accurate and convenient to just measure the voltage across a known resistance and do the calculation. It is also possible the battery in your DMM is dying and you're getting inconsistent (high) readings as a result. You do need to get to the bottom of that measurement issue, and it has nothing to do with the transistor, similar errors would appear if you were just measuring the current through a resistor connected across the supply voltage, and I suggest working with that simplified circuit first until everything is consistent and makes sense.

In any case, I would not be using a 100mA rated transistor to switch 94mA. You can use a much beefier transistor such as 2N4401 (or SS8050 in this case) and get much better and more consistent results with negligible cost difference. Actually I'd probably use a MOSFET unless cost was a huge concern, but that's another story. The BJT with a forced beta of 10 wastes a lot of base current, and that current has to be driven from somewhere - which may not be capable of driving that much with guaranteed output level. Even with a 20 forced beta it's still a lot. An SI2302 can switch 100mA with less than 7mV drop and 4.5V drive, for less than $9 for a reel of 3,000.

\$\endgroup\$
1
  • \$\begingroup\$ Hi, thanks for pointing out my reading error. # I have re-made the calculations but Rc is almost the same =(5-0.2)/0.09 = 53.3ohm. # Meanwhile, my goal for this circuit isn't to make it work efficiently (I could use an Ib=5mA and it would work even better!) but to understand why it's NOT working as supposed at all. # You've said running this BJT at almost the maximum of its capacity isn't recommended, but even so, it should still work, right? Yet it doesn't, yet the manufacturer shows that beta=10 graph (FIGURE 4). # I want to understand why, then of course I could give up after a bit. \$\endgroup\$ Commented 56 mins ago

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.