How to calculate Transistor Rc and Re for proper mosfet Vg - Page 1
Forum section: electronics / beginners

author: Mechatrommer
author_rank: Super Contributor
timestamp: July 30, 2016, 08:39:14 pm
content: as i want to make switching P-MOSFET from 60V supply, i need robust push pull bjt driver as suggested by bktemp here...https://www.eevblog.com/forum/beginners/vref-voltage-amplification-circuit-please/msg980594/#msg980594i'm having trouble understanding his explanation for my particular objective, so here again to make a bit clear.the main objective is to ensure during P-MOSFET ON period, Vsg is no more than 15V (or -15V Vgs) in any state of operation, hence Vg to GND cannot be lower than 45V to avoid damage to mosfet. secondary objective is the push pull NPN+PNP bjts driver must be capable of supplying and draining 1A to/from the mosfet gate for efficient switching. if i can pull Vg to 0V (gnd) without damaging the mosfet then its easy, but this Vg biasing is difficult to calculate. To get the 2 objectives done (Vg biasing and high gate charge/discharge current), we need to find proper Rc and Re value, this is why this thread is all about.to make this easier, there are 3 state in the operation.1) 0.png: initial condition, comparator output in low position, where all I = 0. Vg = push pull base = 60V (no problem here)2) 1.png: when LM393 output goes high (15V), mosfet gate start to discharge from 60V to 45V. i know the push pull need to pull at rate of 1A during that time, so i drew the current flow in the picture, assuming the PNP in push pull driver is in saturation (hFe = 10), ie its Ib = 1A / 10 = 0.1A. the biggest trouble is, what is the leftmost NPN's Ie? and Ic? and hence how to calculate Re and Rc?3) 2.png: when LM393 output goes low (from high), mosfet gate start to charge again from 45V to 60V. since the leftmost NPN is now off, i'm guessing Ie = 0, so i drew the current flow again in the picture. now it become easier, i think i can calculate Rc now, but not sure if my assumption is correct.my assumption:from state 3:Rc = (60v - 45v) / 100mA = 150 ohm. is this correct?but when i go to state 2 again, i'm not sure because Ie = Ic + 0.1A (from push pull's base), Ic is the problem... hence i cannot think on how to calculate Re. and Rc = 150 ohm seems like too low, since bktemp in the other thread suggested the value of 3Kohm, but then 3Kohm wont be able to charge mosfet gate at 1A, this is real conflict! please help. thanks.

author: Mechatrommer
author_rank: Super Contributor
timestamp: July 30, 2016, 10:10:32 pm
content: Quote from: blueskull on July 30, 2016, 09:20:48 pmyes if the leftmost NPN is in saturation. if not... then Rc, Re, and Rt will form the divider, Rt is the mystery... Rt = Vce / Ic of the leftmost NPN... anyway. what i'm interested at is how the current flow in and out of the leftmost NPN... how to make kirchoff analysis on every leftmost NPN's pins... at base, collector and emitter, taking into account hFe and Vce in linear region. or even maybe, analysis in saturation region.

author: Mechatrommer
author_rank: Super Contributor
timestamp: July 30, 2016, 10:29:59 pm
content: or in other word... let say the leftmost NPN is Q1 (picture below) i want to control how much current that pass through Rc = 200mA and then divided to push-pull's base = 100mA, and hence 100mA will go into Q1's collector, and then i can ensure Ie = 100mA (assuming Ib = Ie / hFe = negligible), Ve = Re * 100mA, and then Ve + Vce of Q1 will not go lower than 45V? i can calculate Ve = Re * 100mA = 15V (Vb), but for Vce i need to find the Ib value so that in linear region Vce will be 30V... is this possible?

author: bktemp
author_rank: Super Contributor
timestamp: July 31, 2016, 06:32:58 pm
content: 100mA is too high, it will result in 60V*100mA=6W power dissipation.2SC2383 and 2SA1013 have a current gain of at least 60, typically 100. So for 1A a base current of about 20mA is ok.For 15V gate voltage Rc must be 20mA * 15V = 750 ohms.To get 20mA current, Re must bei (Vbase-Vbe)/Re = (15V-0.7V)/20mA=715 ohms. Because of the voltage drop on the 3k pullup, the base voltage will be a lower, so 680 ohms should be ok.The worst case power dissipation (at 100% duty cycle) will be ~0.28W in Re and 0.3W in Rc, so use 0.5W resistors. The current source transistor will dissipate 0.6W, so it will get quite hot (~84°C above ambient). Maybe a small heatsink will be necessary.By the way: You don't want saturation on any transistor in this circuit, because if they saturate they will get very slow (depending on how much you drive them into saturation it can take several microseconds until they turn of, even with a negative base current). To keep them in linear operation the supply voltage must be at least Vbase + Vgate = ~30V.The 100k pulldown and the diode are not necessary.

author: Zero999
author_rank: Super Contributor
timestamp: July 31, 2016, 07:00:56 pm
content: Quote from: Mechatrommer on July 30, 2016, 08:39:14 pmThe push-pull transistors are emitter followers, so it's impossible to saturate them, without driving the bases outside of the power supply voltage.As mentioned above, assume a minimum Hfe of 60 which would be a maximum base current of 162/3mA.

author: Mechatrommer
author_rank: Super Contributor
timestamp: July 31, 2016, 07:26:32 pm
content: Quote from: bktemp on July 31, 2016, 06:32:58 pmthanks for coming to the rescueyes i will aware on heat dissipation. i'm willing to lower my requirement if heat and hence power loss in the driver is unacceptable. i just need to get around this transistor driving current first. but i think i got it now (from my own drawing) i was just confused how the current is divided between leftmost NPN's collector and into the push pull's base node during state 2 (1.png) but i guess during leftmost NPN is ON, majority of the 20mA will come from the push pull's base into the emitter where current passing through Rc during the time is almost 0, correct? otoh i believe during state 3 (2.png) when leftmost NPN is OFF, Ie = 0 and push pull's base will get the 20mA supply totally from Ic. i hope that will be the case.Quote from: bktemp on July 31, 2016, 06:32:58 pmthanks for this advice i will meditate on this.Quote from: bktemp on July 31, 2016, 06:32:58 pmi think they are, its for the leftmost NPN's reverse Vbe protection since LM393 output will swing to negative rail (-15V) when its internal bjt is switched on (output LOW). the 100k pulldown will improve the diode recovery respond from my breadboard testing and simulation.Quote from: Hero999 on July 31, 2016, 07:00:56 pmi know about the push pull base follower. i mean (i hope) the leftmost NPN, the one who is driving the push pull will go into saturation so it can supply 100mA into the push pull's base. but as bktemp has adviced, i will want to avoid saturation to the leftmost NPN in this case, as fast respond to higher freq PWM input is another concern.Quote from: Hero999 on July 31, 2016, 07:00:56 pmnow as you seconded bktemp, this put much more relief to my brain as it will put lesser burden on the push pull driver (the leftmost NPN), thanks. i'll recalculate.

author: Mechatrommer
author_rank: Super Contributor
timestamp: July 31, 2016, 07:56:40 pm
content: Quote from: bktemp on July 31, 2016, 06:32:58 pmnow i missed this, from your suggestion. Rc = 750ohm, Re = 680ohm. if somehow the leftmost NPN go into saturation (Vce = very close to 0V), this Rc and Re will form static divider of 60V * (Re / (Re + Rc)) = 29V (rounded off), and if the push pull is able to follow this voltage, there will be 29 - 60 = -31V of the mosfet's Vgs, this exceeded the maximum rating spec and will damage the mosfet's gate.how can i put the leftmost NPN into saturation? this is how (from my logic). since its collecting 20mA from the push pull's base, then all i need to do is supplying its Ib with 20/10 = 2mA from LM393 side, is this mentality correct? if so, then this is the dilemma. i need to ensure saturation will never occur, ie Vce of the leftmost NPN must be minimum 60 - 15 - 15 = 30Vce. so i can safe bet the mosfet's gate voltage is Ve + Vce = 15 + 30 = 45V, ie 45 - 60 = -15Vgs. the big question is... what Ib current (from LM393) that will ensure Vce not smaller than 30V? and hence avoiding saturation to the leftmost NPN in any situation? do i have to get it from graph plots in the datasheet? since i cannot rely on some optimistic hFe value and use in Ib = Ic /hFe formula, hFe will change with so many parameters (and Vce too!) that i dont know where to start from the datasheet.

author: bktemp
author_rank: Super Contributor
timestamp: July 31, 2016, 08:07:04 pm
content: Quote from: Mechatrommer on July 31, 2016, 07:26:32 pmFor the steady state voltage and current the right side current booster stage can be omitted, because it only increases the current and after the gate is charged no current flows. It also lowers the voltage by 1 Vbe drop (~0.65V), but this is negligible.When the current source turns on, the voltage drop across Rc will initally be 0V, so no current is flowing here and all the current is flowing into the current booster stage and into the mosfet gate. When the voltage rises, Rc takes more current, reducing the available gate current (this is a small drawback of this simple circuit) until it reaches 15V when all the current is flowing thru Rc.When the current source switches off, the voltage across Rc will initially be 15V, so 20mA are flowing thru Rc. These 20mA are now drawn from the current booster and out of the mosfet gate.To reduce the peak power dissipation in 2SA1013 (60V*1A=60W!), you can add a resistor in collector wire: If you add 30ohms, it will take half of the voltage and therefore half of the power dissipation.QuoteYou are right, I missed the -15V.For such cases I prefer LM311, because it has a seperate emitter pin for the output transistor.

author: Zero999
author_rank: Super Contributor
timestamp: July 31, 2016, 08:19:14 pm
content: Quote from: Mechatrommer on July 31, 2016, 07:26:32 pmYou won't get that transistor into saturation either, not unless you remove Re and get the full voltage across Rc, which would destroy the MOSFET.If you want the voltage across Rc to be 15V, then you need a gain of 1 so Re must equal Rc.

author: Mechatrommer
author_rank: Super Contributor
timestamp: July 31, 2016, 08:36:54 pm
content: ok i think i have a clearer picture now, thanks guys.

author: bktemp
author_rank: Super Contributor
timestamp: July 31, 2016, 08:38:16 pm
content: Quote from: Mechatrommer on July 31, 2016, 07:56:40 pmThe transistor will not go into saturation, because of Re:The transistor adjusts the collector current to get a voltage of Vbase-Vbe on emitter =15V-0.7V=14.3V. Because there is a fixed resistor on emitter, there will be a constant current of 14.3V/680ohms=21mA. So Vce=0 will never happen unless the 60V supply voltage will go below 30V when 15V will be across Rc and 14.3V across Re. You can add a 15V zener diode across Rc just in case to avoid voltage spikes.QuoteThe base voltage of the current source does not need any current limiting, because of Re. The base can be driven directly by a square wave voltage (e.g. microcontroller pin). If your 60V can go below 30V, you can reduce Vbase to get a higher margin by either reducing the 100k pulldown to form a voltage divider or by adding a zener diode to limit the base voltage. For example if you reduce the base voltage to 5.1V by adding a 5.1V zener diode, Re needs to be reduced to about 220 ohms.If you calculate the typical base current needed for supplying 20mA collector current you simply take the typical value from the datasheet. The tolerances are quite high, so you don't care if hfe is 60, 100 or 150. If possible without other drawbacks, the driving circuit should be able to supply 5-10x the necessary current just in case you have a transistor with a lower hfe value. In this case the 3k pullup can supply enough current. The voltage drop will reduce Vbase a bit and therefore the current, but this can be compensated by adjusting Re. You can always fine tune the values if necessary. The calculated values should be a good starting point.