transistor matching

[john_ellis]

Hi all

My point about using different base resistors operating on a 5V supply is that this is fine to match transistors.

Andrew T- it is not so important to obtain hfe at a particular Ic unless you want to check the spec., but for matching you only need to know that two transistors pass the same current for a given base resistor. With a few different base resistors, and I tend to use values from about 10k to 100 ohms in sequence 10-4.7-2.2- and decades- you can check the current for a wide range.

Fotios- I use an ammeter in series with the collector because the voltage drop is likely to be less than about 0.5V for even high currents. If you use 1 ohm resistor and a 5V supply then at 4A IC you have pushed the collector voltage down to 1V. If you keep the ammeter in series with the collector, you only need one ammeter and can make a guess at the base current from the base resistor (If Vbe is about 0.6 for low currents and rises to maybe 1V at high currents the base current is approx. (5-0.6) or (5-1) divided by the R.

If you also want to check two transistors for euqal Vbe - if you want to ensure they match for use in parallel configuration, for example- connect both bases and emitters together. Connect your ammeter in one collector, and take the other collector directly to the PSU. Again using base resistors, this time feeding both at once, measure Ic1 and then swap the meter to the other collector to measure Ic2. If the Vbe's are different, you will get different Ic's. It is important to make sure that both transistors are connected for each measurement, because if one is open circuited, the base current will flow mostly in that device instead of balancing between the two!

If you want to measure at a more specific current, use two 5V power supplies, and use fotios's 10 ohm base resistor to ground. Use resistors in the emitter to -5V (for NPN) and connect the collector to +5V. Use your meter to monitor the base resistor voltage to get the current. THis time the emitter resistor sets Ie so if you want hfe, subtract the base current from Ie to get Ic. The advantage is that you can set Ic quite closely whatever the gain e.g. a 3.9 ohm load will give about 1A, and a 1 ohm load 4A.

But this method needs two PSU's although still only needs one meter

cheers
John

[AndrewT]

Hi John,

it is not so important to obtain hfe at a particular Ic unless you want to check the spec., but for matching you only need to know that two transistors pass the same current for a given base resistor. With a few different base resistors, and I tend to use values from about 10k to 100 ohms in sequence 10-4.7-2.2- and decades- you can check the current for a wide range.

I am not at all sure your method works.
Take my original Iq=100mA and for a moment assume hFE for this transistors are in the range 50 to 100. These will require a base current of between 1mA and 2mA.
Set up the test and arrange for the first transistor to flow 100mA by setting Vbe and measure the base current. Now keep all the settings the same and swap devices.
If any transistors match the Vbe and the hFE of the first then the collector current will be the same and equal to the desired 100mA quiescent current.
However, ALL the others will have a fixed Vbe and base current but have a different Ic, lots of matching sets can be identified but each set are at a different Ic value.
Summarising, yes you can find matching Ic but they do not equal the desired quiescent current value.

Aksa recently posted his quick method (100 power devices in 40minutes) and he also selects on the basis of matching Vbe (to the mV) and aims for +-5% on hFE at the quiescent current.

I repeat,
the matching should be done at a fixed collector (or emitter) current and not on a fixed base current.

In case some are wondering why pairs (or multiples) in the output stage need matching the previous stage (VAS or driver) sends the SAME voltage to the bases of the parallel pairs. Each of the devices in the pair turns that base voltage into a base current and an output current. The output currents should match at quiescent condition and with the help of the emitter resistors they should also track each other as the output current varies both up and down.

[Nordic]

Are you saying drivers and outputs should have similar hFE?

[jacco vermeulen]

The way i read it, AT is saying (again and again) that matching Vbe of parallel output devices is important too.

When there's a fire, it's equally important to open all doors at the same time as having the same number of people going through each individual exit by controlling the flow and having similar exit opening dimensions to obtain the optimal result.

[AndrewT]

Hi,
some builders match Vbe (I am in this camp).
Other builders match hFE.
A few builders (and designers) match both Vbe and hFE.

The third category of builder/designer will have their own priority for which is more important (hFE or Vbe).

Then the question comes down to, for what operating conditions the matching need to be done and to what tolerance.

BTW,
I move across to the third camp when matching LTP input devices.

Nordic,
it is almost impossible for the low to medium power drivers (hFE 50 to 200) to have as poor a gain as the big and beefy output devices (hFE 25 to 100).

[Nordic]

Ah ok, I only started measuring yesterday, and that seems to match my findings so far.

[jackinnj]

Here's the right way to set I(b) == you can use an instrumentation opamp as a programmable gain source -- the positive feedback -- this was shown by Borberly in his tracer article in 1990 (from a Burr-Brown application note) and variants appear on the web (see EDN) -- TI also shows how to do it with their power opamps (OPA549) which have a current limit pin and can source amps:

[jackinnj]

This is even neater -- note that the current is opposite sign to the input voltage -- the circuit appeared in EDN in 2003 -- i goosed up the current capacity with the totem pair -- sorry TI, but I used LinearTech opamps as these are what I had on hand:

[john_ellis]

Hi Andrew T

Let's start by agreeing that a matched pair of transistors will have equal Vbe and hfe for given Ic. There is no other option, but I have found that in a batch of transistors with the same lot number, matching hfe's often indicates similar characteristics and Vbe's tend to also match. Perhaps I should have mentioned this, but I have said so in other threads.

When you said

However, ALL the others will have a fixed Vbe and base current but have a different Ic, lots of matching sets can be identified but each set are at a different Ic value.

the point about the base resistor is that this fixes the base current, very nearly, but it does not guarantee that Vbe's are the same - and there may be slight variations in Ib as a result, but these will be small.

However, to match transistors with equal Vbe the second method I suggested should be fine. By forcing equal Vbe, if the gains or Vbe's are different the collector currents will be different. In fact, this might even be the better approach because it will define equal Ic for Vbe at the current needed for Iq.

The simplicity of this approach is that it uses one power supply, one meter, and a few base resistors. It is conceivable, if the gains are not checked separately, that you might have a pair of transistors whose Vbe's are different but gains compensate, but in the case you mentioned-setting equal Vbe's for the quiescent current- the approach will show matched pairs for given Ic.

I assumed at the start of this thread, we were talking about the simplest way to match transistors. If you build a transistor characteriser then you should be able to get good info.

cheers
John

[fotios]

In this point it is good we clarify each term and each symbol separately, in order to does not become confusion. And in order he does not laugh with us, somebody that knows better from us.
1) The collector characteristic curve is obtained by varying collector to emitter voltage (Vce) and measuring collector current for different values of base current.
2) The transfer characteristic curve is obtained by varying the base to emitter voltage (Vbe or bias) or current at a specified or constant collector voltage, and measuring collector current.
3) The gain parameter most often specified is the current gain (â) from the base to the collector. The power gain of a transistor operated in a common emitter configuration is: Power gain = power output / power input = ib^2.â^2.rl / ib^2.rin = â^2.rl / rin
The parameter rin (input impedance) because of the large number of components of which it is comprised usually is expressed as a maximum base to emitter voltage (Vbe) under specified input current conditions.
A measure of the current gain of a transistor it its forward current transfer ratio and is the ratio of the current flows in the output electrode to the current flows in the input electrode. Because transistors are used most often in the common emitter configuration, characteristic curves that are given by the constructors of semiconductors usually shown for the collector or output electrode.
The current gain (or current transfer ratio) of a transistor is expressed by many symbols. The following three are the most commonly used:
1) beta (â): general term for current gain from base to collector of a transistor in the common emitter configuration.
2) hfe: ac gain from base to collector or ac beta
3) hFE: dc gain from base to collector or dc beta
The common emitter current gain, â , is the ratio of collector current to base current: â = Ic / Ib either as dc or ac expressed.
Finally, useful values of beta (â) are normally grater than ten.
Regards
Fotios Anagnostou