Design Article
Successfully choose complementary bipolar transistors
Peter Demchenko, Vilnius, Lithuania
2/21/2013 11:40 AM EST
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For circuit designs that use 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 1This circuit makes it easy to test and match the current gain of complementary bipolar transistors. Matched transistors will cause the voltmeter to read 0V. |
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 10 μA; 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:
 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.
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Marian Stofka
2/25/2013 6:42 AM EST
The idea is interesting; although the emitter current of PNP is compared with the collector current of NPN and this causes an imperfection, as IE = IC + IB.
The method proposed is suitable for matching a couple of pairs, say 3.
If you had to match n = 10 PNPs with 10 NPNs; you needed to perform at least n.(n + 1)/2 = 55 comparing procedures.
For industrial basis (n equal, or higher than 10), it is preferable to measure β separately for each transistor; to order then PNPs and NPN into two columns with bottom-up increasing β in both columns.
In each row you then have a best-way-matched pair.
In this case you need just 2n measurements (20, for n =10).
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Peter Demchenko
3/4/2013 7:09 PM EST
Marian:
1. Yes, but "beta should be high enough that IE~IC." - and most modern transistors do correspond.
2. This is a simple "ad hoc" circuit having minimum components - 3 resistors only.
3. To form pairs from two sets of NPN and PNP devices you can connect NPN device and then "sweep through" the PNP set while the corresponding pair is obtained; if no (PNP set is exhausted) - simply change NPN device with another one from the NPN set and repeat the procedure!
So I see no obstacles to select as many pair as you want:)
- Peter
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Marian Stofka
3/5/2013 12:34 PM EST
Peter,
As I have told, your idea is interesting; I am not critiquing it.
What you are saying about measurements in general; is enumerated for an example for n = 10 in the previous poster.
By the way, your name sounds Russian.
Marian
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Peter Demchenko
3/6/2013 1:56 AM EST
Marian:
Yes, I agree, the math is OK, but look, let you have 1000 NPN and 1000 PNP, but your needs are 10 pairs only:)
May be optimal would be to fill sorta list (on paper or even better computer program, which order the list) while the search is not completed?
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Marian Stofka
3/6/2013 11:37 AM EST
The case, you are describing now, with having much more transistors, than you need to match - is rather a game than a routine.
Here also a fortune plays a role. Having exceptional fortune, you could match the first 10 pairs, you took at random. In the case of bad luck, even after checking 100 randomly selected "pairs" - you needn't find a sufficiently matched pair.
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