Darlington pair

[malc9141]

Q: about Darlington Pair.

At first sight, it's an obvious "mechanism."

But if you look at a chart for a single (npn) transistor, the current permitted to flow from C to E is a power function of the Voltage applied to B.

But then that current goes to the base (B2) of the second transistor. The voltage at that point is fixed. So how does the rising current applied to B2 allow amplification of the C2>E2 current? The voltage in the second system is fixed and maximal. So we must be reducing resistance in the base B2, but how does that happen by applying more current to B2??? (of course, I know that current is what matters in reality - but it needs voltage to drive it)

 

[ronsimpson]

But if you look at a chart for a single (npn) transistor, the current permitted to flow from C to E is a power function of the Voltage applied to B.

The voltage B-E needs to get to 0.7V (0.65V or what ever it takes). Below that there is no (little) current. The voltage really does not get (much) above 'turn on voltage'. The base just takes current. Current in X current gain = Collector current.

But then that current goes to the base (B2) of the second transistor. The voltage at that point is fixed. So how does the rising current applied to B2 allow amplification of the C2>E2 current? The voltage in the second system is fixed and maximal. So we must be reducing resistance in the base B2, but how does that happen by applying more current to B2??? (of course, I know that current is what matters in reality - but it needs voltage to drive it)

The B to E voltage is 'fixed'. If that is the right word. You really can't pull it up. The B-E voltage is 'one diode drop' or ' about 0.7V'. The B-E-B-E of a darlington is 2D or 1.4V. (all this assumes the E is on ground.

 

[ChrisP58]

If it were a resistor, then a fixed voltage would mean a fixed current. But the B-E junction is not a resistor, it is a PN silicon junction.

You may have a resistor in series with the base to help control the current, but the base current is not directly controlled by the base-emitter voltage.

The B-E voltage does need to be above the minimum 'turn on' threshold, but otherwise, the current can increase independent of the voltage.

 

[crutschow]

The base-emitter voltage does vary with current in an exponential fashion, but over a typical base current range it usually varies no more than about a tenth of a volt, from 0.6V to 0.7V or so. That is why a BJT looks like it is controlled by the base current even though, by semiconductor theory, it is the base-emitter voltage that controls the collector-emitter current.

 

[malc9141]

Thanks. But if you have a fixed power source - say, a small battery of 9v - and a trimmer/pot to vary voltage to the Base of npn1, you can increase the current through npn1.
But what happens when that current (from emitter1) arrives at the base of npn2? Presumably it's now 9 v?

 

[crutschow]

Thanks. But if you have a fixed power source - say, a small battery of 9v - and a trimmer/pot to vary voltage to the Base of npn1, you can increase the current through npn1.
But what happens when that current (from emitter1) arrives at the base of npn2? Presumably it's now 9 v?

No. Remember that the base-emitter voltage is about 0.7V when the transistor is on, so the voltage at the base of npn2 is equal to the emitter voltage of npn1 or 0.7V below the voltage at the base of npn1. At the npn1 emitter npn1 is amplifying current, not voltage. That's why a Darlington has a much higher current gain than a single transistor.

 

[malc9141]

Ah; so the "output" voltage from emitter of npn1 is not the line voltage: it is only what's "gone in", less the "drag" from the base of npn1.
And (question) that current & voltage (power) is applied to the base of npn2. This is further amplified. (?)

 

[crutschow]

Ah; so the "output" voltage from emitter of npn1 is not the line voltage: it is only what's "gone in", less the "drag" from the base of npn1.
And (question) that current & voltage (power) is applied to the base of npn2. This is further amplified. (?)

As an approximation, BJTs can be viewed as current amplifiers. Thus a Darlington looks rather like a single transistor with a current gain roughly equal to the current gain of npn1 times the current gain of npn2 with a combined npn1 base voltage to npn2 emitter voltage drop of about 1.4V. This can be converted to a voltage by adding a resistor from the collector of npn2 to the power supply , which then gives voltage gain, similar to a single transistor.

 

[MikeMl]

This might help:

First is Base voltage and Collector current of a single transistor. Second is what happens when two transistors are connected in a Darlington configuration.

Note the linearity of Collector Current vs Base Current. Also notice the resulting voltages at the Bases of the transistors, especially in the Darlington.

 

[malc9141]

I think where I was stuck was in thinking that when turned on, there was no significant voltage drop across the transistor. So with the Pair, I presumed the second base voltage (in your example) was 2V. But I think you are saying there is a voltage drop from collector to emitter, dependent on base voltage (and associated current). This voltage drop thro' the transistor diminishes as the base voltage is increased.

I had thought that once current was flowing, collector to emitter, through transistor 1, source (=battery) voltage would get applied to base of transistor 2.
But this is not true; rather, it's a controlled, lesser, voltage.

 

[ronsimpson]

rather, it's a controlled, lesser, voltage.

YES.....and maybe.... it is a controlled current. The voltage is 0.7V less than the supply.
Yes the first transistor does not turn on to the point there C-E = 0 volts. (Voltage C to E of the first transistor is supply-0.7V)

 

[crutschow]

You must think of the output of transistor 1 as an emitter-follower output, not as a switch. An emitter-follower output is typically about 0.7V below (follows) the base voltage. The difference is that the current at the emitter 1 output is multiplied by the current gain (beta or Hfe) of transistor 1. Thus the collector-emitter current of transistor 1 is much higher than its base current.

 

[malc9141]

Is it possible to tell me the voltage and current applied to the load in this example? It is the guts of the circuit I plan to build.
Thanks++

 

[alec_t]

~ 16v, 3.5a.

Edit: .....so the power transistor will dissipate ~ 70W ! You'll need a BIG heatsink.

 

[kinarfi]

Here's what I think you drew, is it correct?
I have always thought of pn transistors as being current devices, not voltage devices, that is they control and are controlled by current, the main design & operating specs always refer to IB & IC, HFE.

 

[malc9141]

Thanks v much indeed! It's in the range I had aimed for. The operation is over in <1 msec, but there is a big heat sink. Thanks again.

 

[malc9141]

Thanks, Kinarfi. Your diagram is correct.

I accept that transistors are current devices. But I needed to know what power I was putting in to my Load. And I didn't understand what the Voltage was "downstream" of the Collector.

This all goes back to de-energising a coil. I need a quick switch-off (and reversal of polarity) but not so fast I create too big a reverse current. In that conversation, we were considering hundreds of volts and micro-secs for field collapse.

With about 3 amps (as cited by Alec-T), things look pretty safe.

But besides all that, I hadn't a clue about PD beyond the npn, or what it was at the base.

 

[malc9141]

Here's what I think you drew, is it correct?
I have always thought of pn transistors as being current devices, not voltage devices, that is they control and are controlled by current, the main design & operating specs always refer to IB & IC, HFE.

I like your modelling. Set 1 , Set 0.5 etc.
It's not very sensitive to changing the pot, is it? ! I thought it would be much more sensitive. I might need a bigger R in the Load section.

Thank you!

 

[malc9141]

Hi Kinarfi -
If it's not too much trouble, can you calculate would happen to the values you gave if the Source voltage was dropped from 36 V to 20 V ?

 

[kinarfi]

Following the suggestion of greaneto, I added 100Ω to the base of mj1 and you mentioned inductive kick back so I changed the load to an inductor 5mh with a resistance of 4.5Ω and added a schottky diode, and ran the switch for 1 msec.