Weird behaviour when using Darlington Transistor - Page 1
Forum section: electronics / beginners

author: brendan-petrol
author_rank: Contributor
timestamp: June 11, 2012, 02:29:21 am
content: G'day,Just wondering if someone could help me clear this up. I've made this circuit as a component of a project I'm working on (i am new to electronics, hence the posting in the Beginner forum) and I am noticing some weird behaviour. If I use the voltage I want to use, 12v, it passes the current even if the switch is open. If I bring that 12v down to 5v, it works as intended.Also, how can I tell if pulling 2A through these at 12v is OK? I assume it will be in the datasheet, but I don't know where to look. Obviously I want to build it within spec, because I don't want this thing to fail in the middle of a show (it's for my local theatre)!Thanks!

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: June 11, 2012, 03:16:08 am
content: I am just guessing here, but maybe you have blown one of the transistors.Take them out of circuit and test for shorts.The reason you may have blown one is that you don't always have a resistance on the base. It would depend on the pot setting.I would put a base resister onto the base and not even have a pot.A number out of the sky would be 100 ohms.Probably better to do a calculation or err on the high side and then bring it down until you get enough current.

author: hlavac
author_rank: Frequent Contributor
timestamp: June 11, 2012, 03:23:46 am
content: What is it supposed to be doing? If it's sort of a linear dimmer/motor speed controller it's horribly wrong. 316R down to a full short of +5V thru the bases. And you are dissipating up to 24W in a device that can only take 1W.What you observe could bethermal runaway. BJTs do that.ZTX1049A datasheet. Looks good to up to 4A continous, but only 1W, so make sure Vce is no more than 0.5V when on at 2A. It will burn if you close enough the transistor and Vce rises above the 0.5V.Good enough only if you use it as a ON/OFF switch in this configuration.For 2A you need only about 10mA going through the base of ZTX1049A. If you connected +5V through 470R resistor directly to base of ZTX1049A it would turn it on.If you do darlington like this base current of theP2N2222Ashould be only about 0.133mA.That 316 Ohm trimmer/pot there is totally wrong. You would probably want something like 10k potentiometer connected between +5V and GND, and the middle thru another 27k resistor to the base of the transistor. But you would still need to upgrade the second transistor to one that can take 24W and has an adequate heatsink.

author: hlavac
author_rank: Frequent Contributor
timestamp: June 11, 2012, 03:46:13 am
content: This is what you probably want(Java enabled browser required).

author: brendan-petrol
author_rank: Contributor
timestamp: June 11, 2012, 07:58:28 am
content: Nah, it was going to be part of a PWM circuit for 2A of LEDs. And awh, crap, forgot to look at the max power. That sucks! These are just things I could find in my (not very well stocked) electronics parts drawers.Looks like i'll have to learn how to use MOSFETs. Probably would've been a better idea anyway. And the only reason I had a trimpot is i couldn't find something around that value. I wanted to simulate a small base current and see how it would react.Anyone have any beginner level doco about MOSFETs, particularly when connecting them to an AVR, or preferably a 74hc595 - which can supply (if i remember correctly) only 6 mA of current per pin...

author: amyk
author_rank: Super Contributor
timestamp: June 11, 2012, 08:07:47 am
content: Quote from: brendan-petrol on June 11, 2012, 07:58:28 amMOSFET don't draw any current through the gate except momentarily during switching.

author: brendan-petrol
author_rank: Contributor
timestamp: June 11, 2012, 08:12:29 am
content: Oh, okay! Where in the datasheet would I find how much they draw whilst switching? And if they don't draw any current except when switching, how do they turn on/off?

author: amyk
author_rank: Super Contributor
timestamp: June 11, 2012, 08:23:17 am
content: http://en.wikipedia.org/wiki/MOSFETTheir operation is based on electric field. You should start reading more than just datasheets

author: brendan-petrol
author_rank: Contributor
timestamp: June 11, 2012, 09:10:56 am
content: Ahhh, that makes sense.Now I need to learn about heatsinking!

author: hlavac
author_rank: Frequent Contributor
timestamp: June 11, 2012, 09:16:05 am
content: If it's only for PWM switching, you should be fine with the parts that you have, just put a proper base current limiting resistor there (~27k) and you should be fine with what you have.

author: hlavac
author_rank: Frequent Contributor
timestamp: June 11, 2012, 09:20:48 am
content: Dave hasan episode on heatsinks (#105).But if it's just switching the 2A fully on/off 1W should be fine and you should not need any heatsink.

author: brendan-petrol
author_rank: Contributor
timestamp: June 11, 2012, 09:29:14 am
content: What sort of temperature can the MOSFETs handle? I ran 12V @ 2A through it and flattened part of my fingerprint when I tested the temp! is that normal?

author: hlavac
author_rank: Frequent Contributor
timestamp: June 11, 2012, 09:45:18 am
content: As for how to derive the value, for BJTs to fully switch on certain collector current Ic, you need base current Ib that is Ic/beta or h21e.Take minimum beta/h21e/DC current gain value from the datasheet, divide collector current by it and thats your minimum base current.Base-emitor junction of BJT is basically a forward biased diode, so there is a voltage drop of ~0.6V. Subtract that from the base voltage and then divide by base current to get maximum value of base resistor.For darlington, multiply the two minimum betas together and there will be two base diode drops.ZTX1049A has DC current gain of ~450 at 2A.P2N2222A has h21e of ~75 at 10mA current it needs to feed into the ZTX1049A's base.So the base current should be at least 2A/(450*75) = 0.00005926A = 59.26uABase driving voltage is 5V, subtract base voltage drops of max 1.2V for P2N2222A and max 0.95V for ZTX1049A = 2.85V.Now you need the base current for minimum base voltage of 2.85V to be greater or equal to 59.26uA so 2.85V/0.00005926A = ~ 48k Ohm.So the base resistor could in fact be 47k or a bit less then that.

author: hlavac
author_rank: Frequent Contributor
timestamp: June 11, 2012, 09:55:47 am
content: Quote from: brendan-petrol on June 11, 2012, 09:29:14 amNope. That is not good. Sounds like you are not switching it on enough. The catch is they usually need quite a bit of gate voltage (~10V for most of the power switching ones) to turn on completely. Driving a power MOSFET from a 5V PWM is not so easy.MOSFETs, when properly turned on, have very low resistance.They are basically like a variable resistor, with resistance lower the more the gate voltage exceeds certain threshold voltage (Vgs(th)) up to a certain point (linear region), then they flatten out limiting the current to some maximal value that is based on Vgs(th) (sturation).Gate looks like a small capacitor that needs to charge over the threshold voltage for the MOSFET to turn on.Which MOSFET do you have?

author: MikeK
author_rank: Super Contributor
timestamp: June 11, 2012, 10:26:12 am
content: If you want to switch the MOSFETs from the 74HC595 you can get logic level MOSFETs, such as the MTP3055 or IRL520. There are plenty others.

author: brendan-petrol
author_rank: Contributor
timestamp: June 11, 2012, 10:31:42 am
content: I've got the STP16NF06 (http://www.jaycar.com.au/products_uploaded/ZT2277%20-%20STP16NF06.pdf).Could I use a transistor like the 2n2222 that I have a heap of to switch the higher voltage at low current to switch the MOSFET on? Or have I got the wrong end of the stick completely?I will order some of those logic level MOSFETs, but my local electronics store doesn't have them!

author: MikeK
author_rank: Super Contributor
timestamp: June 11, 2012, 11:19:31 am
content: An NPN transistor needs the base to be at least 0.7V above the emitter. This is the base-emitter threshold to turn that junction on. So if the emitter is connected to the gate of the FET, which needs to be, I dunno...5V?, then the base of the NPN needs to be 5.7V just to switch on. This is why NPNs are considered low-side switches (and PNPs are high-side switches).If you're switching the NPN with 5V, then no, it can't be done. If you can use the same 12V as the FET source, then I think you're fine, but you really have to watch the base current so you don't blow the transistor.If you have an op-amp you could do Dave's constant current dummy load. I think you also might be able to switch the FET with a NPN-PNP push-pull thing (if you have some PNPs) if it needs to run from 5V.But I think I'm now in over my head at trying to help.

author: Mechatrommer
author_rank: Super Contributor
timestamp: June 11, 2012, 12:21:11 pm
content: OP question... i guess the driving NPN base is floating high (capacitance) and turned on. try pulling it down using higher value resistor and retest with the 12V. MOSFET? worst in this regard.

author: hlavac
author_rank: Frequent Contributor
timestamp: June 11, 2012, 01:25:46 pm
content: Or you could build alittle MOSFET driverlike this.It uses one PNP and one NPN transistor.MOSFET models in Falstad's circuit simulator are inadequate for power MOSFETs, don't mind the small current thru the load.

author: Axel1973
author_rank: Regular Contributor
timestamp: June 12, 2012, 06:09:22 am
content: Quote from: brendan-petrol on June 11, 2012, 02:29:21 amId say, since your input as shown is "floating" when the switch is open, the transistor may pick up any wierd LF/HF Noise from the air and start to conduct partly. try to "pull down" the input (base) pin of the transistor with some hight value resistor (like 10K or so) to GND. This may stabilize the input and bring down the transistor to OFF mode.Of course im assuming your transistor can handle at least 12+ volts and 2A.pls reply if this helped any.greetsAxel