Controlling load from 3.3v logic: FET vs BJT vs darlington pair vs H-bridge? - Page 1
Forum section: electronics / beginners

author: poodyp
author_rank: Regular Contributor
timestamp: June 26, 2011, 04:59:06 am
content: I've been trying to find the answer to this for several days now but keep running in circles...I'm working on a project with a 3v motor, a microcontroller running at 3.3v and a battery at 3.7v. Every time I find someone suggesting one solution, I'll search for info and find someone else suggesting a different solution, and I don't know what the "best" one for my situation is.Will a 2N7000 with a Vgs(th) of 3v max be fully saturated at 3.3v? Will my micro be able to source enough current (25ma) to saturate a transistor?I'm quite confused at this point. Any help is greatly appreciated.

author: Psi
author_rank: Super Contributor
timestamp: June 26, 2011, 05:33:59 am
content: As i read the datasheet.... It says the Vgs(th) is normally 2.1V but could be up to a max of 3V. So at this voltage the fet is able to turn fully on.However there is a graph in the datasheet that shows that a 3V Vgs(th) will only work for the lowest of drain current.It looks like at 2A current and 25deg C the Vgs(th) needs to be 8.5V ??Seems rather high to me

author: jahonen
author_rank: Super Contributor
timestamp: June 26, 2011, 07:29:15 am
content: It usually is best to look at specified Rds(on) figures and see if they are specified for the gate-source-voltage one is using. Vgs(th) is the voltage where there starts to be some conduction, in practice you'll need quite a bit more to open the MOSFET fully. But it also depends what are your current requirements?In this case, it seems that there are no less than 4.5 volts specified so 2n7000 is not perhaps the best for the purpose. For example,FDN327(120 milliohms @ 1.8 V Vgs) might be better, or, then the BJT.Regards,Janne

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: June 26, 2011, 09:56:44 am
content: In many cases there are several equally good solutions - depends on the load.MOSFETs are usually the most efficient as there is no static drive current.2n7000 is a pretty poor MOSFET - there are zillions with much better specs, especially when driving from 3.3V.

author: Zero999
author_rank: Super Contributor
timestamp: June 26, 2011, 10:38:22 am
content: How much current does the motor need?

author: Simon
author_rank: Unknown
timestamp: June 26, 2011, 10:51:27 am
content: yes the load requirement will determine a lot

author: ejeffrey
author_rank: Super Contributor
timestamp: June 26, 2011, 11:15:24 am
content: You can easily saturate a BJT with a 3.3 volt drive, they turn on very hard right around 0.6 volts. The important parameter for the BJT will be the beta (current gain). If beta is too low or your load current is too high, you won't be able to drive the base sufficiently. The other problem with BJTs is that they can't pull the collector down below a few tenths of a volt. You can fix the current gain problem at the expense of speed and output voltage drop with a darlington pair.MOSFETs require higher voltages to switch, and the low voltage devices generally have worse on resistance. However, in principle they don't have a minimum drain-source voltage drop. The voltage drop is simply the load current times Rds on.An H-bridge can be built from either BJTs or MOSFETs, and are simply a way of driving a load in a bipolar fashion. Integrated H-bridge chips usually have logic level inputs and have the internal circuitry to make sure that the switches (BJT or FET) is fully saturated. It also takes care of the high-side drive you need for bidirectional control. A dedicated driver (H-bridge or half-bridge) is more expensive and takes more space than a single switch, but it is the way to go if you need to drive a high power load or you need high-side switching.

author: Simon
author_rank: Unknown
timestamp: June 26, 2011, 12:00:16 pm
content: can a small simple voltage boost be used in these cases ? just to supply the gate, or will having a higher gate voltage than drain voltage be an issue?

author: poodyp
author_rank: Regular Contributor
timestamp: June 26, 2011, 12:22:24 pm
content: I'm embarrassed to say unfortunately I don't have the motors with me at the moment, and I have no spec sheet or model numbers. They're normal toy car type motors meant to be driven from 2 AA batteries. I'd be completely guessing if I said 750ma, but hopefully not too far off.I need through hole components as I don't have the materials to make a PCB, and just want to use a breadboard for now, and a protoboard later. I did manage to find theIRF3708, but at $1.95 a piece are pretty darned expensive. I'm finding most FETs I look at only quote Rds at 10v.As for transistors what about theMPSW45? Vbe of 2v when Ic is 1A and Ib is 2mA. But then I'll also have a voltage drop of .7 when using a transistor correct?Thanks for the help so far, I think I'm understanding things little better.

author: Zero999
author_rank: Super Contributor
timestamp: June 26, 2011, 12:43:30 pm
content: You can get high gain, low saturation transistors which hardly loose any voltage and have low very base drive requirements.For this application, the ZTX690 seems perfect. It will only drop 0.5V, when ICis 1A and IBis only 5mA - better than any Darlington pair.http://www.diodes.com.tw/zetex/_pdfs/3.0/pdf/ZTX690B.pdf

author: Simon
author_rank: Unknown
timestamp: June 26, 2011, 12:44:41 pm
content: I would't run such a motor on a 2N7000, if memory serves it takes 200mA Not really enough

author: Psi
author_rank: Super Contributor
timestamp: June 26, 2011, 08:12:47 pm
content: I'm not sure if they will work at 3.3v but i seem to remember opto isolators being available that have high current push/pull outputs, like up to 1A.They would be easy to drive as you'd just be turning an led on/off.

author: poodyp
author_rank: Regular Contributor
timestamp: June 26, 2011, 10:11:11 pm
content: Okay, just to make sure I'm not going insane reading these datasheets, the panasonic2SD2504at 20mA IBand 600mA ICwill have a VCEsat of ~0.15V according to the 5th figure ("IC/IB= 30" is the ratio they used to measure correct?)? And the 6th figure shows a gain of ~500 at an ICof 600mA and VCEof 2v? Gain goes up as the VCEgoes up correct? At 55c a pop those are looking pretty good right now. Is there some horrible flaw I'm missing (other than Panasonic seems to be discontinuing all their transistors)? I may just go with the ZTX690 as suggested by Hero just to save myself any more headaches.Someone should write a book on how to translate spec sheets from one company to another.

author: Simon
author_rank: Unknown
timestamp: June 27, 2011, 05:45:24 am
content: once you get used to datasheets you find they all follow the same pattern and state the same parameters. Some just rearrange things differently.

author: scrat
author_rank: Frequent Contributor
timestamp: June 27, 2011, 06:33:18 am
content: Quote from: Simon on June 26, 2011, 12:00:16 pmBootstrap is the common solution that works if the load is run on PWM and FETs are used (no continuous gate current).In simple words, the gate drive supply is taken from a capacitor connected to a supply referenced to ground (through a diode) and to the source of the high-side FET. The capacitor gets charged when the low-side is ON, and then remains higher than the high-side's source when this goes up (i.e. the low-side in turned OFF).An isolated DC-DC converter can also be used, of course. A boost converter referenced to ground can be an issue, since the high-side's source is floating up and down.http://www.dz863.com/drawsch_1096.html

author: dopplershift
author_rank: Contributor
timestamp: June 27, 2011, 09:20:30 am
content: I think what you're looking for is a FET with a "logic-level gate." You can find this under the heading of "FET feature" in Digikey's parametric search.Something like the2SK4017from Toshiba looks good (if you want an N-channel). According to the datasheet* driving the gate at 3V will give you a 2A drain current capability (page 3). And you can get them for 69c each from Digikey.... not sure about local suppliers though.Also note that you can easily put MOSFETs in parallel to lower the effective Rds_on and improve current capacity.And yes, your micro-controller's ~20mA will be sufficient to drive a FET, since MOSFETs are voltage-controlled devices. That is; they essentially don't require any current to turn on.*I've found MOSFET datasheets to be generally very optimistic (misleading) in terms of actual current capability. Readthis articlesfor a good explanation of why you should take the datasheets with a pinch of salt.

author: Zero999
author_rank: Super Contributor
timestamp: June 27, 2011, 04:31:43 pm
content: Quote from: poodyp on June 26, 2011, 10:11:11 pmYou've read the datasheet correctly. That transistor is perfect for you except it's being discontinued, so if you're building a small number of one offs, it'll be fine as long as you buy plenty of spares but if you're going to build loads of them, forget about it.

author: poodyp
author_rank: Regular Contributor
timestamp: June 28, 2011, 07:48:32 am
content: Quote from: dopplershift on June 27, 2011, 09:20:30 amDang, I've been looking at every FET I could find and couldn't figure out which spec meant I wasn't going to burn up the FET running an amp through it. I guess I just wasn't reading them right.Quote from: Hero999 on June 27, 2011, 04:31:43 pmHeh, I feel like I finished some quest.Thanks, I think I'll pick some up just to have a few options. In the future once I get PCB etching down I'll switch to SMT components as I'd like to get this as compact as possible.One more question, heat for FETs and BJTs, how much will they dissipate? For the FET would it be ID^2*RDS? What about the BJT? IC*VCE?Thanks everyone for all the info, it's much appreciated.

author: scrat
author_rank: Frequent Contributor
timestamp: June 28, 2011, 08:40:06 am
content: P = V*I in DC. So you got the answer on your own...

author: ejeffrey
author_rank: Super Contributor
timestamp: June 28, 2011, 06:51:11 pm
content: Quote from: poodyp on June 28, 2011, 07:48:32 amThe latter is always true (neglecting the base current), and also for FETs: ID*VDS. This is just the definition of electrical power: P(t)=I(t)*V(t). The former is only true when the MOSFET is saturated so that it behaves like a resistor. For instance, with high speed switching circuits (like PWM) the dissipation during the transition can be important.

author: poodyp
author_rank: Regular Contributor
timestamp: June 29, 2011, 06:34:27 am
content: So for the 2SK4017 the datasheet only has one stated number for switching time, measured at 10v, but just to try this out, the total is 69ns, so with a PWM frequency of 20kHz, at 3.3v and 1A the switching loss would be about 1mW? Again for RDS(on)it's only measured at 10v and 4v, but even .2 ohm it'd only me 200mW at 1A. With an Rth (ch-a)of 125 total C/W will be 25 degrees? I assume the switching time would be much slower at 3.3v, but I don't think I'll be switching at more than a few hundred Hz.With the transistor it'd be about 110mW at the same current output, but I can't figure out what the thermal resistance is.So, theoretically, the transistor is twice as efficient, but deliver a whole .1v less, and for 11 cents less?