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Old 27th November 2013, 04:27 AM   #1
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Default Transistors 101 question about 2 transistor Darlington configuration

I am slowly warming up to some design using transistors and am trying to understand the purpose of R1 in this figure of a Darlington pair:
Click the image to open in full size.

Sometimes I see the Darlington pair drawn with R1 and sometimes not. Does the presence of this resistance change the performance of the Darlington or can it be omitted?
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Old 27th November 2013, 07:39 AM   #2
mirlo is offline mirlo  United States
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Without the resistor, the darlington gives a much higher current gain than the single transistor, but the transconductance of the composite is cut in half vs the single bjt and the high frequency performance is severely, almost fatally, compromised by a factor like the current gain of the second transistor.
This is because the transconductance and frequency response are both heavily dependent on collector current, which is much lower in the first transistor.
The resistor biases up the first transistor to speed it up and raise its transconductance.

( transconductance is the small signal ratio of ( delta I_c ) / ( delta V_be ). )

Eric
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Old 27th November 2013, 04:25 PM   #3
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So, if turn-off time is not critical, can the resistor be omitted?

Would the turn off response time show up in a SPICE sim of a circuit that uses the transistors? I could just check it that way...
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Old 27th November 2013, 04:59 PM   #4
Mooly is offline Mooly  United Kingdom
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Quote:
Originally Posted by CharlieLaub View Post
So, if turn-off time is not critical, can the resistor be omitted?

Would the turn off response time show up in a SPICE sim of a circuit that uses the transistors? I could just check it that way...
It depends entirely on the intended application but generally including a resistor is good practice. That said, some specific circuits would not use one but saying that I'm not thinking of audio applications. Try it in simulation, it certainly should show up because its a very real effect.
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Old 27th November 2013, 05:02 PM   #5
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Quote:
Originally Posted by CharlieLaub View Post
So, if turn-off time is not critical, can the resistor be omitted?

Would the turn off response time show up in a SPICE sim of a circuit that uses the transistors? I could just check it that way...
If this is a linear application there is no "turn-off"

If a switch (but why? who needs that gain in a switch?) then yes. Or maybe.

So, for linear, we do need an R, else the first transistor maybe operating at a very low Ic.

Like most engineering, its a compromise.
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Old 27th November 2013, 05:25 PM   #6
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Here is the application:

a Capacitance Multiplier with over-voltage protection

Look for the BD139/TIP3055 and BD140/TIP2955 in Darlington configuration.
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Old 27th November 2013, 06:18 PM   #7
Mooly is offline Mooly  United Kingdom
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I probably would add something like a 1k in that application. Another practical issue could be the rating of the BD139/140 and the maximum allowable collector current. Something like TIP41C/42C might be more appropriate.
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Old 27th November 2013, 07:01 PM   #8
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Quote:
Originally Posted by Mooly View Post
I probably would add something like a 1k in that application. Another practical issue could be the rating of the BD139/140 and the maximum allowable collector current. Something like TIP41C/42C might be more appropriate.
Although I didn't post an updated schematic I changed the driving (first) transistor from BD139/BD140 to 2N4401/2N4403 and am using TIP35C/TIP36C as the second (power) transistor.

In my sims I see the first transistor current not exceeding a few tens of milliamps peak. I happen to have the 2N devices on hand and they are rated to 600mA. The BD139/BD140 were overkill. Vin and Vout are always within a few volts of each other, unlike in a power amp.

Regarding turn off time, the fastest transient that the transistor will be dealing with is 120Hz ripple - nothing very fast at all. I don't see any effect whether the resistor is in the circuit or omitted. I was using a value of 220R-470R.
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Old 27th November 2013, 07:16 PM   #9
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You are confusing signal rise/fall time with turn-off time, I think.

Turn off time, roughly, is how the transistor comes out of its switched on state (ie saturated) to a linear or switched off (no base current ) state.

This is NOT a switching application - the transistors are on all the time after initial switch on.

Linear parameters apply, not switching.
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Old 27th November 2013, 07:23 PM   #10
Mooly is offline Mooly  United Kingdom
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Quote:
Originally Posted by CharlieLaub View Post
Although I didn't post an updated schematic I changed the driving (first) transistor from BD139/BD140 to 2N4401/2N4403 and am using TIP35C/TIP36C as the second (power) transistor.

In my sims I see the first transistor current not exceeding a few tens of milliamps peak. I happen to have the 2N devices on hand and they are rated to 600mA. The BD139/BD140 were overkill. Vin and Vout are always within a few volts of each other, unlike in a power amp.

Regarding turn off time, the fastest transient that the transistor will be dealing with is 120Hz ripple - nothing very fast at all. I don't see any effect whether the resistor is in the circuit or omitted. I was using a value of 220R-470R.
You have to look at the collector current of the drivers under load. With all those paralled devices I assume it a supply of high current output. Try a pulsed load on the output (simulate it) using a FET and load resistor. You could easily need several hundred milliamps or more to drive those output devices. The 2N devices are nowhere near suitable.
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