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Old 21st January 2014, 06:58 PM   #1
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Default What is the Ft of a compound transistor

Can it be derived from the characteritics of the driver and output BJTs ?
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Old 24th January 2014, 07:55 PM   #2
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Yes, basically. You can construct Bode plot for both and multiply. Generally a high frequency driver plus lower fT output will be dominated by the output device.

John
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Old 24th January 2014, 09:54 PM   #3
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It can be worse than that; for example, the classic darlington pair requires some method of pull-down from the OP device base in order to ensure turn-off. Without enough base-charge suckout the composite can be rather worse than either device used.

(For anything you care about it's worth calculating how much you need, to optimise turn-off vs loading on the driver. Choose amongst several intersecting curves...)

Last edited by martin clark; 24th January 2014 at 09:56 PM.
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Old 25th January 2014, 09:06 AM   #4
sbrads is offline sbrads  United Kingdom
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You could have a go at measuring fT Fake transistors?. My own efforts weren't terribly accurate but you can easily tell if fT has changed.
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Old 25th January 2014, 03:04 PM   #5
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Hi Martin

Your comments address two separate issues. fT is defined by current gain, so it is collector current out for base current in. This means high impedance. You can get a higher fT for a Darlington than for a single transistor - even though the power transistor may have a gain less than 1 the driver which may have a high fT could still have a high gain, so the pair have greater than unity gain.

If you have a low impedance drive, as you suggest, then it is no longer "fT" but effectively common collector. In practice, in an audio amplifier or indeed in most practical circuits I agree we want the transistors to switch on and off as fast as possible. Providing a low impedance helps to turn them on and off faster.

In fact, in one of Doug Self's early articles he cited the good old 2N3055 as having a rather alarming characteristic. In typical amplifiers at the time, this was that the supply current begins to rise at frequencies above 10kHz or so. This is related to the 1 MHz fT of those old transistors. One solution I used was to replace the usual 68 or 100 ohm base resistors by 10 ohm resistors. But the drivers get overworked a bit then.

John
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Old 25th January 2014, 07:22 PM   #6
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IMO fT is not as useful as parameter for compounds as it is for single transistors, simply because we are looking at a 2nd-order response. You cannot really infer hFE(f) from it (you can for a single transistor if DC hFE is known).
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Old 25th January 2014, 08:19 PM   #7
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fT is the frequency where the gain is unity, so you can apply that to one or more transistors, even if there is a second order roll off. Yes, you would have to make two or more measurements in the gain-frequency response to find the value. It is easier to determine fT for a single transistor, but the question was asked if you can obtain it from the individual responses of the transistors, which you can.



John
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Old 27th January 2014, 09:37 PM   #8
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I followed the Bode plot route.
One can construct the Bode plot of a transistor from the Ft and the DC current gain, Beta0 given in the data sheets.
Using Beta = Beta0 / ( 1 + j * ( f / Ft ) )
The cut off frequency is Ft / Beta0
Before roll off, the amplitude of Beta is Beta0
After roll off the amplitude of Beta decreases 20dB per decade, intersecting the 0dB line at f = Ft
So it is easy to construct the Bode plot from Ft and Beta0.

For a coumpound BJT
Output transistor Ft1 Beta1
Driver transistor Ft2 Beta2
Ft2 > Ft1


I finally found the following result:
The Ft of the compound transistor is
The geometric mean of Ft1, Ft2 , in other words: Sqrt( Ft1 * Ft2 )
or
Ft1 * Beta2
Whichever is smaller.
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Transistor junction temperature is not transistor case temperature.

Last edited by mchambin; 27th January 2014 at 09:40 PM.
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