I omitted the rest because the rest works fine. But here is the whole circuit.
V6 is a DC source (battery), I'm using a pulse source because spice doesn't simulate a mechanical switch. So when the pulse source V6 goes to zero this simulates the switch open. When the voltage from V2 hits about 8.5 V the zener conducts turning on Q17 and Q20 which shuts down U12 which blocks the battery V6.
The purpose of U11 is to switch off the load D2 and D3, then Q16 conducts and supplies the load represented by D5 and D4. I've built and tested this; the only problem I'm encountering is what I stated in my original question. Both gates won't go to ground it's not a problem for U12 I still get over -5Vgs . U11 's gate stays about
1.2V above ground giving me -3.8VGS .It's not a big deal the NTR0202PL is oversized for the load but it was the right price and package. I was just wondering if I could eek out another 500mV or so, to get Vgs differential in the 4.5V range.
-------You said you used a pulse to simulate a mechanical switch. I assume that means the circuit the way you actually built it has a switch not shown on the drawing. I haven't taken the time to go over the whole circuit, but I think when you turn off V6, then the base of the pnp transistor gets pulled to ground through R23, which might turn it on.
Here are the simulated waveforms if it explains more clearly what I'm doing.
Top trace (RED ID3) the switch is closed : 50mA current pulse for 1mS steady state 20mA.Through the first load D2 and D3.
Second trace ( RED I(U12:4)) is the mosfet drain of U12 conducting from the battery until Q20 is biased and turns it off. The 7805 is now supplied by a DC/DC converter represented by V2 pulsed source in the schematic.
Bottom trace (GREEN I(R12)) shows the second load conducting at 135mS (switch open).
It would help if you explained what you are trying to do. It seems like a lot of extra circuitry to switch among various sources and loads, and I don't understand why you are seemingly trying to turn on a transistor to set Vgs of a PMOS to zero to turn it off, rather than just have a resistor from gate to source and pull the gate to GND to turn it on.
If you can post the circuit in LTSpice ASCII, it might help.
Also, I did not see any power supply or battery (other than the pulse sources), and there is only one capacitor, which seems to be a bypass for a power source that is not identified. The 7805 should have bypass capacitors.
LTSpice has a voltage controlled switch component, but it only works for DC in one direction. And you must set some of the parameters for it to work.
This is for a dual channel photovoltaic driver one load turns on the high side fet (D2 and D3),the other turns it off (D4,D5). This is turning on/off the fly back represented by V2 pulsed source. I've tested this in the flyback and it works well, I used mechanical switches in place of the two FETS to test it. Now i'm doing it using semiconductors.
There really aren't that many components I have drawers full of dual transitors with and without bias resistors as well as single pre-biased BJT's. So it's not as bad as it looks. But I know it can be done with a dual comparator like the lm2903. I just thought now is a good time to use some of the transistors.
If i turn on the transistor and sink current through the gate resistor it raises the gate voltage up to Vce minus Vsat.Shutting the fet off.
to turn it on.
Huh thats what I'm doing the emiter is at the source and the collector is at the gate tied to ground through a resistor, same thing. I want to control when they turn on/off.
I explained this in my previous post spice doesnt simulate a mechanical switch. Even if I use a ridiculously high resitance for Roff spice stiil shows the battery voltage when the switch opens. So when the pulse source goes to zero this reflects the real world there will be no voltage on the unconnected side of the switch. The second pulsed source (V2) represents the start up of a flyback converter. Also spice shows both FET's gate being pulled to ground, in the real world they stay up about 1.2V. I think gearhead may be right.
I dont usually include these unless they impact on what I'm trying to simulate. In this case they dont.The capacitor your referring to isn't for bypass it sets a RC time constant at the base of the Q14 providing a 50mA current pulse to the driver for 1mS . This increases the gate driving capabilities of the driver.You can see my previous post I posted a link to a screen capture of some simulated waveforms.
I do have a question if I do use comparators to switch the FETS. I'm using an LTC1440 to monitor the battery voltage and if it drops to
6.5V raise the gate of the first FET and block the battery, also light an LED indicator. Anyways my question is ; the LTC1440 has a 1.18V reference can I use this reference for my LM2903 . In other words will having a voltage present at the input of an unpowered comparator destroy it?
The datasheet has that info. You can take the inputs above supply. You just have to keep them below the max for your part, the ON Semi LM2903 says 36v for example. And the comparator will give the correct output as long as at least one of the inputs stays within common mode range (a couple volts below Vcc or whatever for that particular comparator). If you take both the inputs above Vcc-1.5 or Vcc-2 whatever, then the output is undefined; but in practice I think it often goes low.
I see what I did wrong. I had Vser set to 0.6 as in the example, and that made it conduct only in one direction. I was using it to analyze switching transients on AC relays. The corrected circuit follows.