TL494 push pull output not going to zero - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: May 26, 2016, 03:10:21 pm
content: I am trying a circuit using the TL494 and the output of which is given to 2 pairs of tranisistors in push pull mode.I am measuring my waveform across the secondary of the gate drive transformer. The GDT is unloaded. But I see that the waveform is non centered around zero.I have check as per the TL494 datasheet the Vce saturation of its internal transistors is around 1.3V so possibly the output trasistors may not be going off fully I haven't been able to exactly zero down to the problem. Any help would be highly appreciated.Regards.ZR

author: Yansi
author_rank: Super Contributor
timestamp: May 26, 2016, 05:53:48 pm
content: Because it is not meant to go to absolute zero and your circuit is seriously flawed.First, what you have to know is the the saturation voltage of the TL494 is quite high, because of internal structures that limit the current through the output transistors. The saturation voltage of the 494 output transistor is not safely low enough for this circuit*, so to dismiss the first flaw, you should work the output stages in the 494 with their collectors on V+ and use the emitters as outputs, otherwise ur simply not going to the ground, not enough! The Vcesat of TL494 outputs is merely 1V, together with the Ube of BD140 it is about 1,6V. Not very good.*Thats why PC power supplies with 494s have those two transformer driving transistors sitting wit their emitters on a pair of diodes.The second serious flaw is you are missing diodes on the outputs of your complementary pairs. They will conduct during demagnetization of the primary inductance of the transformer. A schottky type diodes are recommended, as only then the voltage resting on the mosfet gates during dead time will be minimum possible.The third flaw is you are missing a DC blocking cap in series with the primary. Put there about 470n to 3uF foil cap (depending on the secondary load).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 26, 2016, 06:07:25 pm
content: Ah yes, my old circuit symbols, in fact I dare guess this circuit was based off an old schematic of mine.The GDT magnetizing current acts to pull the emitter followers below GND for part of the cycle. The "slop" voltage (range between positive and negative conduction) for this circuit is dangerously wide, like 2V, leaving you with little to no room for safe gate voltages. This comes first from the emitter followers, which don't conduct until the emitter is more than 0.6V above or below the base voltage; the 494 outputs themselves don't pull down terrifically strongly, as mentioned.If you don't need much drive current: consider the TL598. Use schottky clamp diodes on the outputs, to prevent the voltage from going beyond -0.3V negative or VCC+0.3V positive.If you need more current, consider using a dual gate drive IC instead of emitter followers.Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: May 28, 2016, 12:45:47 pm
content: Thanks Yansi and T3sl4co1l,I realize from your comments that the flaws were serious in the current implementation of the driver.So I made another circuit with a mosfet driver this time its a TC4420 with 6A peak drive.When I check the output of the GDT with out any load (open circuit) then I get perfect waveforms with steep rising and falling edges.But when I connect a capacitive load to it 2.2nf (this is close to the mosfets gate capacitance) then I get distorted waveform as per the attachment this waveform is across the 2.2nF capacitor at the output of the GDT.I am wondering whether this distortion is caused due to the GDT pushing current into the schottkey diodes long after the driver as switched off?The GDT is trifillar wound with 11 turns and is a ferrite toroid.The mosfet driver is well decoupled with 0.1uF and 10uF caps (not shown in the schematic)The waveforms are measured using a div 10 setting on the oscilloscope probe.Any comments would be highly appreciated.

author: Yansi
author_rank: Super Contributor
timestamp: May 28, 2016, 02:39:49 pm
content: Thats what leakage inductance together with the load capacitance does: rings and oscillates. Your transformer might not be wound "tightly enough" with low enough leakage inductance. Also specifying a transformer only by number of turns doesn't really say much.What is important, is to calculate the flux density in the transformer. It is not acceptable, to just take a random ferrite core, put there some turns and think it will work. It might and it might just not.Usually, there is more into it: not only must be the operating flux density kept within a safe level, but also a magnetizing current should be checked. Mostly the transformer can work with a very low turn counts even with safe level of flux density, but such low number of turns implies low magnetizing inductance, thus high magnetizing current. That current is what bothers the gatedriver (integrated or transistorized).For example if it is enough to use only a few hundred miliamps to drive the mosfets directly, it does not make sense to design the transformer with 2 amps peak magnetizing current. (remember, the driver has to deal with the magnetizing current too).But that was just a side note about the transformer design.The magnetic flux density can be calculated using standard formulas, like the Farady's law U = -dPhi/dt and the magnetizing peak current can be easily obtained from the differential formula of voltage on an inductor: U = L * di/dt.Note you have to design in in a way the transformer will work correctly in the whole range: You have to count with the highest permissible supply voltage, lowest possible drive frequency, maximum/full duty cycle, etc.To dampen the ringing on the outputs, I would recommend discarding the resistor from the primary completely (it shouldn't be there anyway if the driver does provide correct dead time generation which here it does) and putting it on each of the secondary. That will dampen the LC circuit (leakage inductance with the load capacitance).Small amount of ringing can be tolerated on the waveform, but it have to be ensured, none of those peaks can switch any of the mosfets on any time. So the peaks on each gate have to be safely lower, than the mosfet gate threshold voltage, which might be as low as 2V.PS: You can't simulate a mosfet gate with a plain capacitor. The mosfet gate is a nonlinear capacitor and all sorts of problems (like the miller capacitance Cdg must be taken in account).PPS: You can also try to twist the wires together and then wind the toroid with it, if you haven't done it that way. That will also lower the leakage. You can also measure the leakage. Short one of the windings and measure the transformer inductance on another free winding. The uH you will see is the leakage mostly. The lower value measured, the better. A coupling coefficient can be also calculated from the ratio of magnetizing and leakage inductance.P3S: The schottky diodes shouldn't be neccessary for the TC4420, as the driver has mosfet output stage with their natural substrate diodes. The schottky can only help lower the forward drop.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: May 28, 2016, 03:22:18 pm
content: Thanks Yansi,I will comment on the issues you bought up a bit later.But just for starters.The total gate charge of the mosfet is 74nCand if I am intending to charge this capacitor in 50nS then with I = Q/t i get 1.5Amps.Wouldn't this amount of current have to be supplied by the GDT apart from the magnetizing current?ZR

author: Yansi
author_rank: Super Contributor
timestamp: May 28, 2016, 03:34:49 pm
content: Magnetizing current is independent of the load current. It's just there. Given by the primary voltage and magnetizing/primary inductance. Also phase shifted against the load current. (it loads the driver with apparent power, VA).Gate charge is usually specified for 10V gate voltage. For your 17V or thereabout the nC figure will be higher.The peak current required to switch on/off the mosfet is not supplied by, but through the transformer. The load current is supplied by the driver. Theoretically, it is an instantaneous process. The same current flowing to the mosfet gate must be visible on the primary side. (Remember how current transformers work...). And in addition to that, there is the phase shifted magnetizing current you could see in the primary circuit, if you have used a current probe (and no load on secondaries). The magnitude of the magnetizing current is independent of the load current.