# P-channel Mosfet switch?

H

#### Hammy

Jan 1, 1970
0
I'm trying to pull this P-channel mosfet gate to ground (U11). The
best I can get is to within 1.1 to 1.2V above ground this only gives a
-3.8Vgs. Somehow current is flowing through R27 enough to pull the
gate up 1.2V; about 400uA. If the PNP is off where is the current
coming from?

I'm trying to turn the mosfet off when the 9Vdc source is switched
out, but I would also like to get the -Vgs to at least 4.5V.

Schematic
http://i25.tinypic.com/14dzhua.png

Thanks

G

Jan 1, 1970
0
I'm trying to pull this P-channel mosfet gate to ground (U11). The
best I can get is to within 1.1 to 1.2V above ground this only gives a
-3.8Vgs. Somehow current is flowing through R27 enough to pull the
gate up  1.2V; about 400uA. If the PNP is off where is the current
coming from?

I'm trying to turn the mosfet off when the 9Vdc source is switched
out, but I would also like to get the -Vgs to at least 4.5V.

Schematichttp://i25.tinypic.com/14dzhua.png

Thanks

The circuit as drawn has no purpose and you have numbers like R27 and
U11, so this looks like part of a larger circuit. What happened to
the other twenty resistors and ten semiconductors?

H

#### Hammy

Jan 1, 1970
0
The circuit as drawn has no purpose and you have numbers like R27 and
U11, so this looks like part of a larger circuit. What happened to
the other twenty resistors and ten semiconductors?

I omitted the rest because the rest works fine. But here is the whole
circuit.

http://i28.tinypic.com/2pobac4.png

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.

G

Jan 1, 1970
0
I omitted the rest because the rest works fine. But here is the whole
circuit.

http://i28.tinypic.com/2pobac4.png

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.

-------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.

When the voltage from V2 hits

H

#### Hammy

Jan 1, 1970
0
Here are the simulated waveforms if it explains more clearly what I'm
doing.

http://i31.tinypic.com/15sao2g.png

Top trace (RED ID3) the switch is closed : 50mA current pulse for 1mS

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).

G

Jan 1, 1970
0
-------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.
Excuse me, R28.

P

#### Paul E. Schoen

Jan 1, 1970
0
Hammy said:
Here are the simulated waveforms if it explains more clearly what I'm
doing.

http://i31.tinypic.com/15sao2g.png

Top trace (RED ID3) the switch is closed : 50mA current pulse for 1mS

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.

Paul

H

#### Hammy

Jan 1, 1970
0
It would help if you explained what you are trying to do.

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.
It seems like a lot of extra circuitry to switch among various sources and loads,

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.
and Idon't understand why you are seemingly trying to turn on a transistor to
set Vgs of a PMOS to zero to turn it off,

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.
rather than just have a resistor from gate to source and pull the gate to GND 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.
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),

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.
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.

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.
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.

Paul

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?

Thanks

G

Jan 1, 1970
0
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?

Thanks

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.

P

#### Paul E. Schoen

Jan 1, 1970
0
John Fields said:
---
That's not true. It's fully bidirectional.
---

---
Yeah, but it's easy.

Go to:

and look on page 164.

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.

Thanks,

Paul

==========================================================================

Version 4
SHEET 1 880 680
WIRE 64 -16 -112 -16
WIRE 16 16 -208 16
WIRE 64 16 64 -16
WIRE -16 64 -48 64
WIRE 0 64 -16 64
WIRE 128 64 80 64
WIRE 256 64 128 64
WIRE 256 112 256 64
WIRE -208 128 -208 16
WIRE -48 144 -48 64
WIRE 128 144 128 128
WIRE 128 160 128 144
WIRE -208 240 -208 208
WIRE -112 240 -112 -16
WIRE -112 240 -208 240
WIRE -48 256 -48 224
WIRE 128 256 128 240
WIRE 128 256 -48 256
WIRE 256 256 256 192
WIRE 256 256 128 256
WIRE -48 288 -48 256
FLAG -48 288 0
FLAG -16 64 Vin
FLAG 128 144 Vres
FLAG -208 240 0
FLAG 256 64 Vrly
SYMBOL res 112 144 R0
SYMATTR InstName R1
SYMATTR Value 270
SYMBOL cap 112 64 R0
SYMATTR InstName C1
SYMATTR Value 0.47µ
SYMBOL voltage -48 128 R0
WINDOW 3 103 175 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 22 174 Left 0
SYMATTR Value SINE(0 360 60 0 0 0 150)
SYMATTR SpiceLine Rser=1
SYMATTR InstName V1
SYMBOL sw 96 64 R90
WINDOW 3 -36 -113 VRight 0
WINDOW 0 24 92 VRight 0
SYMATTR Value MySwitch
SYMATTR InstName S1
SYMBOL ind 240 96 R0
SYMATTR InstName L1
SYMATTR Value 5
SYMATTR SpiceLine Rser=50 Rpar=1Meg
SYMBOL voltage -208 112 R0
WINDOW 3 -219 162 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value PWL(0 0 .1041667 0 .104167 5 .5041667 5 .504167 0 1.00 0
1.001 5 1.50 5 1.501 0 2 0)
SYMATTR InstName V2
TEXT -88 496 Left 0 !.tran 2 startup
TEXT -392 344 Left 0 !.model MySwitch SW(Ron=.1 Roff=100Meg Vt=2 Vh=-.5
Lser=10n Vser=0)

H

#### Hammy

Jan 1, 1970
0
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.

Thanks for clearing that up. I've never had to consider it before. It
will save me a couple of components, by using the LTC1440 refrence.

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