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Given the above configuration and Vcc = 5V, Ie = 0.5mA, β = 100 and Vc = 1.7 how can i calculate the values of the two resistors? By approximating Ie = Ic, I have calculated Rc = 6.53 KΩ but I don't know how to calculate the base voltage to find Rb. Thanks in advance for any help!

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    \$\begingroup\$ No need to approximate Ie = Ic + Ib, Rc is 6k6. \$\endgroup\$ – sstobbe May 30 '17 at 23:51
  • \$\begingroup\$ Yes but isn't Ic = β/(β+1) * Ie? That's why I made that approximation. Is it wrong? \$\endgroup\$ – Peter May 30 '17 at 23:55
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    \$\begingroup\$ @Peter In this specific case the emitter current is first split into a base and collector current, but then these are just recombined back into the current in \$R_C\$. So the current in \$R_C\$ is the exact same value as the emitter current. It has to be. You can see that, can't you? \$\endgroup\$ – jonk May 30 '17 at 23:57
  • \$\begingroup\$ @jonk Yes, I can see that. I have used this equation (Ie = Ic + Ib) to calculate Ib. But I need to calculate the resistance Rb and for that I also need Vb. How can I find that? \$\endgroup\$ – Peter May 31 '17 at 0:03
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    \$\begingroup\$ @Peter Estimate a base voltage. You can call it out of thin air. But I'd guess it is \$630\pm 20\:\textrm{mV}\$. That's just based on "small signal" and "room temp" and a vague notion I have about what to expect. If that's taken away from me, I'd just use the usual \$700\:\textrm{mV}\$ usually picked out of the air, otherwise. Can you get \$R_B\$ from that, then? \$\endgroup\$ – jonk May 31 '17 at 0:05

I gather you'd like to see all the calculations for something like this.

I have to precede any discussion by pointing out what everyone else will point out: "BJTs vary widely. Even with the same part number. You can't calculate exact values and expect to see them if you wire up a real circuit. Reality impinges and what you will observe will be different from what you calculate from theory."

That said, there actually is theory. So one can at least use modest theory to make calculations. I think it helps to know how these things can be done, not because you can calculate exact numbers. But because you can then make reasoned estimates about the boundaries of those values and what "shapes" their behavior.

In that light, let's look at the schematic:


simulate this circuit – Schematic created using CircuitLab

From your given values, you can compute \$R_C=\frac{5\:\textrm{V}-1.7\:\textrm{V}}{500\:\mu\textrm{A}}=6.6\:\textrm{k}\Omega\$. And if you want to assume a fixed value for the base-emitter, such as \$V_{BE}=700\:\textrm{mV}\$, then it's really easy to compute \$R_B=\left(100+1\right)\cdot\frac{1.7\:\textrm{V}-0.7\:\textrm{V}}{500\:\mu\textrm{A}}=202\:\textrm{k}\Omega\$.

But \$V_{BE}\$, if known, determines \$I_C\$. Or conversely, \$I_C\$, if known, determines \$V_{BE}\$. So, since you know \$I_C\$ shouldn't it be the case that you can compute \$V_{BE}\$ from it? And the answer is "yes," but with all those caveats about reality and actual variations with real BJT parts.

In general, in the forward active region, the relationship between the collector current follows the mathematical pattern (model) shown by the Shockley equation (often cited for diodes, but applicable here):


In this case, \$V_T\$ is just the thermal voltage and is about \$26\:\textrm{mV}\$ at room temperature. (\$V_T=\frac{k T}{q}\$, so you can see that it is directly proportional to absolute temperature.) The above formula can be turned around to get:


(In most cases, the \$+1\$ term is tiny by comparison and can be ignored.)

So you can see that \$V_{BE}\$ can be computed. However, you need to know a value for \$I_{SAT}\$, which is the model parameter called the "saturation current." (This is just the y-axis intercept of a log-lin plot.)

Unfortunately, \$I_{SAT}\$ varies a bit. (It's value is quite small. For small signal BJTs, expect it to be in the area of about \$10^{-14}\:\textrm{A}\$.)

There is another way, if you don't know the value of \$I_{SAT}\$. If you happen to have an idea about what \$V_{BE_{CAL}}\$ is at some collector current \$I_{C_{CAL}}\$, then you can compute \$V_{BE}\$ for any other collector current:

$$V_{BE} \approx V_{BE_{CAL}}+V_T\cdot\operatorname{ln}\left(\frac{I_C}{I_{C_{CAL}}}\right)$$

So, if I know that my BJT exhibits \$V_{BE_{CAL}}=700\:\textrm{mV}\$ at \$I_{C_{CAL}}=5\:\textrm{mA}\$, then I would find for your collector current about this: \$V_{BE}=700\:\textrm{mV}+26\:\textrm{mV}\cdot\operatorname{ln}\left(\frac{500\:\mu\textrm{A}\cdot\frac{100}{101}}{5\:\textrm{mA}}\right)\$ or about \$640\:\textrm{mV}\$.

Of course, a different part from the box might give a different value because their saturation currents are different. But they only vary by about a factor of 2, one way or another. So the voltage variation can be bracketed.

So why go to all this trouble when any single BJT will have a different saturation current than another and where that saturation current might vary by a factor of 2 or 3 throughout a batch of the same part number?

Because it helps you to understand how these variations operate and how they might affect results. There are times when you really do need to know that there is a logarithm behavior going on and to know how that impacts questions you may have from time to time.

Let's take the case where a BJT family has \$I_{SAT}=3\times 10^{-14}\:\textrm{A}\$ but where this intercept value might vary from three times that much to only one third that much over a batch of parts. Then for \$I_C=10\:\textrm{mA}\$ you might compute:

$$\begin{align*}26\:\textrm{mV}\cdot\operatorname{ln}\left(1+\frac{10\:\textrm{mA}}{9\times 10^{-14}}\right)\le V_{BE}&\le 26\:\textrm{mV}\cdot\operatorname{ln}\left(1+\frac{10\:\textrm{mA}}{1\times 10^{-14}}\right)\\\\\\661\:\textrm{mV} \le V_{BE}&\le 718\:\textrm{mV}\end{align*}$$

That's \$\pm 29\:\textrm{mV}\$ over nearly a full order of magnitude change in the saturation current.

Doing this calculation shows why it is important to understand how this particular aspect of the BJT works. It teaches something else that actually is very important to gather up. Namely, that \$V_{BE}\$ will only vary by about the same amount for an order of magnitude change in its collector current. (What applies for variation in the saturation current, with the collector current held constant, also then applies to variation in the collector current with the saturation current held constant -- with it is for any specific device you have in hand.) So, this also tells you that even if the collector current were to increase by a factor of 10, you would not expect the base voltage to increase by more than about \$58\:\textrm{mV}\$. Huge collector current changes require only very small changes in the base-emitter voltage.

And that is an important lesson one can learn from theory, even if practice means that there are variations between parts.

  • \$\begingroup\$ Thank you very much for your analytical explanation! Much appreciated you cleared up so many things for me. \$\endgroup\$ – Peter May 31 '17 at 0:54
  • \$\begingroup\$ nice job +2.... \$\endgroup\$ – Sunnyskyguy EE75 May 31 '17 at 8:43

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