Path: EDN Asia >> Design Ideas >> IC/Board/Systems Design >> Selecting complementary bipolar transistors
IC/Board/Systems Design Share print

Selecting complementary bipolar transistors

08 Mar 2013  | Peter Demchenko

Share this page with your friends

For circuit designs that utilise complementary bipolar transistors, you sometimes need to sort the NPN and PNP transistors to have matching dc-current gains (β). One example of a circuit requiring matching is the output stage of an amplifier. The circuit in figure 1 shows a simple test fixture to achieve this match.

Figure 1: This circuit makes it easy to test and match the current gain of complementary bipolar transistors. Matched transistors will cause the voltmeter to read 0V.


Transistors Q1 and Q2 are the devices being tested to see if they are matched. In the test fixture, Q1 and Q2 share the same base current (IB); since there is no additional path where the current can flow, no additional compensation is needed. Note, however, that β should be high enough that IE≈IC. With this detail in mind, resistors R1 and R2 should be equal.

To give the transistors a bit more headroom, an additional voltage drop is introduced between the transistors' base connections. A voltage differential of a few volts is desirable, so a blue LED is a good choice for D1. Its presence helps to set the base voltage for Q1 (VB1) to about half of the supply voltage (VS). Using an LED in the place of D1 is preferable to using a zener diode due to the sharper knee at the low currents. Moreover, you can see the glow of many blue LEDs at currents below 10µA; the glow indicates the presence of base current, which means the circuit is working properly. Equation 1 is used to determine the needed supply voltage:


A typical blue LED will have a forward voltage of about 3.5V; assuming VBE1=VBE2=0.7V, you get a value for VS of about 9.8V.

Resistor R1 sets the emitter current of Q1; it is calculated using equation 2:


You should select an emitter current that matches the application in which the transistors will be used, because beta varies with emitter and collector current. With matching transistors (β12) installed in the test fixture, the voltage drops across R1 and R2 are equal, and the voltmeter will show 0.

Figure 2: For a simpler version, replace the voltmeter with inverse-parallel-connected red LEDs.


The circuit in figure 2 is functionally equivalent but uses a simpler method to indicate when the circuit is in balance. With matched gains, neither of the red LEDs (D2 and D3) will be on.


About the author
Peter Demchenko is from Vilnius, Lithuania.

To download the PDF version of this article, click here.




Want to more of this to be delivered to you for FREE?

Subscribe to EDN Asia alerts and receive the latest design ideas and product news in your inbox.

Got to make sure you're not a robot. Please enter the code displayed on the right.

Time to activate your subscription - it's easy!

We have sent an activate request to your registerd e-email. Simply click on the link to activate your subscription.

We're doing this to protect your privacy and ensure you successfully receive your e-mail alerts.


Add New Comment
Visitor (To avoid code verification, simply login or register with us. It is fast and free!)
*Verify code:
Tech Impact

Regional Roundup
Control this smart glass with the blink of an eye
K-Glass 2 detects users' eye movements to point the cursor to recognise computer icons or objects in the Internet, and uses winks for commands. The researchers call this interface the "i-Mouse."

GlobalFoundries extends grants to Singapore students
ARM, Tencent Games team up to improve mobile gaming


News | Products | Design Features | Regional Roundup | Tech Impact