simple npn question

[TK doublee]

Hi,

I'm stuck on a really simple problem here:

I need to drive high-side line driver (ir2104) control pin with a 5V
signal. The 2104 is operating at 15V (and is switching 80V on the
high-side!), and the input needs to be 15V for the output to turn on.

The control signal is coming from a PIC, at 5V.

I thought, hey, just use an NPN common collector to switch the 2104.
Not so fast: I can't seem to get enough base current to drive the
collector for 0 to 15V, or something.

My BJT theory is pretty poor, I'm a digital guy , can someone
enlighten me here? I'm using your generic 2n3904 device as the switch
(or 2n3906). Feels like I'm trying to build a level shifter to switch
[0,5V] to [0,12V]...?

thanks in advance,
tk

 

[Tim]

On Sat, 16 Jul 2011 21:25:23 -0700, TK doublee wrote:

Hi,

I'm stuck on a really simple problem here:

I need to drive high-side line driver (ir2104) control pin with a 5V
signal. The 2104 is operating at 15V (and is switching 80V on the
high-side!), and the input needs to be 15V for the output to turn on.

The control signal is coming from a PIC, at 5V.

I thought, hey, just use an NPN common collector to switch the 2104. Not
so fast: I can't seem to get enough base current to drive the collector
for 0 to 15V, or something.

My BJT theory is pretty poor, I'm a digital guy , can someone
enlighten me here? I'm using your generic 2n3904 device as the switch
(or 2n3906). Feels like I'm trying to build a level shifter to switch
[0,5V] to [0,12V]...?

IIRC you can get 10 or 20mA out of a PIC -- something more impressive
than the 2mA from some other parts I've worked with. But even 2mA should
be enough base drive to get a reliable 100mA of collector current (that's
counting on an HFE of 50, which is a bit much, but not _too_ unrealistic)
from a 2N3904, and with a 7.5K resistor to VCC, you should be able to get
by with 2mA of collector current -- so I don't know what your problem is.

You are using a base resistor to limit the base current?

BUT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

That's all moot -- the data sheet that I have for the 2104 gives a high-
going logic threshold of 3.0V, which is plenty for a CMOS processor
running off of 5V.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

 

[Rich Grise]

TK doublee wrote:

I'm stuck on a really simple problem here:

I need to drive high-side line driver (ir2104) control pin with a 5V
signal. The 2104 is operating at 15V (and is switching 80V on the
high-side!), and the input needs to be 15V for the output to turn on.

The control signal is coming from a PIC, at 5V.

I thought, hey, just use an NPN common collector to switch the 2104.
Not so fast: I can't seem to get enough base current to drive the
collector for 0 to 15V, or something.

My BJT theory is pretty poor, I'm a digital guy , can someone
enlighten me here? I'm using your generic 2n3904 device as the switch
(or 2n3906). Feels like I'm trying to build a level shifter to switch
[0,5V] to [0,12V]...?

Use the NPN in common-emitter, with the collector connected to a load

resistor, maybe 4K7~10K, and the 2104, with that resistor returned to
the +15V.

From the output of the PIC, put a 1K resistor to the base, and from that
base-resistor junction, put a 10K to ground.

Since the NPN will invert, you'll have to change the polarity of the
output from your PIC.

+15V +15V
| |
| |
[4K7] ------------
| | |
+-------o 2104 o------ out
| | |
| ------------
/ c
[PIC]--[1K]--+-----| 2N3904
| > e
[10K] |
| |
[gnd] [gnd]

Have Fun!
Rich

 

[Jon Kirwan]

On Sat, 16 Jul 2011 21:25:23 -0700 (PDT), TK doublee
<tkdoublee@gmail.com> wrote:

I need to drive high-side line driver (ir2104) control pin with a 5V
signal. The 2104 is operating at 15V (and is switching 80V on the
high-side!), and the input needs to be 15V for the output to turn on.

As Tim points out, the datasheet appears to state that 3V is
enough, even with Vcc=15V. See V_IL and V_IH on page 3.

The control signal is coming from a PIC, at 5V.

Which seems okay for the specs.

I thought, hey, just use an NPN common collector to switch the 2104.
Not so fast: I can't seem to get enough base current to drive the
collector for 0 to 15V, or something.

A common collector (emitter follower) configuration??? That's
not going to get you 15V from 5V. Do you actually have the
collector tied to Vcc (15V) and a resistor connected between
emitter and ground?

My BJT theory is pretty poor, I'm a digital guy , can someone
enlighten me here? I'm using your generic 2n3904 device as the switch
(or 2n3906). Feels like I'm trying to build a level shifter to switch
[0,5V] to [0,12V]...?

Well, as Tim points out you probably don't need an external
BJT. But if you still wanted to use one, you would NOT use
it in the emitter follower config.

Interesting device. Although I'm sure it's very old news to
electronics designers (I'm not), I'd not considered using a
capacitor like that to boost the base drive for an N-channel
high side FET. Tricky and cheap. But I think you need to
drive the output a solid LOW first to make certain that cap
charges up to slightly under Vcc via the external diode.

Jon

 

[Tim]

On Sat, 16 Jul 2011 23:38:41 -0700, Jon Kirwan wrote:

On Sat, 16 Jul 2011 21:25:23 -0700 (PDT), TK doublee
tkdoublee@gmail.com> wrote:

I need to drive high-side line driver (ir2104) control pin with a 5V
signal. The 2104 is operating at 15V (and is switching 80V on the
high-side!), and the input needs to be 15V for the output to turn on.

As Tim points out, the datasheet appears to state that 3V is enough,
even with Vcc=15V. See V_IL and V_IH on page 3.

The control signal is coming from a PIC, at 5V.

Which seems okay for the specs.

I thought, hey, just use an NPN common collector to switch the 2104. Not
so fast: I can't seem to get enough base current to drive the collector
for 0 to 15V, or something.

A common collector (emitter follower) configuration??? That's not going
to get you 15V from 5V. Do you actually have the collector tied to Vcc
(15V) and a resistor connected between emitter and ground?

My BJT theory is pretty poor, I'm a digital guy , can someone
enlighten me here? I'm using your generic 2n3904 device as the switch
(or 2n3906). Feels like I'm trying to build a level shifter to switch
[0,5V] to [0,12V]...?

Well, as Tim points out you probably don't need an external BJT. But if
you still wanted to use one, you would NOT use it in the emitter
follower config.

I missed that common collector part -- hopefully the OP was messed up in
his terminology.

Interesting device. Although I'm sure it's very old news to electronics
designers (I'm not), I'd not considered using a capacitor like that to
boost the base drive for an N-channel high side FET. Tricky and cheap.
But I think you need to drive the output a solid LOW first to make
certain that cap charges up to slightly under Vcc via the external
diode.

That's used everywhere. Tricky, cheap, and ubiquitous. You can't run
down to a 0% duty cycle, and you have to maintain a minimum switching
frequency, or your upper-side gate drive goes away. But aside from that
it doesn't take many components.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

 

[Jamie]

Tim wrote:

On Sat, 16 Jul 2011 23:38:41 -0700, Jon Kirwan wrote:


On Sat, 16 Jul 2011 21:25:23 -0700 (PDT), TK doublee
tkdoublee@gmail.com> wrote:


I need to drive high-side line driver (ir2104) control pin with a 5V
signal. The 2104 is operating at 15V (and is switching 80V on the
high-side!), and the input needs to be 15V for the output to turn on.

As Tim points out, the datasheet appears to state that 3V is enough,
even with Vcc=15V. See V_IL and V_IH on page 3.


The control signal is coming from a PIC, at 5V.

Which seems okay for the specs.


I thought, hey, just use an NPN common collector to switch the 2104. Not
so fast: I can't seem to get enough base current to drive the collector
for 0 to 15V, or something.

A common collector (emitter follower) configuration??? That's not going
to get you 15V from 5V. Do you actually have the collector tied to Vcc
(15V) and a resistor connected between emitter and ground?


My BJT theory is pretty poor, I'm a digital guy , can someone
enlighten me here? I'm using your generic 2n3904 device as the switch
(or 2n3906). Feels like I'm trying to build a level shifter to switch
[0,5V] to [0,12V]...?

Well, as Tim points out you probably don't need an external BJT. But if
you still wanted to use one, you would NOT use it in the emitter
follower config.


I missed that common collector part -- hopefully the OP was messed up in
his terminology.


Interesting device. Although I'm sure it's very old news to electronics
designers (I'm not), I'd not considered using a capacitor like that to
boost the base drive for an N-channel high side FET. Tricky and cheap.
But I think you need to drive the output a solid LOW first to make
certain that cap charges up to slightly under Vcc via the external
diode.


That's used everywhere. Tricky, cheap, and ubiquitous. You can't run
down to a 0% duty cycle, and you have to maintain a minimum switching
frequency, or your upper-side gate drive goes away. But aside from that
it doesn't take many components.

I think he needs to use a basic Opto-coupler.

Jamie

 

[Jon Kirwan]

On Sun, 17 Jul 2011 12:55:32 -0500, Tim
<tim@seemywebsite.please> wrote:

On Sat, 16 Jul 2011 23:38:41 -0700, Jon Kirwan wrote:

On Sat, 16 Jul 2011 21:25:23 -0700 (PDT), TK doublee
tkdoublee@gmail.com> wrote:

I need to drive high-side line driver (ir2104) control pin with a 5V
signal. The 2104 is operating at 15V (and is switching 80V on the
high-side!), and the input needs to be 15V for the output to turn on.

As Tim points out, the datasheet appears to state that 3V is enough,
even with Vcc=15V. See V_IL and V_IH on page 3.

The control signal is coming from a PIC, at 5V.

Which seems okay for the specs.

I thought, hey, just use an NPN common collector to switch the 2104. Not
so fast: I can't seem to get enough base current to drive the collector
for 0 to 15V, or something.

A common collector (emitter follower) configuration??? That's not going
to get you 15V from 5V. Do you actually have the collector tied to Vcc
(15V) and a resistor connected between emitter and ground?

My BJT theory is pretty poor, I'm a digital guy , can someone
enlighten me here? I'm using your generic 2n3904 device as the switch
(or 2n3906). Feels like I'm trying to build a level shifter to switch
[0,5V] to [0,12V]...?

Well, as Tim points out you probably don't need an external BJT. But if
you still wanted to use one, you would NOT use it in the emitter
follower config.

I missed that common collector part -- hopefully the OP was messed up in
his terminology.

I was looking for a reason to explain why the OP felt he
wasn't getting enough voltage. Given the light (near zero)
current requirements of the IR2104 I had a very hard time
understanding why a simple common emitter config wouldn't
give close to 15V output. It was obvious once I reinspected
his wording -- he'd NEVER get there with an emitter follower
and that then had to be the explanation. He actually did use
the right terminology! Just used the wrong config.

Interesting device. Although I'm sure it's very old news to electronics
designers (I'm not), I'd not considered using a capacitor like that to
boost the base drive for an N-channel high side FET. Tricky and cheap.
But I think you need to drive the output a solid LOW first to make
certain that cap charges up to slightly under Vcc via the external
diode.

That's used everywhere. Tricky, cheap, and ubiquitous. You can't run
down to a 0% duty cycle, and you have to maintain a minimum switching
frequency, or your upper-side gate drive goes away. But aside from that
it doesn't take many components.

Thanks. That's about what I'd have guessed about it. I've
seen charge pumps to do the same job, but this one has
another advantage over those in that you can decide what your
Vcc is, which will source your gate drive, and this means you
have better control over what gate drive level you want to
use in an application. The charge pump devices I've seen for
N-channel high sides provided a fixed "lift" that is set by
design. Kind of nice, as well as cheap and tricky.

The capacitor can be sized to drive longer (it has to drive
those common emitter translators and following stuff as well
as leakage and FET gate charge, I guess) but then requires
more charge time too. So I guess you decide your basic
operating rate and try and stick to it.

What's the name for that capacitor in this configuration? I
can see a lot of what it does, but I don't know what to call
it so that others in the field would know what I meant
without having to write a long sentence about it. It is very
much like being 'picked up by your bootstraps', but that's an
already overused term I think.

Jon

 

[John Larkin]

On Sat, 16 Jul 2011 21:25:23 -0700 (PDT), TK doublee
<tkdoublee@gmail.com> wrote:

Hi,

I'm stuck on a really simple problem here:

I need to drive high-side line driver (ir2104) control pin with a 5V
signal. The 2104 is operating at 15V (and is switching 80V on the
high-side!), and the input needs to be 15V for the output to turn on.

The datasheet says you need below 0.8 for a low, and over 3.0 for a
high. Your 5-volt PIC output should be fine.

John

 

[Tim]

On Sun, 17 Jul 2011 11:21:11 -0700, Jon Kirwan wrote:

What's the name for that capacitor in this configuration? I can see a
lot of what it does, but I don't know what to call it so that others in
the field would know what I meant without having to write a long
sentence about it. It is very much like being 'picked up by your
bootstraps', but that's an already overused term I think.

Jon

I think it's usually boost cap or bootstrap cap -- but I think boost is
more common.



--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com