Finding the Real $V_{BE}$

In a BJT in common emitter configuration, what formula can you use to calculate the real $V_{BE}$ so that you don't have to approximate it as $0.7V$? The problem below asks for the actual value of $V_{BE}$ to see the difference it makes in the other parameters when $V_{BE}$ is not assumed to be $0.7V$. However, I can't find a way to calculate the actual $V_{BE}$ because it is usually just assumed to be $0.7V$. 1) Find the collector current, $I_{C}$ , base current, $I_{B}$ , and the output voltage, $V_{OUT}$ using $V_{BE}=0.7V$. State all necessary assumptions.

2) Solve for the real value of base-emitter voltage, $V_{BE}$ @ T = 300K, and the corresponding $I_{C}$, and $V_{OUT}$. Assume that the transistor parameters given were measured at 300 K. State all other necessary assumptions.

• use the data sheet of the BJT, it will be specified as VBE (sat) you will find minimum, typical and maximum values for it, use typical value for normal calculation, also you might see graphs plotted over temperature changes – Raj Feb 5 '17 at 8:08
• Does "real" imply a transistor in the real world, where you don't know the exact temperature, or anything about manufacturing variations? – CL. Feb 5 '17 at 8:28
• You should know it for the rough category of device, by heart, and know what parameters affect the value ($I_S$ for example.) (I know, for example, that I tend to get about $700\:\textrm{mV}$ with $I_C=3\:\textrm{mA}$ for parts similar to 2N3904 and 2N2222A at room temps.) Then you should design to handle the range of device variations you will purchase, temperatures in operating environments you need to support, and long term drift variations. And some situations will be more picky than others. So what efforts you go to, will depend on the application, too. – jonk Feb 5 '17 at 8:55
• You should know that $V_{BE}$ varies by $60\:\textrm{mV}$ per decade change in $I_C$ and that you should also expect about $\frac{-2.2\:\textrm{mV}}{^\circ C}$ change, as well, though that doesn't vary linearly over too wide a range and it can be as little as perhaps $\frac{-1.7 \: \textrm{mV}}{^\circ C}$ and perhaps as much as $\frac{-2.5 \: \textrm{mV}}{^\circ C}$ for surprisingly similar application devices. – jonk Feb 5 '17 at 8:59
• I suggest you do this exercise: solve that CE circuit for both Vbe = 0.7 V and Vbe = 0.75 V. Then look at the resulting change in behavior of the whole circuit. The conclusion should be that it does not change much. Circuits which would rely on a certain precise value of Vbe are impractical. The value of Vbe does not matter much, it is pointless to try to determine an accurate value. – Bimpelrekkie Feb 5 '17 at 11:19

2) Solve for the real value of base-emitter voltage, VBEVBE @ T = 300K, and the corresponding ICIC, and VOUTVOUT.

The Ib/Vbe from solution 1) will unlikely to be consistent. So you will iterate between the calculations:

2.1) From the Ib, calculate a new Vbe, using the schockley's equation;

2.2) from the new Vbe, calculate the corresponding Ie/Ic/Vout;

2.3) from the new Ic/Ie, calculate the Ib;

2.4) go back to 2.1) until the solution converges sufficiently.

That's basically how a spice simulator would have done.

you will find that your solution is sufficiently close to 0.7v (most trnsistors will have a vbe between 0.6-0.7v).

• I see. So I would use the $I_{B}$ that resulted from the assumption that $V_{BE}=0.7V$ – John Smith Feb 5 '17 at 17:19

The 'real' VBE cannot be 'calculated', without making some assumptions, it is what it is at any given temperature and current. The reason we approximate to 700mV, is because that's what it is at room temperature, for a handful of mA, when it's permissible to approximate.

It varies with temperature, about 2mV/C IIRC, and it also varies with current. At low currents it varies according to the diode equation, and at higher currents the residual series resistance of the junction creates an additional drop. I've recently done some measurements of VBE for making a logamp with a transistor, from pA to >100mA, and it varied between 200mV and 900mV.

• Thanks, I edited the question to add more detail to the nature of the problem. – John Smith Feb 5 '17 at 15:58
• I don't see anywhere in the parts of the question you posted where it asks you to calculate VBE, I do see where it tells you to use 0.7v. It might be useful for your own benefit to assume values of VBE of 0.6v and 0.8v, and recalculate the bias conditions. In a well biassed device, the collector current/voltage will not change much. In my first year in industry I was asked to debug an amplifier with a temperature dependent frequency response. They'd taken RB to rail instead of the collector, and as VBE changed with frequency, so did the collector volts and Miller capacitance collector to base. – Neil_UK Feb 5 '17 at 16:39
• Okay, I decided to paste the specific questions that the problem asked. Sorry about the vagueness. You can view the question again. – John Smith Feb 5 '17 at 16:52
• You are given the Is of the transistor as 20fA, this is the saturation current in the diode equation, which you will have been given somewhere else in your handouts, the transistor beta, the Early voltage, all at 300K, so you don't have to worry about temperature. What assumptions should you make, and state? Perhaps that the residual BE resistance is zero so you don't need to add an IR drop to the diode-calculated VBE, or that that resistance is some small finite value, which you state. – Neil_UK Feb 5 '17 at 17:56

The "real" VBE value - for a given collector current Ic - can be calculated only - based on Shockleys equation Ic=Is*[exp(Vbe/Vt)-1] - if you know the value of the saturation current Is. However, this current has very large tolerances and a rather large dependency on temperature. Hence, this value is normally unknown.

Therefore, for calculation purposes (design of BJT stages) we make use of approximate values (0.65...0.7 volts). However, the enclosed graph shows that - in spite of such an approximation - the impact on the desired collector current (deviation from the design value) can be kept within tolerable limits. As we can derive from the graph, for a good design the slope of the stabilization line should be as low as possible.

The graph shows two exponential Ic-VBE functions for two different temperatures - however, the same principle (shifting of the curves) applies for two different Is values (tolerance influence).

• Thanks, I edited the question to add more detail to the nature of the problem. – John Smith Feb 5 '17 at 15:59

Having done both discrete and integrated-circuit design, I've learned to use the stabilized-current approach, where the V_base_ground is set, and an emitter resistor inserted. Thus 1volt on base, at 1mA, assuming 0.7v Vbe, leaves 0.3v for the emitter resistor, hence 300 Ohms is needed there.

That same transistor, to operate at 1uA, or 1,000X less current, will need 0.058v less per decade across the Vbe. This 0.058 (or 0.060 if you prefer to use that) times 3 decades, tells us the transistor needs 0.7 - 0.18 = 0.52 volts. That leaves 0.48v across the current-setting Remitter; at 1uA, which I model as 1MegOhm-per-volt, we need 480,000 ohms; the resistor companies will sell us a 470K[yellow/violet/orange].

On silicon, high value resistors become very expensive area-wise, and current-density methods are used, where you can manipulate the Vbe by using 1000 bipolars in parallel to reduce the Vbe by 3 * 0.06. Run 1mA thru that group, setting up your Vbe, and place a single transistor nearby on the silicon to provide your 1uA. Run that 1uA up to +5v, duplicate with a PNP current mirror back to GND, and you have an incoming 1uA. Repeat this, for 1nA, Repeat this, for 1pA. The Vbe continues to drop, by 0.06v per decade, or 0.18volt per 3 decades. At a picoAmp, we are 3*0.18 = 0.54 volts below where we started, if the bipolar transistors continue that pure exponential-current logarithmic-current behavior and you still use the same size (emitter area) transistors.

• Thanks, I edited the question to add more detail to the nature of the problem. – John Smith Feb 5 '17 at 15:59