Darlington pairs resistors.  - Page 1
Forum section: electronics / beginners

author: sureshot
author_rank: Regular Contributor
timestamp: July 26, 2018, 09:43:29 am
content: Hope this is ok to ask, might seem obvious but i can't find a definitive answer to it.So i would like to try a building a darlington pair from two power transistors, they would be the TIP35C npn bipolar transistor. They will be used as a series pass element in a linear voltage regulator circuit, or zener regulator circuit. I have used single darlingtons before pnp and npn of various power ratings. As the darlington transistors have built in resistors, would i need to use them in this application of a series pass element ? I know they assist in turn off of the transistors, and prevent leakage capacitance. But i don't know if i need them for this linear application.Any help appreciated, and thank you for reading.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 26, 2018, 10:45:15 am
content: The often found build in resistors in a Darlington pair help to reduce leakage current and speed up the transistor in some applications. For a power supply it is a good idea to have at least the resistor B-E for the output transistor. This can be a rather small resistor for an TIP35, e.g. 10-100 Ohms.Usually one does not combine 2 equal size transistors, but has a smaller transistor at the input. This could be something like an BD139 (slightly on the small side) or D44H... (slightly on the large side).

author: CJay
author_rank: Super Contributor
timestamp: July 26, 2018, 10:55:21 am
content: Quote from: Kleinstein on July 26, 2018, 10:45:15 amOhh, Motorola D44H01, my current (intended) go to transistor for linear PSU repairs!

author: sureshot
author_rank: Regular Contributor
timestamp: July 26, 2018, 11:04:10 am
content: This is from a TIP142 i think, are these values appropriate ? And I'm unsure on the diode, unless is protection use. Thank you for the replys.

author: sureshot
author_rank: Regular Contributor
timestamp: July 26, 2018, 11:09:24 am
content: I googled that Motorola D44H01 found nothing on it. Yes i forget things some times, yes a small drive transistor to the final power transistor.

author: CJay
author_rank: Super Contributor
timestamp: July 26, 2018, 11:29:53 am
content: Ugh, my bad, D44H11 and it's ST, not Moto.

author: sureshot
author_rank: Regular Contributor
timestamp: July 26, 2018, 12:27:48 pm
content: Oh ok, looked it up, what makes that transistor a good choice for power supplys ? Or is just replacement as per schematic.

author: exe
author_rank: Unknown
timestamp: July 26, 2018, 12:42:18 pm
content: Quote from: sureshot on July 26, 2018, 11:04:10 amThis diode is parasitic. Not sure it is suitable for protection. This topic was discussed not so long ago (but I wasn't find it). Try googling "darlington parasitic diode".Concerning resistor values, looks fine to me. You can play with values to see how behavior changes. Basically, there is a tradeoff between speed and power loss. Disclaimer: I'm no expert in this stuff.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 26, 2018, 01:21:05 pm
content: The D44H11 is a relatively fast medium power transistor. For the Darlington it sometimes helps to have the smaller transistor faster than the large one. This can help with stability.AFIAK the D44H11 is made not only by ST, from memory more like MOT first and ST as a 2nd source.

author: CJay
author_rank: Super Contributor
timestamp: July 26, 2018, 02:25:50 pm
content: Quote from: Kleinstein on July 26, 2018, 01:21:05 pmThat makes me feel a little better, I thought it was Moto and was a little surprised to find my go to supplier says they're from ST.

author: sureshot
author_rank: Regular Contributor
timestamp: July 26, 2018, 07:14:57 pm
content: Thank you for the replys, yes it all makes sense. As my application is linear, do I really need the resistors ? Looking at the LM723 circuit there is a darlington pair from the IC with no resistors. I'm wandering if there really necessary for a series pass element.Not sure what kind of leakage current, as in the numbers involved. I'm only asking this as it's not a switching application where completely on to completely off is important. What do you think ?Thanks again.

author: exe
author_rank: Unknown
timestamp: July 26, 2018, 07:41:10 pm
content: Quote from: sureshot on July 26, 2018, 07:14:57 pmIn my opinion they are essential for performance. They help removing stored charge -- making circuit faster. Darlington / Sziklai pairs are very slow without them as power transistor doesn't have path to quickly "discharge" base. Of course there is some power loss on resistors, they all may heat up quite a bit. So, you trade efficiency for speed.PS I'm no expert, but I had same question in the past. I did some research.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 26, 2018, 07:57:57 pm
content: The resistor from the larger transistors base to emitter really helps to speed up the transistor. Even if a linear application a power supply also has to handle transients and this can include turning off transistor. The resistor mainly shifts loss from the large transistor to the smaller one - so there is not much disadvantage from the resistor. For a linear PS keep in mind the FBSOA. The actual useful power handling capability can be quite a bit smaller than P_tot.The other transistor may not be needed or may be part of the circuit anyway.

author: exe
author_rank: Unknown
timestamp: July 26, 2018, 08:31:29 pm
content: Quote from: Kleinstein on July 26, 2018, 07:57:57 pmYeah, under short-circuit condition instantaneous current can be quite high and power supply may not react fast-enough or current limiting can be slow. Normally there is a separate bjt across shunt resistor that pulls the base current. Like Tr3 here:http://www.learnabout-electronics.org/PSU/images/Series-with-current-limit.jpg.Also, my observation, pass transistor has much higher rating (2-3x) than expected maximum power dissipation (may be because of SOA). Also good idea to put a reasonably beefy diode (reverse-biased under normal conditions) between collector and emmiter (to protect from input voltage higher than output if power supply charges a battery or big cap).

author: Simon123
author_rank: Regular Contributor
timestamp: July 26, 2018, 08:56:27 pm
content: You can calculate the value for each resistor across base and emitter. Since Ube=0.6V and current leaking from collector into the base (Collector Cutoff Current in datasheets) is about 1mA for 2n3055, the resistor value is about 600 ohms.

author: sureshot
author_rank: Regular Contributor
timestamp: July 26, 2018, 10:57:44 pm
content: Thank you for the replys. I will use them, as well as the diode, I have some 6 Amp axial diodes. Just one more question, I'm thinking these resistors, especially the lower values need to be a couple of watts to dissipate the heat ? Like 0.5 watt resistors is just not enough power handling.

author: sureshot
author_rank: Regular Contributor
timestamp: July 27, 2018, 09:24:13 am
content: Guess i measure the voltage across these junctions with a multimeter, then do the maths and select a value power rating.Anyway thanks again for all the help with this.

author: exe
author_rank: Unknown
timestamp: July 27, 2018, 09:53:20 am
content: In my opinion these resistors shouldn't dissipate much because it's a waste of power, reduces efficiency and make cooling more complicated. I'd limit dissipation to 100-200mW. But again, it's a design decision. Less value -- more power dissipation, but better performance. I'm designing a linear PSU now and I assembled a prototype where I change value and observe performance.Simulators can be useful to some extend, but models not always accurate. So, better make a real device and measure it.

author: sureshot
author_rank: Regular Contributor
timestamp: July 27, 2018, 10:38:35 am
content: Oh ok thanks for getting back to me. So looks like 0.25 to 0.5 watts is more than enough for resistors power rating. Thanks again.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 27, 2018, 11:29:56 am
content: The voltage for the resistors is small (about 0.8 V maximum, as there is the base emitter junction in parallel). So even with 10 Ohms resistor this would be some 80 mA max and thus 64 mW. So nothing to really worry about.With a relatively small resistor at the output stage one may have to check the power dissipation for the smaller transistor, as that extra current needs to be driven by the small one.

author: David Hess
author_rank: Super Contributor
timestamp: July 27, 2018, 04:25:39 pm
content: When driving a Darlington, the driver circuit can remove charge from the drive transistor but not the output transistor through the base-emitter junction so the base-emitter shunt resistor on the output transistor is absolutely needed both to remove charge and leakage. In practice Darlingtons are intended to operate with driver circuits which cannot remove charge so they include base-emitter shunt resistors on both transistors and really that is a good idea either way.

author: Simon123
author_rank: Regular Contributor
timestamp: July 27, 2018, 04:42:45 pm
content: Well power you can also calculate, for example 2N3055 Current is about 1mA, Ube (which is constant 0.6V), so you do the math, 0.6V*1mA=0.6mW ...

author: exe
author_rank: Unknown
timestamp: July 27, 2018, 06:09:15 pm
content: Quote from: Simon123 on July 27, 2018, 04:42:45 pmThat's just to remove leakage current. But to make it switch-off faster current should be bigger than 1mA. In practice resistor is about 100 Ohm or so (sometimes it's built-in 60 Ohm). So, it's a bit bigger than that. On my Sziklai pair it's noteceably warm (I'm building this PSU:http://www.bramcam.nl/NA/NA-01-PSU/NA-PSU-50.GIF).

author: sureshot
author_rank: Regular Contributor
timestamp: July 27, 2018, 06:39:50 pm
content: Thank you for the help, I'm going to go with R1 5K and R2 60R that seems like a place to start. If it acts odd I might have to change something. That psu looks like a real task lol, not one for me at this stage.

author: David Hess
author_rank: Super Contributor
timestamp: July 28, 2018, 01:23:49 am
content: Quote from: exe on July 27, 2018, 06:09:15 pmI usually figure on diverting at least 1/10th of the base current through the base-emitter shunt resistor. In a Darlington this just results in the drive transistor conducting that much more which is not a problem.