I've created the circuit shown. I'm using a 9V battery (actually throwing out 9.53V) and 5V coming from an Arduino to test with both 9 and 5 volts. The transistor is a BC 548B (datasheet I'm using is here).


simulate this circuit – Schematic created using CircuitLab

I've conducted a number tests changing the values of Rb and Rc with the following results, no idea if they're actually right though.

Ref  Rb     Rc     Ib (μA)   Ic (mA)   Beta
1    160k   560    50        15.6      312
2    470k   1.2k   18        6.15      342
3    220k   1.2k   41        7.5       183
4    180k   1.2k   51        7.5       147

Ref  Rb     Rc     Ib (μA)   Ic (mA)   Beta
1    160k   560    24        7.7       321
2    82k    330    52        14.1      271
3    470k   1.2k   9         2.89      321

My questions are as follows;

  1. I understand that from the datasheet, the range for this transistor can range from 200 to 450. I think that the reason that there are values less than 200 in 9V table ref 3 and 4 is because the collector emitter circuit has saturated, and can't rise any higher, causing the beta to drop as the Ib current increases. Is that correct?

  2. In all the textbooks I've looked at, the beta is a static value. "If the beta is X, work out the resister in the base needed to create a current of Y in the collector". I've since read that the beta will fluctuate with temperature and collector current (I think it's collector current). Where do I actually find this data? Where is the table the tells me the beta vs Ic? If the beta is constantly varying, how do you actually select a resister that will always work, and/or have too much current in the what would be load on the collector?

  3. Figure 1 from the datasheet, shows that with a 50μA current in the base, the collector current shouldn't exceed about 11mA REGARDLESS of the voltage between the collector and the emitter. But given 9V ref 1 and 5V ref 2, which both have Ib ~ 50μA I have a higher Ic than stated. Why is this? What is Figure 1 actually telling me?

  4. Figure 3 from the datasheet shows that the hFE is 200 for Ic < 40mA given Vce = 5V. That is obviously not happening given all the results in the 5V table in this post. So again, what is this graph?

  5. I tried to connect the circuit up so that I would have my 9V battery running from the collector to the emitter, and my 5V Arduino powering the base, essentially what a transistor switch is for. I think that's going to short out the Arduino. How do I have the 9V battery running from C to E and 5V on the end of the base? How do I actually wire this?

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    \$\begingroup\$ +1 for carefully researched and not actually very noob question. \$\endgroup\$
    – pjc50
    Commented Feb 17, 2015 at 14:06
  • 1
    \$\begingroup\$ Measure Vce (or compute from Vcc,Ic,Rc) and add it to the tables. That will answer some of the questions. \$\endgroup\$
    – user16324
    Commented Feb 17, 2015 at 14:13
  • \$\begingroup\$ Your data looks fine to me. As you say in 1.) all your voltage drop is across Rc... Transistor is saturated. Beta is not a very well controlled parameter and you should design your circuits for some minimum beta value. \$\endgroup\$ Commented Feb 17, 2015 at 14:16
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    \$\begingroup\$ These are great questions. I wish my classmates thought about this stuff beyond the plug-n-chug formulas. \$\endgroup\$
    – Greg d'Eon
    Commented Feb 17, 2015 at 14:18
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    \$\begingroup\$ Experimental investigation of a 1 transistor amplifier stage with different base and collector resistor values and two supply voltages in order to investigate a basic transistor parameter in order to gain a better understanding of circuit operation and design AND it's 2015. Can it be ? :-) - +10. Alas, only +1 possible. And a bonus "Get Olin to answer politely and at length while mentioning using an Arduino, and not even have him mention the fact, derisively or otherwise" - another +10. Alas still only +1 total possible. | Welcome to Stack Exchange EE anyway ! :-). \$\endgroup\$
    – Russell McMahon
    Commented Feb 17, 2015 at 16:33

3 Answers 3


Your question seems to be about beta or hFE. Yes, this can vary significantly between parts, even from the same production batch. It also varies somewhat with collector current and collector voltage (using emitter as the 0 V reference). However, for any one transistor, its gain actually varies rather little as a function of collector current across a reasonable range, and assuming the collector voltage is held high enough.

The big point you seem to be missing is that you shouldn't be worrying about the exact gain. A good circuit with bipolar transistors works with the minimum guaranteed gain over the intended operating region, but otherwise works fine with the gain being anywhere from there to infinite. It's not out of line for any one transistor at a particular operating point to have 10x more gain than the minimum guaranteed by the datasheet. After taking that into account in the circuit design, it's really just a minor step to make sure the circuit works with the transistor's gain all the way to infinity.

Designing for such a wide range of gain may sound difficult, but it's actually not. There are basically two cases. When the transistor is used as a switch, then some minimum base current, computed from the minimum guaranteed gain, will drive it into saturation. If the gain is higher, then the transistor will just be more into saturation at the same base current, but all the voltages across it and currents through it will still be pretty much the same. Put another way, the rest of the circuit (except for unusual cases) won't be able to tell the difference between the transistor driven 2x or 20x into saturation.

When the transistor is used in it's "linear" region, then negative feedback is used to convert the large and unpredictable gain to a smaller but well controlled gain. This is the same principle used with opamps. The DC and AC feedback may be different, with the first setting the operating point, sometimes referred to as biasing the transistor, and the second controlling what happens when the desired signal is passed through the circuit.


Here is a example circuit that is tolerant of a wide range of transistor gain. It will amplify small audio signals by about 10x, and the output will be around 6 V.

To solve this manually, it's probably easiest to do it iteratively. Start by assuming OUT is 6V, and work from there. Since the gain is infinite, there is no base current, and the base voltage is set directly by the R1-R2 divider from whatever OUT is. The divider has a gain of 1/6, so the base is at 1.00 V. Minus the B-E drop of 600 mV, that puts the emitter at 400 mV, and the emitter and collector currents at 400 µA. The R1-R2 path draws 50 µA, so the total drawn from OUT it 450 µA, so the drop across R3 is 4.5 V, so OUT is at 7.5 V. Now go through the above calculations again assuming OUT is 7.5 V, and maybe one more time after that. You will see the results converge rapidly.

This is actually one of the few cases a simulator is useful. The main problem with simulators is that they give you very accurate and authoritative looking answers despite the input parameters being vague. However, in this case we want to see the affect of changing just the transistor gain, so a simulator can take care of all the drudge work for us, as performed above. It's still useful to go through the process in the previous paragraph once to get a feel for what is going on, as apposed to just looking at the results of a simulation to 4 decimal places.

In any case, you can come up with the DC bias point for the circuit above assuming infinite gain. Now assume a gain of 50 for the transistor and repeat. You will see that the DC level of OUT only changes a little.

Another thing to note is that there are two forms of DC feedback, but only one for the AC audio signals.

Since the top of R1 is connected to OUT, it provides some DC feedback that makes the operating point more stable and less sensitive to the exact transistor characteristics. If OUT goes up, the current into the base of Q1 goes up, which makes more collector current, which makes OUT go down. However, this feedback path does not apply to the audio signal. The impedance looking into the R1-R2 divider is R1//R2 = 17 kΩ. The high pass filter rolloff frequency formed by C1 and this 17 kΩ is 9.5 Hz. Even at 20 Hz, R1//R2 isn't much of a load on the signal coming through C1, and it gets more irrelevant proportional to frequency. Put another way, R1 and R2 help set the DC bias point, but don't get in the way of the intended audio signal.

In contrast, R4 provides negative feedback for both DC and AC. As long as the gain of the transistor is "large", then the emitter current is close enough to the same as the collector current. This means whatever voltage is across R4 will appear across R3 in proportion to their resistances. Since R3 is 10x R4, the signal across R3 will be 10x the signal across R4. Since the top of R4 is at 12 V, OUT is 12 V minus the signal across R3, which is 12 V minus 10x the signal across R4. This is how this circuit achieves a fairly fixed AC gain of 10 as long as the transistor gain is significantly larger than that, like 50 or higher.

Go ahead and simulate this circuit while varying parameters of the transistor. Look at both the DC operating point and what the overall transfer function from IN to OUT of a audio signal is.

  • \$\begingroup\$ This is exactly what I was thinking - if your circuits don't depend on an exact value of beta, they'll be a lot more robust. +1 for the precise details. \$\endgroup\$
    – Greg d'Eon
    Commented Feb 17, 2015 at 15:06
  • 1
    \$\begingroup\$ @OlinLathrop,just for the sake of accuracy: During calculation of the input resistance you have forgotten the Miller effect. This reduces the effective input resistance to app. 6.25 kohms (assuming a voltage gain of app. 10). \$\endgroup\$
    – LvW
    Commented Feb 17, 2015 at 17:27
  • \$\begingroup\$ @LvW: Good point. That would put the high pass rolloff frequency at 25 Hz. So the circuit as presented would be not quite "HiFi", which could be fixed by making C1 larger, like 2 uF. \$\endgroup\$ Commented Feb 17, 2015 at 19:01

1. What causes the apparent beta decrease as the base current increases?

Beta isn't really changing. The collector current is limited by Rc. With Rc = 500 Ω, the maximum collector current is about 18 mA. With Rc = 1.2 kΩ, the maximum current is about 7.5 mA. This comes from Ohm's Law -- 9V / 1.2kΩ = 7.5 mA. With beta > 300, you only need 25 uA of base current to max out the collector current. Adding extra base current doesn't change anything.

2. Where does the datasheet describe the behavior of beta vs. temperature and \$I_C\$?

This datasheet does not give any information on how beta varies with temperature. Beta vs. Ic is discussed in question 4 below. I checked a few other datasheets and didn't see any temperature variation there, either. According to this app note, beta increases by about 0.5% per degree C. A more detailed understanding might require use of the Ebers-Moll model, which includes temperature in the form of the thermal voltage (kT/q). I'm not a BJT master, so perhaps someone else can clarify this.

3. How can \$I_C\$ be greater than what's shown in the datasheet's Figure 1?

Figure 1: Typical Static Characteristics

This section of the datasheet gives typical performance characteristics. These are average values that do not show the variation from unit to unit. A typical graph gives you an idea of an average unit's behavior, but it does not in any way give actual limits on that behavior. That's what the Electrical Characteristics table is for.

4. How can beta be greater than what's shown in the datasheet's Figure 3?

Figure 3: DC Current Gain

Two things are happening here. First, your Vce is not actually 5V in your 5V table, since some of the voltage is being dropped across Rc, so this figure does not represent your actual circuit. Second, this is another diagram showing typical behavior. What it shows you is that beta typically starts to drop off at around Ic = 100 mA. Since the absolute maximum Ic is 100 mA, this means you should expect beta to be roughly constant across the current range of the device. The figure uses 200 as a typical beta, but as you can see from the hFE Classification table, the beta for an individual BC548B could be anywhere from 200 - 450.

5. How can an Arduino be used to drive the base of this transistor?

First, you'll need to get the maximum continuous output current from the Arduino's datasheet. This will probably be in the milliamp range. Your base current must be less than that, which shouldn't be a problem since beta > 200 and Icmax < 100 mA. If you know how much collector current you need (which you should), you can figure out the minimum base current:

$$I_B = \frac{I_C}{\beta_{min}}$$

That will let you pick a base resistor. According to the transistor's Electrical Characteristics table, the Vbe should be around 0.7 V. You know your Arduino outputs 5V, so now you can use Ohm's Law:

$$R_B = \frac{V_o - V_{BE}}{I_B}$$

Connect this resistance between the Arduino IO and the transistor's base. Connect the transistor's emitter, the 9V battery's negative terminal, and the Arduino's ground together.


Supplementing the information given in O. Lathrop´s answer I like to give a short example that may surprise you:

Let`s assume you have designed a simple gain stage (as shown in your post) using a transistor with a current gain of beta=200. The quiescent dc current is Ic=1mA and the measured voltage gain (Rc=2.5kohms) is G=-100. Now - if you change the transistor having a lower value beta=100 you will observe that the voltage gain G will NOT change - provided you have tuned the bias resistor RB to a lower value that allows the same quiescent current Ic=1mA. (This is necessary for a fair comparison).

The reason is as follows: The voltage gain is determined by the transconductance gm of the transistor (slope of the Ic=f(Vbe) characteristic). That means: The "current gain" plays no role - lowering the beta value from 200 to 100 increases the input current only, without influencing the voltage gain (as long as the operating point does not change).


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