2020. 1. 25. 02:10ㆍ카테고리 없음
Totem pole driver is only neccesary when you want very low transition time. As it's a DC-DC converter you must want that. This circuit will not invert the original signal, and will give a high current boost.
Hi folks, I suspect this has been asked and answered many times but I couldnt find anything definitive when I searched. I am trying to understand how this works, I am using an N-Channel FET I sort of expected, when I started looking that the PNP to be at the top with its collector to the gate and the NPN below it in a similar fashion. How wrong can you be with just the one head? I found several examples, pulled a couple of GP transistors from my box (2N3906 / 2N3904) And shoved them onto a breadboard. Well it works, but it occurred to me that it might be a really good idea to find out why and how.
Particularly as I want to make sure it keeps working. I would appreciate a leg up and some recommended reading, BJTs for dummies perhaps!! Thanks in advance looking forward to learning. This is the power board for an AVR I am trying to build, the origional is toast. The 317 puts out about 6V at startup and as soon as the fet begins to turn on that will quickly rise to 10.4v The FET hold its source at about 25v, zener - VGS(th) In theory at this point the PWM would have kicked in (b) is the output of an opamp with a max 50mA sink capability, no source to speak of. This board isnt finished, no fet driver yet, but I have an earlier itteration with the fet gate pulled up by resistors which is whats driving the genny now, badly I might add.
Click to expand.Doesn't invalidate anything I said, the design is flawed from the gitgo. I will repeat, MOSFETs are not linear devices. If you try to use them as such you will not have good results. I could be wrong, but I don't think I am. BJTs are linear, not all the way across, but for wide region of their curve.
This is why they work well for emitter followers (the other name for common collectors). It is obvious you are trying to use it as a voltage follower feeding a LM317.
Funny thing, a BJT would be a drop in replacement, a Darlington BJT even more so. Have you trying wiring this concept? The LM317T must have a 120Ω minimum resistance between the output and adjust, to satisfy the 10ma minimum current requirement. What is the target output of the LM317 regulator? And do you have the bridge circuit connected to the mains?
This is not a good idea, but I can't interpret the schematic that far. Quick side lesson. The first illustration shows a classic Common Collector configuration (AKA, voltage follower, emitter follower). A emitter follower will match the voltage of the base minus the BE drop.
The input resistance at the base will be the gain of the transistor (beta, or β) times the emitter resistance. So if β = 150 (not an uncommon value) then the input resistance of this circuit is 150KΩ. This is a common voltage regulator for many designs. Darlington Pairs have the gain of both transistors multiplied. So if both transistors have a β of 150, the total gain is 22,500, and the input resistance of the circuit is 22.5MΩ.
This kinda makes up for the extra BE drop. With BJTs the arrow points to the negative side, always. If the bias is reversed they are turned off, otherwise they look a lot like diodes with extra bonus points. You'r right the MOSFET is a lousy follower, I even think I understand why now thanks to all the input here. That said I have a FET, whereas I dont have any high voltage high power BJT's. Definatly a requirement of the next itteration I think. Its offset starts at about 2V and increases with input voltage to about 5V.
The circuit is built, all but the totem-pole. The lm317 has a 150/1100Ω divider giving about 10.5V once its input is at a couple of volts above that and the FET is limiting that input voltage to about 25V. Both caps are 0.1F and I have added an LED limited to 5mA at 10V across the output. This application has a power supply of 10 to 12VAC when it first powers up and as much as 130VAC once it running. Thats a real world problem that I cant avoid, if it doesn't work at 10V then it cant ever reach 120V and my generator will not work.
Lets put the supply part of this to bead for the time being, I know it needs changing but frankly thats not as important, right now, as getting the PWM gate drive to work on the second FET. Am I correct with these asumptions:- 1.
Both transistors are working in comon collector mode. They can only be ON when the gate is charging or discharging because Ibe is what turns them on.
Ib will be inherrently limmited because once the turn on threshold is reached any aditional 'potential' current will be Iec. The final phase of gate charge / discharge will always rely on Ibe because the transistor will be off. I am desperatly in need of education here. Please help if you can.
I'll say it again, no one is born knowing this stuff. I would suggest some side experiments before getting into high power circuits. The latter tend bite harder, and the sparks are bigger.
BJT stands for Bipolar Junction Transistor (hence the name BJT transistor is an oxymoron). It has two junctions, like a sandwich. You can put the N material on the inside, or the P material on the inside, so there are two kinds, NPN and PNP. They act just alike except they are opposite in polarity (like antimatter to matter). They have three leads, base, emitter, and collector. Our text book is very good on the subject.
It is worth reading and rereading. Any time you see Ibe, or Ice, or Vbe drop it is a good bet we are talking a BJT. MOSFETs are much newer devices, and spun off of conventional FETs. They have excellent turn on resistances (very low values), and can usually handle more power and current than the leads.
Given how nonlinear they are they tend to be used quite effectively for digital designs, stuff that is either one or off but not between. They have a gate, a source (which is usually the currents exit, go figure), and a drain (again, opposite the name). They also have polarities, a N channel and a P channel, but for technical reasons the N channel tends to work better. BJT transistors handle analog quite well, which includes audio and analog power supplies (while MOSFETs handle SMPS (switching mode power supplies, which is digital)) quite well. I've messed with BJTs both professionally and for pleasure for well over 30+ years.
I've messed with MOSFETs for around 20, and for pleasure for about 5 or so years. Click to expand.I wish I had a choice. MY generator AVR is dead. Parts arn't available. No water, because I cant pump any and power only when the sun shines. Not good all in all.
I am just about to update the thread on AC Sensing and generator control with my current circuit and build. I do understand the very basic principles and therfore differences between BJT's and FET's I even looked up JFET's which appear to work in depletion mode, wondering if they may be a better choice for the totem-pole MOSFET driver. My problem is real world implimentation of my extreemly sketchy knolidge. Are my asumptions about how the BJT totem-pole correct particulaly in respect of controlling the base currents? I expect it will soon die if I am wrong.
The FET I am mis-using in a linier fashon is getting very hot even with a substantial heatsink. Is this likly to be a product of its misuse or is it simply disipating too much power?
Totem Pole Mosfet Driver For Mac
Do you have it mounted on a heat sink? It is absolutely required. Here is what I am talking about.
Totem Pole Driver
I would like to find a higher voltage transistor, but TIP102 is all I can come up with for the moment. I'll ask SgtWookie to check in and suggest better.
There is an assumption on my part this will work. Truth is, I don't have a clue. I don't have the total picture, and if I did I'm not sure I'd know how to fix it. The 50Ω resistor (47Ω actually, since that is the nearest value) is needed, and should be soldered as close to the gate as possible.
Ok, I got Thor's MOSFET switching today, albeit very slowly. Here is what I came up with for a simple driver circuit: there IS a 2.2KΩ base resistor on the NTE2343; I just forgot to draw it, PWM is 5V 500Hz, 50% duty cycle, coming from my arduino. I generate an inverse PWM waveform with the comparator.
When the upper (PNP darlington) is on, the lower (NPN darlington) is off, and vise versa. This allows 12V to come in and charge the gate, then go out. Here are my 5V complementary signals to the 2 darlingtons: Here is the output of the darlingtons (the 12V MOSFET drive signal): Here is output of a (Qg = 170nC) being driven by my driver circuit. Seems to have a decent wave form.
Here is the output of (Qg = 1710nC) being driven by my driver circuit: Really pitiful turn-on time. This is running @ 500Hz.
I plan to run at 20Khz. If I don't improve my driver circuit before going to 20KHz, the FET will never even reach full ON. What can I do to improve this circuit? I thought this would be superior to the commercial driver ICs, but I guess not. According to my math it should be able to source & sink 5A, which should switch the FET on/off in 342nS, so I don't get why its talking so long to switch.
I currently have jumper wires from my breadboard to the gate of the FET, maybe that's the problem. I'm also not using the 'Kelvin source' terminal. Once I figured out that what I have made here is called a 'totem pole', started googling that and what I found looks like this: They have the NPN and the PNP reversed compared to mine, and they aren't using a seperate inverted signal like I am. How does that work? You're having so much trouble because you're using the transistors as saturated switches instead of emitter followers, and you are turning both transistors on at the same time.
(Don't believe me? Look at the emitter current for the NTE2343; you'll probably see 500mA.) You're going to have a hard time if you insist on using the transistors as saturated switches. Once they turn ON, it'll be hard to get them to turn OFF quickly.
You'll have a much easier time of it to just use a dedicated gate driver IC. I AM ordering a dedicated driver IC, but my brain won't move on until I understand this totem pole emitter follower thing. In the diagram at the bottom of post #1 shows the usual and correct way to do this, but it makes no sense to me. If you applied the same PWM signal to both transistors as they show, would you not turn on both transistors at the same time? And how does it work without a base resistor?
And why are they using a high side switch on the low side and a low side switch on the high side? You're trying to use the transistors as common emitter saturated switches. They're showing the transistors used as common collector emitter followers.
When you use transistors as saturated switches, you can get the Vce very low, but it can take a LONG time to get the transistor to recover from being deeply saturated. If you use a transistor as an emitter follower, you wind up with a voltage drop from the base to the emitter, but since the transistor never enters saturation, it responds much more rapidly - there is no 'recovery' from a saturated state. I'll throw a simulation together for you. OK, here you go; see the attached. 'In' is a 50kHz 50% duty cycle nearly-square wave. Your circuit (with the base drive inversion corrected; the comparator is no longer needed though) is on the left, a totem-pole emitter follower on the right.
It was necessary to use a common emitter front-end to get the required gate voltage swing; otherwise a logic-level input would have only raised the gate to 4.3v, even if the logic went all the way to 5v. Notice that Gm1, the gate signal for your driver is nearly a triangle wave. The gate signal for Gm2 is nearly a square wave. Click to expand.The TIP42 is not a Darlington, but the NTE2343 is (the TIP142 is not exactly the same as an NTE2343, but 'good enough' for what I was trying to show.) Darlingtons have very low bandwidth compared to a standard BJT.
The Vce(sat) varies, but with a light load, it'll go as low as 0.63v. It's the Vbe that's doubled over a standard bjt, as there are essentially two base-emitter junctions in series. The Sziklai pair has a lower Vbe than a Darlington does, with a similar gain. Sziklai pairs use both a PNP and NPN transistors, instead of two of one kind as a Darlington does.
Ok so Q3 is an upside down PNP and the emitters of Q3 & Q4 are tied together? And when voltage is applied to the bases of Q3 and Q4 (Q5 OFF), Q4 conducts and charges the gate capacitance and Q3 does not conduct, keeping the charge in? Then when voltage is removed from the bases of Q3 & Q4 (Q5 OFF), Q3 conducts and allows the gate charge to bleed out?
I thought a transistor (NPN or PNP) needed current at it's base to conduct. When a PNP is used upside down, it acts like a NPN except it conducts with no current at the base, and does not conduct when current is at the base?