tell me about darlington transistors

[birt]

Can you do this with any type of transistor? jfets? mosfets?

For example, is it possible to make a switch in the buffer of a ruby to put another jfet after the first one to have a little boost?
what happens if you put a darlington configuration in a simple pnp or mosfet booster? how many darlington stages can you use (say an LPB with 5 switches for 5 transistors)

[robbiemcm]

I too would like to know about Darlingtons. I've seen schematics with the second tranny coming off the collector and some coming off the emitter.. I don't get it.

[R.G.]

Back when transistors were first invented, they were pretty sorry devices compared to what we have now. Gains of ... ten... were good.

A fellow named Darlington conceived the idea that the emitter of a bipolar transistor could drive the base of another one directly and get a gain that was the product of the two gains. I think it may have been patented, but I haven't done the search. This was one of th a whole class of two-transistor connections that were invented shortly after transistors became available. The Darlington connection not only gives higher gain, it improves input impedance by the same factor.

Can you do this with any type of transistor? jfets? mosfets?

Strictly speaking, "Darlington" applies just to bipolars. You can hook up JFETs and MOSFETs with source to gate, but the biasing gets tricky, as the gate of a JFET must be more negative than its source by some volts, and the gate of a MOSFET has to be more positive than its source by some volts. In addition, there is not much to gain with such connections, because the FETS have essentially open circuit input impedances, and so the resulting currents in the output device are the same as if you only had one device. You don't get any advantage, only expense and complication. It's bipolars (both NPN and PNP) that can take advantage of the darlington connection.

what happens if you put a darlington configuration in a simple pnp or mosfet booster?

It acts like a single transistor with a very high current gain and a two diode-drop input offset. This may or may not have *any* effect on the operation of the circuit in question. If the circuit uses emitter feedback, the gain has already been lowered from what one device could do. Using a darlington merely increases the input impedance with no change in gain at all.

There are two places to use darlington connections - when you need high current gain, and/or when you need high input impedance and can't get it some other way. The output stages of solid state power amps have to have current gains of typically over 1000. Darlingtons are quite popular there. Used as an amplifier, the input impedance of a darlington is the transistor's gain times higher input impedance than a single transistor. For instance, if you have a 2N3904 with a gain of 200 and a 100 ohm emitter resistor, the input impedance is close to 20K (hfe*Re). If you add a second 2N3904 in a darlington connection, the input impedance goes up to about 4 megohms (hfe*hfe*re).

how many darlington stages can you use

As it turns out, not too many. The problems are temperature sensitivity and leakage. All transistors leak some from collector to base. With silicon, it's on the order of nanoamperes. So if you have a transistor stage biased at 1ma, a nanoampere of leakage times a current gain of 200 is a leakage current of 200nA, which is 1/5000th of the normal current, so the leakage and the temperature effect of leakage are pretty trivial. If you go to a two stage darlington, the first transistor's leakage gets multiplied by the second transistor's gain, so you now have a leakage of 40uA, a ratio of 1/25th of the normal bias, and you can start to see leakage/temperature effects, but the're not too bad. A third stage gets the leakage up to 8 ma, and the leakage now swamps the normal current and it's hard to get anything done. The leakage is so bad that it seriously affects your biasing and varies all over the map as temperature changes.

Simply stacking darlinton stages gets impractical and quickly.

I've seen schematics with the second tranny coming off the collector and some coming off the emitter.. I don't get it.

The term "darlington" for transistor connections refers ONLY to one bipolar device with its emitter connected to the base of a second device of the same polarity. If it doesn't have an emitter connected to a base, it isn't a darlingto.

In EE classes we had to study all the two-device compounds. If you have one transistor, you can put signal into either the base or the emitter. The possible circuits are Common Collector (CC, or emitter follower), Common emitter (CE, signal into base, output at collector), and common base (CB signal into base, output on collector). These can then be cascaded in two-device cascades in all the combinations:
CC->CC; a darlington follower
CC->CE which is a follower driving a common emitter stage, useful for higher input impedance, and is a darlington amplifier
CC->CB used for radio frequency work
CE->CC, buffered amplifier, keeps gain high by unloading the collector of the amplifier stage
CE->CE, cascaded gain stage for high gain
CE->CB, also called a cascode
CB->CC, massively useful in RF design, not audio
CB->CE, useful for isolating the drive signal from the output signal with gain in some situations - but rare.
CB->CB, never seen one used this way.

In each setup, either transistor can be of either polarity, and sometimes making it an NPN->PNP or PNP->NPN offers advantages in biasing by folding the biasing across the power supply and sometimes making extra biasing parts not needed. The study of the two-transistor compounds is fascinating.

Of course, this is all textbook theoretical learning, and for those of you who reject textbook theory, it can't possibly be of any use or practical application, because the textbooks are always wrong. Right?   

[birt]

there goes my idea for a 5 stage mosfetbooster

[R.G.]

There's nothing that says you can't cascade five - or ten, or a hundred - stages of MOSFET or any other gain stages. You just can't practically do it without breaking the DC connection between stages with coupling capacitors, or using overal DC feedback to stablize the thing somehow in the face of all that variation.

As a sidelight, the modern integrated circuit operational amplifier *is* the logical result of that line of thinking. The input is a differential CE stage, driving a second CE stage, which drives either a CC or darlington CC stage for output. The open loop gain is on the order of 100K to 10M(!) at DC at least, and overall feedback stabilizes the whole mess.

[birt]

what do you get when you don't stabilize it? it's not big amounts of gain that i want, i just like crazy effects

[R.G.]

Well, you do get gobs of gain. That's kind of the problem. Like putting a kid fresh off a fat tires bicycle onto a hardtail harley. How do you control it?

In many cases, you can't feed it a small enough signal to keep the output from being plastered hard to ground or the power supply. In a lot of cases it oscillates either at low frequencies (motorboating), or at RF, or both, emitting squarks of radio nastiness. Neither of these are particularly affected by your input signal. There are easier ways to make noisemakers that ignore your signal.

As a practical matter, without some feedback stabilization, a purely DC coupled device consisting of over three devices cascaded is going to be darned difficult to get any signal through because it's so sensitive. Not impossible, but quite difficult. If it worked perfectly, it would be hard to tell.

[robbiemcm]

Thanks for those replies RG, they were really helpful! And I know chances are that you probably have this on your site already, but do you have a link to anywhere that shows how to bias transistors?

[R.G.]

Partially. I started a very basic beginner's guide to electronics in general, but never got it finished. It might be some help, though. Look at:
http://geofex.com/Article_Folders/How_It_Works/hiw.htm down near the end.

The real secret is that the emitter must be 0.6V more negative than the base, no matter what. So you decide what voltage goes on your emitter, then put the base at 0.6V higher than that, and it all works. There are some applications where that won't work, like with a zero emitter resistor, or as an output in a class B power amp, or in a Fuzz Face, but those kinds of circuits are relatively  rare. The gain of a common emitter (CE) amp is always Rc/Re where Re is the unbypassed part of the emitter resistor. If the emitter resistor is all bypassed, then Re is the Shockley resistance Rs=25mv/Ie, and is usually in the 10 to 100 ohm range for normal signal biasing.

However, the DC bias point cannot ignore the emitter resistor. So if you want a CE stage, gain of 10, have 9V to play with, and can afford 1ma of battery current in the collector. For maximum signal swing, you want the voltage across the collector emitter and collector resistor Rc to be equal, and you want the emitter resistor to be 1/10 of the collector resistor. That means the 9V supply is divided into 21 parts: 10 on the collector resistor, ten on the Vcd, and one on the emitter resistor. Then the voltage at Rc is (10/21)*9V = 4.28V the voltage across the emitter resistor is 0.428V. The current is 1ma, so the collector resistor has to be 4.28V/1ma = 4.28K and the emitter resistor is428 ohms. These are not standard values, so you can use 4.7K and 470 ohms. The emitter voltage has to be (1/21)*9 = 0.428V, so the base **must** be approximately 0.6+0.428 = 1.028V to make that emitter voltage work. You can use a resistor divider, a single resistor that drops 9V to 1.028V at the specific base current, a corrent source, anything that makes the voltage at the base be 1.028V.

The usual is a two resistor divider. The divider needs to handle at least ten times the base current to not be loaded down by the base current significantly. If we get a transistor with an HFE of 100, then the base current is 1/100 of the emitter current, or 10uA. We then need a resistor divider that divides 9V down to 1.028V and trickles 100ma through it.

You can write down the two equations in two unknowns: Vb = 1.028V = 9V*R2/(R1+R2) for the voltage divider part, and 9V/(r1+R2) = 100uA for the current condition, and solving them you get R2 = 10.28K and R1 = 79.72K. It'll probably work OK if not exactly with 10K and 75K resistors.

You see the trick though - where does the emitter have to be, and how much current does it need. From that you know where the base has to be and how much current it has to get. After that, it's all Ohm's law.

[Transmogrifox]

Can you do this with any type of transistor? jfets? mosfets?

For example, is it possible to make a switch in the buffer of a ruby to put another jfet after the first one to have a little boost?
what happens if you put a darlington configuration in a simple pnp or mosfet booster? how many darlington stages can you use (say an LPB with 5 switches for 5 transistors)

I haven't thought about what you'd gain with it, or what degradation of performance may exist, but you could use a MOSFET or a JFET to drive the base of a BJT in a "Darlington" configuration.  The FET would just be biased at the BJT base current.  Perhaps this could be used as a high impedance input to a BJT power amplifier stage since the input impedance of FETs is essentially independant of their bias currents.

I would expect that using a FET in this way would make an emitter follower more distinctly nonlinear, thus adding a certain amount of distortion to the signal that would not be as severe with a normal BJT-BJT darlington configuration.  I don't know if it would be a desirable signal coloration or not.