What determines the resistance needed on the base/gate of a transistor? - Page 1
Forum section: electronics / projects

author: PhilippeASP
author_rank: Contributor
timestamp: July 26, 2017, 03:55:40 pm
content: I remember reading that, for a given transistor, you want to have the right amount of resistance on the middle pin. Too much and the current drops and the transistor acts like an amplifier, boosting the tension it sees on the gate. So for example putting 12 volts on the drain, you'd be able to put an alternating analog signal of 2 to 4 volts on the gate, and you'd get 6 to 12 volts analog coming out of the source. But if you want the transistor to just act like a switch, you want to reduce the resistance, increasing the current going into the gate. This puts the transistor into "Saturation" and means that the analog signal you'd put on the gate won't give you an analog signal out. At a certain threshold, the transistor turns open, and it goes (not quite instantly, but pretty fast) from 0v to 12v. (Please correct me if I'm remembering any of this incorrectly.)I know there was an equation that allowed you to know what resistance you need for saturation... something to do with gain B, but I don't remember what it is. I know it also depends on the type of transistor. PNP or NPN. or N-channel MOSFET or P-channel MOSFET.My project involves opening the transistor using a control card that can produce 5v. It will allow a 25v tension to pass onto a lamp that needs 80 mA. I think this transistor is well suited for that.https://www.infineon.com/dgdl/Infineon-I45P03P4L_11-DS-v01_00-en.pdf?fileId=db3a30431ddc9372011e07ee39732802But nowhere in the specs do I see mention of the gain. If I’m reading the document properly, it looks like an N-channel MOSFET. All I really need to know is that equation that allows me to determine how much current I need to pass through the gate to saturate the transistor.When in doubt I can just stick a 10K on there and play around with the values until it works, but I'd like to do this a bit more professionally if I can.EDIT:Correction, in case future forum users come to this thread looking for answers and don't read past the first post of the thread. The way I described a transistor being used as an amplifier works for BJT type transistors, not MOSFET ones. The gain thing I mentioned is applicable to PNP or NPN, but a MOSFET one won't have it in it's specs sheet.

author: rs20
author_rank: Super Contributor
timestamp: July 27, 2017, 04:33:43 am
content: Firstly, when you say "tension", I'm going to assume you mean "voltage". "Tension" is a highly antiquated term at best (although it does sound cool!)Secondly, the terminals of a BJT are called Collecter, Base, Emitter; the terminals of a MOSFET are called Drain, Gate, Source.As long as you replace the word "gate" with the word "base", Most of what you're describing applies to BJTs (NPNs and PNPs), but doesn't apply to MOSFETs so well. In particular, take any well-designed/working MOSFET-based switch circuit, and you can add a 1 megaohm resistor to its gate, and it'll still switch between perfectly off and fully saturated -- it'll just take a lotlongerto switch. This is because MOSFETs are controlled by the voltage on their gate, and virtually no current goes in and out of the base except during switching (the gate looks like a capacitor, in other words).The MOSFET you selected (IPB45P03P4L) is a P-channel MOSFET, not an N-channel MOSFET. This will make it very difficult to wire in to a mixed-voltage system like this; you should get an N-channel MOSFET. Also the absolute maximum VDSvoltage of 30V seems a little bit too close for a 25V lamp. I'd select a 40V VDSMOSFET at least; just a bit more margin for error. Also, that's a really chunky, big transistor for a little 80mA load. I'd suggest something along the lines of the2N7002instead. I'll refer to this datasheet below.Anyway, in order to turn the MOSFET on, you just have to take the gate to a high enough voltage -- as mentioned before, any gate resistor only controls how rapid the switching occurs, not how "on" the switch ultimately becomes. You can take a look at Figure 6 to help you see how the transistor will perform:- You are driving the gate using a 5V control card, so (assuming the output is truly, fully 5V: verify this assumption) your Vgs in the on state will be 5V. So we look at the curve labelled VGS= 4V (since that's a conservative estimate.)- Plug in 0.08A (80mA) on the Y axis, and read off the voltage across the switch (VDS) off the X-axis: it's something tiny, like 0.2V. Hooray, your transistor switches well enough, and this part is suitable.You should also calculated the power dissipated in the transistor (ID* VDS) and check it against the power dissipation limit in the datasheet; and then also look at the thermal resistance of the part to calculated the temperature rise, and check it against the max temp. But, spoiler: the power dissipation is so low in your particular case that these checks will certainly come out OK.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: July 27, 2017, 06:56:14 am
content: I would suggest looking at some of the basic introductory webpages on transistors. You seem to remember some of the basics but are mixing them up (for example, for amplifiers we like to keep our FETs in saturation, for switching we use triode/linear region). FETs are also at low frequency voltage controlled and not current controlled whereas bipolar junction transistors are current controlled.Places like allaboutcircuits and such will prove a good starting point. I would usually sugfest Wikipedia but unfortunately the wiki page on MOSFETs is controlled by solid-state physics people and is too complicated to provide an introductory read on the topic.Prof. Behzad Razavi also has some university courses on introduction to analog design which can be viewed on YouTube. It might be a place to start, although I think he focuses on CMOS. I haven't watched them myself and only heard from colleagues that it is a good starting point.

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: July 27, 2017, 08:18:31 am
content: The b-e junction of a BJT is effectively a diode, therefore it exhibits a volt drop of about 0.7v which increases only slightly with larger currents. This is what we mean when we say that a BJT is a current amplifier.For switching with a BJT, you need a base current that will saturate the BJT at the lowest gain likely to be encountered So, if we assume 100mA max load and a worst-case gain of 20, then that means 5mA base drive. A resistor to provide 5mA at 4.3v (5-0.7) would be 0.86k. Call it 820R.The dissipation of the resistor will be only 22mW (4.3 *4.3 / 820) so any standard size will do.In practice the gain will be more like 50 so this will be easily enough. When designing you try to allow for marginal-quality components though, that way you get less rejects.To switch a MOSFET you typically just need to apply a gate voltage of between 5v and 10v. It really is that noncritical. Don't exceed 20v or gate insulation breakdown may occur. If you want to protect your MOSFET against gate overvoltage, put a 12v (or thereabouts) zener between gate and source, and a resistor of a few k in series with the gate. 10k will be just fine.The MOSFET you mention is rated to saturate fully with a gate voltage (Vgs) of 4.5v so it should perform satisfactorily with 5v logic. Provided there is no loss of voltage swing, of course.It is a P-channel though, which is not so suitable as the control circuit would have to be on the positive rail of the 25v supply. Probably not what you want.HTH.

author: PhilippeASP
author_rank: Contributor
timestamp: July 27, 2017, 01:36:41 pm
content: @rs20I see... so MOSFETs are never really meant to be used as amps. It's the other, bi-junction ones that do that.@TheUnnamedNewbieYeah, that's a good idea. I just need to find tutorials that have moved past the utmost basics (which I mostly have covered) but not rush to far into the university-level theory and get completely lost (like I often feel when I just dive into a specs sheet or a wikipedia article).Quote from: blueskull on July 27, 2017, 07:10:39 amThat's correct, I've done some college-level electronics courses, all in French. Why? I'm still not sure, but that's what I ended up doing.@IanMacdonaldSo... the P-channel MOSFETs turn on when I bring the gate to ground? And I need to keep it at 25v to keep the current from flowing down to the source?