How to know output voltage of a emitter follower circuit ? - Page 1
Forum section: electronics / projects

author: drakejest
author_rank: Regular Contributor
timestamp: July 03, 2021, 07:51:12 pm
content: I am driving a mosfet at 5v, themosfet i am usingturns on at Vgs = -4.5v Since i have a supply voltage of as low as 5v, i need to be careful that the mosfet gets as close to 5v as possible, hence it will cause temperature problems at higher current loads.the totem pole circuit is made up ofNSS40301andNSS40300How do i calculate for the output voltage at the gate of the mosfet. If possible i would not want a voltage beyond 4.5v on the gate since i had done my temperature calculations based on the Vgs= -4.5v resistance

author: SiliconWizard
author_rank: Super Contributor
timestamp: July 03, 2021, 08:05:02 pm
content: As Q2 is a bipolar transistor, and its base is driven at the supply voltage through R2, won't Q4's gate voltage (thus Q2's emitter voltage) be at worst 1V below the supply voltage (VBE_sat is 1V max according to the datasheet), thus @5V, it might be as low as 4V, not meeting your requirement if I got it right.

author: Pulsepowerguy
author_rank: Regular Contributor
timestamp: July 03, 2021, 08:21:31 pm
content: The base to emitter voltage drop of Q3 is approximately 0.636V at room temperature but the voltage on the drain of Q1 is not exactly zero so the output can only approach 0.636V relative to common. Instead of a follower with BJTs one could use a FET to pull the gate of Q4 to common. This requires what I think of as a true totem pole circuit (the one shown is, in my mind, a follower).I would consider an integrated circuit as a discrete solution takes a bit more work, unless you are more interested in the journeyHere is an interesting app note on gate drivers:https://www.ti.com/lit/ml/slua618a/slua618a.pdf?ts=1625342067638Look at figure 11 specifically for your needs.

author: TimFox
author_rank: Super Contributor
timestamp: July 03, 2021, 08:28:08 pm
content: If you Google for that circuit, it is a “complementary emitter follower”, not a totem pole. The output stage of a 7400 TTL gate is called a “totem pole”.Search now, before the language becomes politically incorrect.

author: SiliconWizard
author_rank: Super Contributor
timestamp: July 03, 2021, 09:07:14 pm
content: Yeah. The classic totem pole with complementary transistors is done with MOSFETs. Implementing a "totem pole" with bipolar transistors is typically done with two transistors of the same type. Typically, two NPN transistors. Like this:https://fr.wikipedia.org/wiki/Push-pull_(%C3%A9lectronique)#/media/Fichier:7400_Circuit.svg

author: drakejest
author_rank: Regular Contributor
timestamp: July 04, 2021, 07:21:32 am
content: I see i was wrong to call it a totem pole circuitIm stuck yet really close to finishing the projectSo all i have to do is find a transistor whos Vbe is at max 0.5v. time to go digging for parts then

author: xavier60
author_rank: Super Contributor
timestamp: July 04, 2021, 08:41:08 am
content: I put a resistor between B-E so that the Gate eventually charges to the full voltage, like 470Ω.

author: xavier60
author_rank: Super Contributor
timestamp: July 04, 2021, 08:55:11 am
content: Most of the time is spent moving the Gate through the plateau voltage level. After that, the Gate is easier to charge and discharge.

author: Marco
author_rank: Super Contributor
timestamp: July 04, 2021, 10:58:01 am
content: Quote from: drakejest on July 03, 2021, 07:51:12 pmMax Vgsth is -2.5 according to that datasheet?Any way, the simplest rule of thumb (0.6V Vbe) puts you over your design constraint. Make something with more margin for error.

author: drakejest
author_rank: Regular Contributor
timestamp: July 04, 2021, 11:52:46 am
content: So i cant find a transistor that has low Vbe so i tried going on using all mosfets, with a push pull mosfets that takes significantly lower atleast 10x lower gate charge needed. and since they mosfet they should drop only very little voltage (i think).I might find better mosfets for the push pull but for now i think these two are best that is availablePMV213SNandDMP3050

author: drakejest
author_rank: Regular Contributor
timestamp: July 04, 2021, 11:59:41 am
content: Quote from: Marco on July 04, 2021, 10:58:01 amI ment that at -4.5v is when the resistance is acceptable to prevent heat problems for my application.Quote from: Marco on July 04, 2021, 10:58:01 amIt has been so long that i have used discrete transistors that i have forgotten about them. The lowest i could find are even 1v, and i could not source better mosfets that wont go overbudgetI have tried making the push-pull using mosfets with significantly lower gate charge so it would be easy to turn on with a level shifter, what do you think ?

author: drakejest
author_rank: Regular Contributor
timestamp: July 04, 2021, 12:02:44 pm
content: Quote from: xavier60 on July 04, 2021, 08:41:08 amI do not follow how that works, can you please elaborate?

author: xavier60
author_rank: Super Contributor
timestamp: July 04, 2021, 12:09:11 pm
content: Quote from: drakejest on July 04, 2021, 12:02:44 pmIf you put a resistor between Base and Emitter of Q2, Q4's Gate voltage will settle to the same as Q1's Drain voltage.

author: drakejest
author_rank: Regular Contributor
timestamp: July 04, 2021, 12:51:58 pm
content: Quote from: xavier60 on July 04, 2021, 12:09:11 pmThe voltage does increase when i simulated it , although not fully to 5v but its something, im also not using the same transistors (no model for the one in the question).https://ibb.co/vJYwshGI am also not getting the spike edgeshttps://ibb.co/MgST53v

author: Zero999
author_rank: Super Contributor
timestamp: July 04, 2021, 12:59:10 pm
content: Quote from: Pulsepowerguy on July 03, 2021, 08:21:31 pmOnly whilst there's a significant emitter current. Once the MOSFET gates have been charged, the emitter current will be near zero and the base voltage will be near 5V, or 0V, depending on the input voltage.

author: drakejest
author_rank: Regular Contributor
timestamp: July 04, 2021, 01:02:37 pm
content: Quote from: Zero999 on July 04, 2021, 12:59:10 pmSooo.. the original circuit will work fine?...but when i tried to simulate a circuit, although not using the same transistors, there is really a significant voltage drop on the gate relative to the souce.Circuit

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: July 04, 2021, 02:13:24 pm
content: There are MOSFET drivers made for doing exactly this, but I don't know if they meet your voltage needs.

author: Peabody
author_rank: Super Contributor
timestamp: July 04, 2021, 02:13:53 pm
content: I'm confused. Your later all-mosfet model has the P-channel on top and the N-channel on the bottom. But the original bipolar model had the NPN on top and the PNP on the bottom. So which is it? If you put the PNP on top and the NPN on the bottom, won't that provide the full 5V and 0V range to the mosfet gate, or very close to that? Can you simulate that? The control polarity would have to be reversed, but that's just software.

author: drakejest
author_rank: Regular Contributor
timestamp: July 04, 2021, 02:48:49 pm
content: Quote from: Peabody on July 04, 2021, 02:13:53 pmto be honest i dont really know anymore, the first circuit was based on a "totem pole circuit" but it seems that is not the correct term for it, if you google that keyword circuit like my first model shows up, for the mosfet version i am basing it on an answer to thisthreadthat said, simulating using 2n3904 on the bottom and 2n3906 on the top i am getting a much higher voltage on the gatebeforeandafterso i am now very confused on how do i proceed haha, i could go back to the first model i have and swap places the transistor, or use the mosfet version. both of them should be now capable of pushing the gate voltage near the source.Quote from: Conrad Hoffman on July 04, 2021, 02:13:24 pmI really did try to look for one, but buying from the specific vendor that have it is just to expensive to ship in my area.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 04, 2021, 03:32:02 pm
content: Just how fast do you need this poor thing to go? 3A bipolars at full poop will do it in about 18ns. 120R is a quite strong pullup, consuming 42mA continuously while on (and more at higher supply voltage).The discrete inverter (doing it with MOSFETs) is not a bad way to do it, but it does suffer from shoot-through during the period when the gate voltage is above (N-ch Vgs(th)) and below VDD + (P-ch Vgs(th)). Which for most parts is maybe a volt away from the supply so the 1...4V range will draw more or less the full current those transistors are capable of (which is at least as much as peak gate current will be).There's really no penalty to using BJTs, so it might be that a pair of 3904/6 or 4401/3 are used to drive the MOSFETs, which drive the final big gate, or yet more BJTs for an even bigger gate (5-10A BJTs are commonly used in this way, for large industrial IGBT modules with 100s of nC gate charge and +/-15V swings).If you're just turning stuff on and off, a 10k pullup and optional 3904/6 follower (i.e., capable of 100-200mA) will do. This is already in the 100s of ns range, good for certain PWM applications. The pullup merely sips current, maybe not something you'd want for battery operation but easy to ignore elsewhere. 4401/3 or PBSS303NX/PX or the types shown, will do fine for faster speed while using a corresponding lower pullup value, or some other means of drive (a 74HC or CD4000 family logic gate can do well here, translating level from the 3.3V input to up to 6V or 15V respectively).And by 3904 and 4401 and friends, I mean 2N- for classic TO-92, MMBT for SOT-23, or various other manufacturer-specific designations, alternate packages, etc..For higher supply voltage (over 10-15V), I strongly recommend adding a voltage limiter, to avoid applying excessive gate voltage. Vgs(max) might be rated 20V or whatever, but that doesn't mean you want to hang at that forever, you're left with no headroom for supply transients, and the excessive gate swing costs a huge amount of drive power and switching time. Simple limit is just a series resistor from the first (level shifting / common source) transistor, then a zener from gate to VDD. A 5-10V zener will do.An alternate option: use a fixed current from the common-source transistor, by driving it with a known Vg(on) input level, and adding source degeneration (a resistor between source and GND). The source voltage Vs rises to a bit below Vg(on) - Vgs(th), and the drain current is set by the resistor value, Id = Vs / Rs. Vgs(th) is pretty sloppy, so this isn't great with MOS; it's perfectly serviceable with BJTs, with the tighter spread in Vbe (say 0.5~0.8V over commercial temperature range). Note that the drain/collector won't saturate to a voltage lower than Vs, so there is some tradeoff. Anyway, once the current is fixed, simply apply that current to a specific value pullup, and the pullup drops a voltage V = Id * Rpu. Thus, Vgs(on) for the output, is defined by input Vg(on) and the resistor ratio.Tim

author: drakejest
author_rank: Regular Contributor
timestamp: July 04, 2021, 04:02:38 pm
content: Quote from: T3sl4co1l on July 04, 2021, 03:32:02 pmAbout 1MHz, of course if it can go beyond i would not mind it.Quote from: T3sl4co1l on July 04, 2021, 03:32:02 pmcan you show me an example circuit how that is done? So you mean like a voltage follower made up of BJTs driving a voltage follower made up MOSFETs which then drive the final mosfet?Quote from: T3sl4co1l on July 04, 2021, 03:32:02 pmIf you still remember me from my previous post, it mentioned that i will be having upto 24 volts supply. although all the mosfets i mention are rate for Vg= +- 25v, i do plan to add some protections after i resolve which circuit i should use.Quote from: T3sl4co1l on July 04, 2021, 03:32:02 pmIm sorry i dont quite get this one, can please show me an example circuit so i can better understand.

author: magic
author_rank: Super Contributor
timestamp: July 04, 2021, 05:34:02 pm
content: How about: replace Q3 with a Schottky diode and use a more powerful FET (or NPN?) for Q1.

author: Zero999
author_rank: Super Contributor
timestamp: July 04, 2021, 09:20:08 pm
content: Quote from: drakejest on July 04, 2021, 01:02:37 pmIs that once it's settled to the steady state? During switching there will be a significant voltage drop, as the gate charges and discharges.How much current is the MOSFET passing? The on resistance will be a bit higher, if there isn't quite so much gate drive, but it might still be acceptable.

author: drakejest
author_rank: Regular Contributor
timestamp: July 04, 2021, 10:14:14 pm
content: Quote from: Zero999 on July 04, 2021, 09:20:08 pmyes its already steady state.I switch up the resistors PNP on top, NPN on bottom as suggested by @peabody, i got a higher voltage at the gate this time,circuit is on steady state.

author: xavier60
author_rank: Super Contributor
timestamp: July 04, 2021, 10:25:35 pm
content: In a practical circuit, Q4 and Q6 will be damaged by excessive B-E current.Best to use small MOSFETs in that topology. I sometimes use a SO8 package with 2 complimentary MOSFETs inside. It's AO branded.I put a low ohm resistor or ferrite bead in series with VCC to reduce the shoot through current.http://www.aosmd.com/pdfs/datasheet/ao4606.pdf

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 05, 2021, 01:30:29 am
content: Quote from: drakejest on July 04, 2021, 04:02:38 pmWith what kind of load? You're dreaming if you think you can get any kind of power out of an SMT P-ch at that frequency.You must also understand the switching loop, and load reactance, at that kind of frequency. An anonymous "LOAD" block is no longer sufficient, nor the skyhook to +5V.What will you build this on? Ground plane I hope.QuoteNo, not voltage follower, not by MOS. What you have here,https://i.ibb.co/X25MN6Q/image.pngis a resistor-loaded inverter cascaded into a complementary (CMOS) inverter.MOS follower (sources together, output) loses too much voltage to be practical here (a bit over Vgs(th)).An example looks like this,typical input (from the left) would be a resistor divider from a logic level signal, possibly with a capacitor in parallel with the top resistor to speed up transition. For 3.3V CMOS levels, 2.2k from CMOS pin to base, and 1k from base to GND, would be typical. With probably up to 47pF in parallel with the 2.2k; MPSH10 doesn't need much, it's fairly fast as is (MMBTH10/81 being the SMT equivalents, those are just the SPICE model names in the screenshot).Notice L1 || R5 providing supply impedance, to reduce shoot-through current, at some expense to pull-up speed and "top" stability.Waveform:Into 10nF + 2.2R or 1R, I forget which. So, equivalent to a 100nC gate, more or less. Notice the "top" is going all over, due to flyback from the inductor (it rises and overshoots due to Q8 B-C forward bias, settles, then settles again with a damped ringdown determined by Q5+Q7+Q8 capacitances, L1 and R5).And here's the physical build. The bottom side is contiguous ground plane, and there are some vias painstakingly added to top side GND to ensure low ground inductance and high current capacity. Signal input on the right, with a differential level-shifting circuit (not shown) that uses 5 transistors (and hence allowing the lower level, somewhat slower input waveform, top trace). The pullup CCS is closer to the middle, with the "470" chip resistor.Note three 1uF bypass caps on +V (green wire) (one hidden under L1, one by Q8, one bottom-left).Note the use of a CCS for sharper pullup than a resistor. A pullup resistor's current drops as voltage rises, giving an exponential curve. The CCS keeps the current stable much longer, giving a straighter ramp, faster for the same on-current.Note the emitter follower into the CMOS inverter: even small MOSFETs have big thick gates (~30pF+), and significant advantage is had by emitter-following the mere ~10mA BJT inverter. Again, transition time into the CMOS inverter costs shoot-through, raising their dissipation significantly and wasting supply current (or needing more snubbing, at the expense of output speed).QuoteLike this, of course the pullup resistor should be emitter-followed for speed, and optionally replacing the NMOS with a NPN.This also shows a local gate drive supply:https://www.seventransistorlabs.com/tmoranwms/Circuits_2010/Regen_Motor.pngNote the supply is hanging below +Batt, as I briefly mentioned in the gate driver thread ("could connect Vs to source, if not for..." or something like that). Pass would be better than zener shunt regulated, but that requires a negative regulator (or building one with an emitter follower).Basically, this is a proper P-ch drive, albeit not a very efficient or compact one (as discrete circuits tend to be).Tim

author: gman76
author_rank: Contributor
timestamp: July 16, 2021, 06:54:33 pm
content: Quote from: SiliconWizard on July 03, 2021, 09:07:14 pmYou're right. That's because on a planar process of the 70's, NPNs were the only available type that had any beta and BW. All early 7400 logic gates used only NPNs.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: July 17, 2021, 12:22:01 am
content: Quote from: drakejest on July 03, 2021, 07:51:12 pmNo it's not.Threshold voltage is between -1 and -2.5V according to page 2.Just because they specify some drain resistance at some specified voltage on the first page (which is mostly for marketing) does not mean that is the threshold voltage.In your case this changes the problem from how to turn it on well enough, to how to turn it off quick enough and reliably.