Is this transistor correct?

[ST]

Hello,

I have a 7660 DC-DC converter which I would like to turn on and off from a
microcontroller. The microcontroller runs at 5V and the 7660 runs at 9V (it
doubles 9V to 18V). I was thinking of using a transistor to do this. In
particular, I was hoping to use a BS108 transistor as its the only type I
have. My questions are:

1) Will this transistor be able to do what I want?
2) I was going to connect the emitter to ground and the collector to the
ground of the 7660. The V+ pin of the 7660 would then goto the 9V supply.
Is this the correct way of doing it?
3) What value resistor do I use on the base of the transistor? Does the
value really matter?
4) I have more than 1 BS108 transistor, so I could make a darlington pair.
Would this be better? If so, how do I wire it up?

Thanks for any help.

 

[John Popelish]

ST wrote:

Hello,

I have a 7660 DC-DC converter which I would like to turn on and off from a
microcontroller. The microcontroller runs at 5V and the 7660 runs at 9V (it
doubles 9V to 18V). I was thinking of using a transistor to do this. In
particular, I was hoping to use a BS108 transistor as its the only type I
have. My questions are:

http://www.semiconductors.philips.com/acrobat/datasheets/BS108_3.pdf

1) Will this transistor be able to do what I want?

Possibly. You don't say exactly what you want. Quite likely, when
you disconnect the negative rail pin of the 7660, the 18 volt output
will not go to zero, but to a diode drop or two below 9. Is this what
you want? I am assuming that your 9 and 18 volt measurements are made
with respect ot the negative side of the 5 volt microprocessor supply.

BS108 has a typical on resistance or 2.7 ohms with 2.8 gate volts,
which should waste very little of the 9 volt supply to the 7660 with a
few milliamp load.

2) I was going to connect the emitter to ground and the collector to the
ground of the 7660. The V+ pin of the 7660 would then goto the 9V supply.
Is this the correct way of doing it?

Except that this is a mosfet, not a junction transistor, so the
emitter is actually source, base is gate and collector is drain.

3) What value resistor do I use on the base of the transistor? Does the
value really matter?

No gate resistor is needed.

4) I have more than 1 BS108 transistor, so I could make a darlington pair.
Would this be better? If so, how do I wire it up?

Bad idea with mosfets. Paralleling two would lower the losses by
half, though.

Thanks for any help.

--
John Popelish

 

[ST]

"John Popelish" <jpopelish@rica.net> wrote in message
news:4114F191.7452B05@rica.net...

ST wrote:

Hello,

I have a 7660 DC-DC converter which I would like to turn on and off from
a
microcontroller. The microcontroller runs at 5V and the 7660 runs at 9V
(it
doubles 9V to 18V). I was thinking of using a transistor to do this.
In
particular, I was hoping to use a BS108 transistor as its the only type
I
have. My questions are:

http://www.semiconductors.philips.com/acrobat/datasheets/BS108_3.pdf

1) Will this transistor be able to do what I want?

Possibly. You don't say exactly what you want. Quite likely, when
you disconnect the negative rail pin of the 7660, the 18 volt output
will not go to zero, but to a diode drop or two below 9. Is this what
you want? I am assuming that your 9 and 18 volt measurements are made
with respect ot the negative side of the 5 volt microprocessor supply.

BS108 has a typical on resistance or 2.7 ohms with 2.8 gate volts,
which should waste very little of the 9 volt supply to the 7660 with a
few milliamp load.

2) I was going to connect the emitter to ground and the collector to the
ground of the 7660. The V+ pin of the 7660 would then goto the 9V
supply.
Is this the correct way of doing it?

Except that this is a mosfet, not a junction transistor, so the
emitter is actually source, base is gate and collector is drain.

3) What value resistor do I use on the base of the transistor? Does the
value really matter?

No gate resistor is needed.

4) I have more than 1 BS108 transistor, so I could make a darlington
pair.
Would this be better? If so, how do I wire it up?

Bad idea with mosfets. Paralleling two would lower the losses by
half, though.

Thanks for any help.


--
John Popelish

Thanks for the reply.

I was hoping that the 18V output would go to zero. If I connected the
source to the V+ pin of the 7660 and the drain to the 9V supply, would that
allow me to turn the 7660 device completely off?

 

[John Popelish]

ST wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:4114F191.7452B05@rica.net...
ST wrote:

Hello,

I have a 7660 DC-DC converter which I would like to turn on and off from
a
microcontroller. The microcontroller runs at 5V and the 7660 runs at 9V
(it
doubles 9V to 18V). I was thinking of using a transistor to do this.
In
particular, I was hoping to use a BS108 transistor as its the only type
I
have. My questions are:

http://www.semiconductors.philips.com/acrobat/datasheets/BS108_3.pdf

1) Will this transistor be able to do what I want?

Possibly. You don't say exactly what you want. Quite likely, when
you disconnect the negative rail pin of the 7660, the 18 volt output
will not go to zero, but to a diode drop or two below 9. Is this what
you want? I am assuming that your 9 and 18 volt measurements are made
with respect ot the negative side of the 5 volt microprocessor supply.

BS108 has a typical on resistance or 2.7 ohms with 2.8 gate volts,
which should waste very little of the 9 volt supply to the 7660 with a
few milliamp load.

2) I was going to connect the emitter to ground and the collector to the
ground of the 7660. The V+ pin of the 7660 would then goto the 9V
supply.
Is this the correct way of doing it?

Except that this is a mosfet, not a junction transistor, so the
emitter is actually source, base is gate and collector is drain.

3) What value resistor do I use on the base of the transistor? Does the
value really matter?

No gate resistor is needed.

4) I have more than 1 BS108 transistor, so I could make a darlington
pair.
Would this be better? If so, how do I wire it up?

Bad idea with mosfets. Paralleling two would lower the losses by
half, though.

Thanks for any help.


--
John Popelish

Thanks for the reply.

I was hoping that the 18V output would go to zero. If I connected the
source to the V+ pin of the 7660 and the drain to the 9V supply, would that
allow me to turn the 7660 device completely off?

A high side switch would do the trick, but the BS108 is probably not
the device to use. You would have to pull its gate a few volts above
the 9 volt rail to turn it on.

I would try a PNP transistor (emitter to +9, collector to 7660, 10k
base to emitter) driven by a common base NPN (base connected to the
+5 rail, emitter driven by active low logic through a current limiting
resistor that sets the output current, collector to base of PNP). Do
you have any small PNP and NPN transistors available, like 2N3904,
2N3906, etc.?

--
John Popelish

 

[colin]

"John Popelish" <jpopelish@rica.net> wrote in message
news:411517B4.6E91EBE@rica.net...

ST wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:4114F191.7452B05@rica.net...
ST wrote:

Hello,

I have a 7660 DC-DC converter which I would like to turn on and off
from
a
microcontroller. The microcontroller runs at 5V and the 7660 runs
at 9V
(it
doubles 9V to 18V). I was thinking of using a transistor to do
this.
In
particular, I was hoping to use a BS108 transistor as its the only
type
I
have. My questions are:

http://www.semiconductors.philips.com/acrobat/datasheets/BS108_3.pdf

1) Will this transistor be able to do what I want?

Possibly. You don't say exactly what you want. Quite likely, when
you disconnect the negative rail pin of the 7660, the 18 volt output
will not go to zero, but to a diode drop or two below 9. Is this what
you want? I am assuming that your 9 and 18 volt measurements are made
with respect ot the negative side of the 5 volt microprocessor supply.

BS108 has a typical on resistance or 2.7 ohms with 2.8 gate volts,
which should waste very little of the 9 volt supply to the 7660 with a
few milliamp load.

2) I was going to connect the emitter to ground and the collector to
the
ground of the 7660. The V+ pin of the 7660 would then goto the 9V
supply.
Is this the correct way of doing it?

Except that this is a mosfet, not a junction transistor, so the
emitter is actually source, base is gate and collector is drain.

3) What value resistor do I use on the base of the transistor? Does
the
value really matter?

No gate resistor is needed.

4) I have more than 1 BS108 transistor, so I could make a darlington
pair.
Would this be better? If so, how do I wire it up?

Bad idea with mosfets. Paralleling two would lower the losses by
half, though.

Thanks for any help.


--
John Popelish

Thanks for the reply.

I was hoping that the 18V output would go to zero. If I connected the
source to the V+ pin of the 7660 and the drain to the 9V supply, would
that
allow me to turn the 7660 device completely off?

A high side switch would do the trick, but the BS108 is probably not
the device to use. You would have to pull its gate a few volts above
the 9 volt rail to turn it on.

I would try a PNP transistor (emitter to +9, collector to 7660, 10k
base to emitter) driven by a common base NPN (base connected to the
+5 rail, emitter driven by active low logic through a current limiting
resistor that sets the output current, collector to base of PNP). Do
you have any small PNP and NPN transistors available, like 2N3904,
2N3906, etc.?

--
John Popelish

my thoughts exactly but you cld stil use your bs108 in place of the npn
transistor John sugested to turn on the PNP transistor.
the comon base(or gate) config means the operation is inverted ie high op
turns the 18v off.

with comon emiter npn (or comon source bs108) to drive the pnp a high op wil
turn the supply on. u stil need a resister from drain to the base of the pnp
to limit the curent.

Colin =^.^=