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
John
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...)
(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...)
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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.
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
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
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).
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
John
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.
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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