1

I need to choose a value for R1 to get the bulb shining. I was aiming to deliver 2 A current in there.

In the circuit below I am using NPN BJT transistor with beta value of 100 and built-in B-C potential value of 0.7 V.

My approach to figuring the resistance is

$$I_b=\frac{I_c}{\beta}=0.02 A$$

$$R_1=\frac{V_{CC}-V_{BE}}{I_B}=\frac{5-0.7}{0.02}=215 \Omega$$

However, in simulation, this resistance yields 1.87 A collector current.

P.S: R1 of 200 Ohms would yield 2 A collector current in simulation

Essentially, my question is why are my calculations of resistance are wrong/flawed?

Kamran Poladov
  • 125
  • 1
  • 7
  • Following your simulator logic, gain would be: 18A/20mA = 900 approx. – Marcelo Coronel Apr 24 '20 at 20:50
  • Have a look [here](https://electronics.stackexchange.com/a/481317/38098). I actually targeted a device at the current you are looking to achieve.... though the circumstance there is a bit different. The process for developing a result is similar, though. – jonk Apr 25 '20 at 07:24
  • For a \$2\:\text{A}\$ load, it's usually better to select an NFET (though they may be more expensive and have their own idiosyncrasies.) This is because a single BJT current gain figure is rarely (except for super-beta) much more than a few hundred, or so. And that's assuming "active mode," which here you probably would prefer a "saturated switch" mode where the current gain is an order of magnitude lower. So for BJTs, you usually must stack up two BJTs, not one. This will increase the voltage drop across the "switch," reducing the remaining voltage available for the lamp. – jonk Apr 27 '20 at 02:30

2 Answers2

1

The \$\beta\$ value for a BJT applies to operation in forward active operation. In this circuit you want to operate the BJT in saturation. For saturation you should assume an \$I_c/I_b\$ ratio of maybe 1/10 of \$\beta\$.

That will lead you to use a lower value for R1. You probably still won't get exactly 2 A of \$I_c\$. Rather, \$I_c\$ will depend more strongly on the power supply voltage and the effective resistance of the load.

The Photon
  • 126,425
  • 3
  • 159
  • 304
  • I need BJT operating in active mode though. Why would I *want* it to be saturated? – Kamran Poladov Apr 24 '20 at 19:52
  • Also, in its current state, BJT is in active mode. My question is why the current supplied right now is not 2 A. – Kamran Poladov Apr 24 '20 at 19:54
  • 2
    Saturation will result in the least wasted power in the BJT. If you keep the BJT forward-active you won't be able to accurately control the load current, because \$\beta\$ varies from part to part, with temperature, etc. You can add some feedback to minimize that effect but it will make your circuit more complicated. – The Photon Apr 24 '20 at 19:54
  • @KamranPoladov, most likely your BJT is **not** fully in forward active, but on the verge of saturation. What do you measure for \$V_{ce}\$? – The Photon Apr 24 '20 at 19:55
  • $$V_{CE}=1.2V$$ which is >>0V, thus I assume BJT is in active mode – Kamran Poladov Apr 24 '20 at 19:56
  • Then the reason is that, as you can see from the voltage probe in your schematic, the \$V_{be}\$ for your transistor model is not 0.7 V but almost 1 V. – The Photon Apr 24 '20 at 19:59
  • That is exactly what I suspected but I checked the model details, it's built-in B-C potential is 0.75 V – Kamran Poladov Apr 24 '20 at 20:01
0

Simply put, your calculations are ok, your simulation is wrong.

What went wrong?. Please, don't blame the circuit simulator. The problem is that you're assuming that all transistors behave the same. That's why we insert models in circuit simulators. What model are you using for the transistor?. I think none. Your transistor is using default values for parameters(e.g gain, Vbe, I(s), etc.) which you don't know and you are assuming other values for those parameters.

Conclusion, insert a model for that BJT and simulate again. Cheers! :D

PD: the insertion of a model is dependent of the simulator you're using. Can't help you with that unless you say what simulator you're using.

Marcelo Coronel
  • 166
  • 6
  • Thanks for your answer! Let me clarify some things to you, I am using a virtual BJT of NPN type, it's not exactly a real life model, you are right. However, I have checked the default params for this transistor, and, as I said earlier, it's Vbe is 0.7 V. Something else is making base voltage higher than it's supposed to be which is so confusing to me... – Kamran Poladov Apr 25 '20 at 00:55
  • Vbe usually gets higher when more current is flowing through the collector of the BJT. In other words, it's dependant of Ic. There's not such a thing as a fixed value for Vbe alone, but there's a value for Vbe given a determined Ic with the BJT at a fixed temperature. – Marcelo Coronel Apr 25 '20 at 03:47