5v 20mA max output driving a 5v 500mA max circuit.

[Daniel Pitts]

I have a signal at 0 or 5v. The source of that signal can only handle
20mA of current.

I have another circuit which needs to have the same voltage level with
reference to ground, but it could have up to 500mA going through it.
The actual current is controlled by a current sink, and it may vary over
time.

In other words, when my signal is low, I want no current. When my signal
is high, I want to source as much current as possible for the sink to
take in.

I have some BJTs laying around, both PNP and NPN, both can support the
max Ic I need. From what I figure, I'll need to hook my signal up to
the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v for the
resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually have 8
of these signals (although, only one is active at a time), so I was
hoping to avoid adding the 8 resistors and 8 transistors to my project,
though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

 

[petrus bitbyter]

"Daniel Pitts" <newsgroup.nospam@virtualinfinity.net> schreef in bericht
news:IbIft.1739$Ue.1387@newsfe31.iad...

I have a signal at 0 or 5v. The source of that signal can only handle 20mA
of current.

I have another circuit which needs to have the same voltage level with
reference to ground, but it could have up to 500mA going through it. The
actual current is controlled by a current sink, and it may vary over time.

In other words, when my signal is low, I want no current. When my signal
is high, I want to source as much current as possible for the sink to take
in.

I have some BJTs laying around, both PNP and NPN, both can support the max
Ic I need. From what I figure, I'll need to hook my signal up to the
base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v for the
resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually have 8
of these signals (although, only one is active at a time), so I was hoping
to avoid adding the 8 resistors and 8 transistors to my project, though
that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

It's not clear to me what the directions of the currents are both for output
and input. Modern digital outputs can often handle 20mA both sink and source
but an input that requires op to 500mA sink or source? Is it a digital
input? What are the voltage limits of that input? Crucial things to know for
providing a usefull answer.

petrus bitbyter

 

[George Herold]

On Apr 30, 12:58 am, Daniel Pitts
<newsgroup.nos...@virtualinfinity.net> wrote:

I have a signal at 0 or 5v. The source of that signal can only handle
20mA of current.

I have another circuit which needs to have the same voltage level with
reference to ground, but it could have up to 500mA going through it.
The actual current is controlled by a current sink, and it may vary over
time.

In other words, when my signal is low, I want no current. When my signal
is high, I want to source as much current as possible for the sink to
take in.

I have some BJTs laying around, both PNP and NPN, both can support the
max Ic I need.  From what I figure, I'll need to hook my signal up to
the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v for the
resistor. 4.3v/20ma=215ohms).

An emitter follower? That should be fine. The transistor might use
less than the whole 20mA. (depends on the current gain.) Do you care
about speed? (it might take some time to turn off.) I'm a bit
confused about the load. What are you driving?

George H.

Will a circuit like that work? Is there an easier way? I actually have 8
of these signals (although, only one is active at a time), so I was
hoping to avoid adding the 8 resistors and 8 transistors to my project,
though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

A

 

[Daniel Pitts]

On 4/30/13 6:03 AM, George Herold wrote:

On Apr 30, 12:58 am, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
I have a signal at 0 or 5v. The source of that signal can only handle
20mA of current.

I have another circuit which needs to have the same voltage level with
reference to ground, but it could have up to 500mA going through it.
The actual current is controlled by a current sink, and it may vary over
time.

In other words, when my signal is low, I want no current. When my signal
is high, I want to source as much current as possible for the sink to
take in.

I have some BJTs laying around, both PNP and NPN, both can support the
max Ic I need. From what I figure, I'll need to hook my signal up to
the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v for the
resistor. 4.3v/20ma=215ohms).

An emitter follower? That should be fine. The transistor might use
less than the whole 20mA. (depends on the current gain.)
That's more than fine, 20mA is the maximum the signal source can handle,

but it could be pico-amps for all I care.

Do you care
about speed? (it might take some time to turn off.)
I care a little about the speed. are we talking more than 10's of

milliseconds?

I'm a bit
confused about the load. What are you driving?
The load is several parallel LEDs, all controlled by another chip (a

shift register with latch constant-current sink). Each "led" will get
20mA, and I can have up to 24 on at the same time, (that is, between 0
and 24), so a maximum of 480mA (rounded up to 500ma)

I've posted about this project in the past, its basically my own LED
display, based on an RGB LED matrix.
<http://www.seeedstudio.com/depot/datasheet/2088RGBMatrix.pdf>

I need to switch on exactly one of the rows, and set up the column shift
registers. I switch rows at about 8KHz, so I suspect most transistors
can handle that no problem.

There might be another alternative. On digikey, I found "PMIC - MOSFET,
Bridge Drivers, Internal/External Switch", which look like they do what
I want as well, with fewer discrete parts. It costs a bit more, and the
interface is "serial" rather than a simple 3-to-8 decoder I currently
have. Overall, I'm not sure which is the better approach. I think
overall I'd use a little less power when using the mosfet chip, which
could be useful if I make this a battery powered device.

What's the difference between internal and external for these "bridge
driver" chips? Also, what's the difference between High-Side and
Low-Side in these chips?

Thanks,
Daniel.

 

[Tim Wescott]

On Tue, 30 Apr 2013 09:01:52 -0700, Daniel Pitts wrote:

On 4/30/13 6:03 AM, George Herold wrote:
On Apr 30, 12:58 am, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
I have a signal at 0 or 5v. The source of that signal can only handle
20mA of current.

I have another circuit which needs to have the same voltage level with
reference to ground, but it could have up to 500mA going through it.
The actual current is controlled by a current sink, and it may vary
over time.

In other words, when my signal is low, I want no current. When my
signal is high, I want to source as much current as possible for the
sink to take in.

I have some BJTs laying around, both PNP and NPN, both can support the
max Ic I need. From what I figure, I'll need to hook my signal up to
the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v for
the resistor. 4.3v/20ma=215ohms).

An emitter follower? That should be fine. The transistor might use
less than the whole 20mA. (depends on the current gain.)
That's more than fine, 20mA is the maximum the signal source can handle,
but it could be pico-amps for all I care.

Do you care
about speed? (it might take some time to turn off.)
I care a little about the speed. are we talking more than 10's of
milliseconds?

I'm a bit
confused about the load. What are you driving?
The load is several parallel LEDs, all controlled by another chip (a
shift register with latch constant-current sink). Each "led" will get
20mA, and I can have up to 24 on at the same time, (that is, between 0
and 24), so a maximum of 480mA (rounded up to 500ma)

I've posted about this project in the past, its basically my own LED
display, based on an RGB LED matrix.
http://www.seeedstudio.com/depot/datasheet/2088RGBMatrix.pdf

I need to switch on exactly one of the rows, and set up the column shift
registers. I switch rows at about 8KHz, so I suspect most transistors
can handle that no problem.

There might be another alternative. On digikey, I found "PMIC - MOSFET,
Bridge Drivers, Internal/External Switch", which look like they do what
I want as well, with fewer discrete parts. It costs a bit more, and the
interface is "serial" rather than a simple 3-to-8 decoder I currently
have. Overall, I'm not sure which is the better approach. I think
overall I'd use a little less power when using the mosfet chip, which
could be useful if I make this a battery powered device.

What's the difference between internal and external for these "bridge
driver" chips? Also, what's the difference between High-Side and
Low-Side in these chips?

That is almost a "if you have to ask, don't use the chip" sort of
question.

"Internal" usually means that the chip includes the drive FET.
"External" usually means that the chip is designed to drive a FET gate.
"High side" is what you're doing, but not necessarily what you want (and
I ain't gonna 'splain -- it takes too long. See my "if you have to ask"
comment).

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

 

[Tim Wescott]

On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:

I have a signal at 0 or 5v. The source of that signal can only handle
20mA of current.

I have another circuit which needs to have the same voltage level with
reference to ground, but it could have up to 500mA going through it. The
actual current is controlled by a current sink, and it may vary over
time.

In other words, when my signal is low, I want no current. When my signal
is high, I want to source as much current as possible for the sink to
take in.

I have some BJTs laying around, both PNP and NPN, both can support the
max Ic I need. From what I figure, I'll need to hook my signal up to
the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v for the
resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually have 8
of these signals (although, only one is active at a time), so I was
hoping to avoid adding the 8 resistors and 8 transistors to my project,
though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then a PNP
with the emitter connected to +5V, a resistor to your 20mA driver, and
the collector connected to your load, is what you want.

You need a PNP that has an HFE of at least 25 when it is saturated, which
may be hard to find if your junkbox has older parts in it. This _may_
work at the 8kHz that you state elsewhere in the thread, particularly if
your driver can source to the base of the transistor; that'll help it
turn off.

A P-channel logic-level MOSFET would be perfect here, again assuming that
you can stand the inversion. A gate resistor is still a good idea to
protect your source against current surges, but it's much easier to get
suitable PMOS parts than it is to find suitable high-HFE PNP parts.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

 

[Tim Wescott]

On Tue, 30 Apr 2013 13:17:28 -0700, George Herold wrote:

On Apr 30, 1:31 pm, Tim Wescott <t...@seemywebsite.com> wrote:
On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:
I have a signal at 0 or 5v. The source of that signal can only handle
20mA of current.

I have another circuit which needs to have the same voltage level
with reference to ground, but it could have up to 500mA going through
it. The actual current is controlled by a current sink, and it may
vary over time.

In other words, when my signal is low, I want no current. When my
signal is high, I want to source as much current as possible for the
sink to take in.

I have some BJTs laying around, both PNP and NPN, both can support
the max Ic I need.  From what I figure, I'll need to hook my signal
up to the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v
for the resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually
have 8 of these signals (although, only one is active at a time), so
I was hoping to avoid adding the 8 resistors and 8 transistors to my
project, though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then a PNP
with the emitter connected to +5V, a resistor to your 20mA driver, and
the collector connected to your load, is what you want.

Tim, stupid question, why is that better?

I didn't qualify my statement, did I? Because the transistor will pull
up a lot closer to 5V -- with an emitter follower you have the B-E drop
in series with whatever drop your driving electronics has at the base
current. That not only means less voltage, but higher driver impedance
and hence more variation with current.

Come to think of it, maybe the OP is better off to use NPN emitter
followers and accept that the emitters will be somewhere between 4 and
4.3V (and look at the data sheet of whatever it is that delivers 20mA!!!).

So it's not a stupid question.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

 

[Tim Wescott]

On Tue, 30 Apr 2013 13:14:01 -0700, George Herold wrote:

On Apr 30, 12:01 pm, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
On 4/30/13 6:03 AM, George Herold wrote:



On Apr 30, 12:58 am, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
I have a signal at 0 or 5v. The source of that signal can only
handle 20mA of current.

I have another circuit which needs to have the same voltage level
with reference to ground, but it could have up to 500mA going
through it. The actual current is controlled by a current sink, and
it may vary over time.

In other words, when my signal is low, I want no current. When my
signal is high, I want to source as much current as possible for the
sink to take in.

I have some BJTs laying around, both PNP and NPN, both can support
the max Ic I need.  From what I figure, I'll need to hook my signal
up to the base, with at a minimum of 215ohms (Vbe = .7v, leaving
4.3v for the resistor. 4.3v/20ma=215ohms).

An emitter follower? That should be fine.  The transistor might use
less than the whole 20mA.  (depends on the current gain.)

That's more than fine, 20mA is the maximum the signal source can
handle, but it could be pico-amps for all I care.

Hi Daniel, First I may say something wrong.. and then hopefully those
with more knowledge will correct me.


Do you care
about speed?  (it might take some time to turn off.)

I care a little about the speed. are we talking more than 10's of
milliseconds?

No. when you saturate a bjt it takes a microsecond or so to come out of
saturation.
(hmm I'm not real sure about that number... maybe 100's of ns?)

It depends to a great extent on the BJT, and to a lesser extent on the
way you drive it.

Basically, the base is conductive for as long as there are minority
carriers in there. Under normal circumstances those carriers need to
recombine -- and how long that takes depends on the carrier lifetime and
(I think -- JT can correct me if he's listening) on the base geometry.

If you actively suck carriers out of the base by bringing it negative
with respect to the emitter (in the case of an NPN -- positive if it's a
PNP) then you can speed up the turn-off, sometimes to an astonishing
degree.

(Purpose made "switching transistors" used gold-doped bases because it
killed the minority carrier lifetime. I can't remember numbers, but
they're out there. Schottky logic (74S, 74LS, etc.) used Schottky diodes
between base and collector, to hold the transistors out of full
saturation and speed things up).

At any rate, if you've got an NPN with the collector and base at 5V it's
not going to saturate -- you need to forward-bias the C-B junction to
saturate the base, and you only do that on an NPN by bringing the base
voltage above the collector voltage.

Emitter followers are, in general, pretty fast.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

 

[George Herold]

On Apr 30, 12:01 pm, Daniel Pitts
<newsgroup.nos...@virtualinfinity.net> wrote:

On 4/30/13 6:03 AM, George Herold wrote:



On Apr 30, 12:58 am, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
I have a signal at 0 or 5v. The source of that signal can only handle
20mA of current.

I have another circuit which needs to have the same voltage level with
reference to ground, but it could have up to 500mA going through it.
The actual current is controlled by a current sink, and it may vary over
time.

In other words, when my signal is low, I want no current. When my signal
is high, I want to source as much current as possible for the sink to
take in.

I have some BJTs laying around, both PNP and NPN, both can support the
max Ic I need.  From what I figure, I'll need to hook my signal up to
the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v for the
resistor. 4.3v/20ma=215ohms).

An emitter follower? That should be fine.  The transistor might use
less than the whole 20mA.  (depends on the current gain.)

That's more than fine, 20mA is the maximum the signal source can handle,
but it could be pico-amps for all I care.

Hi Daniel, First I may say something wrong.. and then hopefully those
with more knowledge will correct me.

Do you care
about speed?  (it might take some time to turn off.)

I care a little about the speed. are we talking more than 10's of
milliseconds?

No. when you saturate a bjt it takes a microsecond or so to come out
of saturation.
(hmm I'm not real sure about that number... maybe 100's of ns?)

I'm a bit
confused about the load.  What are you driving?

The load is several parallel LEDs, all controlled by another chip (a
shift register with latch constant-current sink). Each "led" will get
20mA, and I can have up to 24 on at the same time, (that is, between 0
and 24), so a maximum of 480mA (rounded up to 500ma)

I've posted about this project in the past, its basically my own LED
display, based on an RGB LED matrix.
http://www.seeedstudio.com/depot/datasheet/2088RGBMatrix.pdf

I need to switch on exactly one of the rows, and set up the column shift
registers.  I switch rows at about 8KHz, so I suspect most transistors
can handle that no problem.

There might be another alternative. On digikey, I found "PMIC - MOSFET,
Bridge Drivers, Internal/External Switch", which look like they do what
I want as well, with fewer discrete parts. It costs a bit more, and the
interface is "serial" rather than a simple 3-to-8 decoder I currently
have.  Overall, I'm not sure which is the better approach. I think
overall I'd use a little less power when using the mosfet chip, which
could be useful if I make this a battery powered device.

Yup you could use a FET too. You've got I*R losses in the FET where R
is the on resistance. In the BJT you'll have (maybe) 300-500 mV of
Vce drop... that might be an issue if you need all of the 5 volts for
the LEDs.

What's the difference between internal and external for these "bridge
driver" chips? Also, what's the difference between High-Side and
Low-Side in these chips?

Well low side is when the switch (transistor) is on the ground side of
the load, (so the load floats above ground) and high side is when the
switch sits between the power supply and load with the load
grounded.

George H.

Thanks,
Daniel.- Hide quoted text -

- Show quoted text -

 

[petrus bitbyter]

"Daniel Pitts" <newsgroup.nospam@virtualinfinity.net> schreef in bericht
news:RVRft.1163$hl7.126@newsfe14.iad...

On 4/30/13 6:03 AM, George Herold wrote:
On Apr 30, 12:58 am, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
I have a signal at 0 or 5v. The source of that signal can only handle
20mA of current.

I have another circuit which needs to have the same voltage level with
reference to ground, but it could have up to 500mA going through it.
The actual current is controlled by a current sink, and it may vary over
time.

In other words, when my signal is low, I want no current. When my signal
is high, I want to source as much current as possible for the sink to
take in.

I have some BJTs laying around, both PNP and NPN, both can support the
max Ic I need. From what I figure, I'll need to hook my signal up to
the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v for the
resistor. 4.3v/20ma=215ohms).

An emitter follower? That should be fine. The transistor might use
less than the whole 20mA. (depends on the current gain.)
That's more than fine, 20mA is the maximum the signal source can handle,
but it could be pico-amps for all I care.

Do you care
about speed? (it might take some time to turn off.)
I care a little about the speed. are we talking more than 10's of
milliseconds?

I'm a bit
confused about the load. What are you driving?
The load is several parallel LEDs, all controlled by another chip (a shift
register with latch constant-current sink). Each "led" will get 20mA, and
I can have up to 24 on at the same time, (that is, between 0 and 24), so a
maximum of 480mA (rounded up to 500ma)

I've posted about this project in the past, its basically my own LED
display, based on an RGB LED matrix.
http://www.seeedstudio.com/depot/datasheet/2088RGBMatrix.pdf

I need to switch on exactly one of the rows, and set up the column shift
registers. I switch rows at about 8KHz, so I suspect most transistors can
handle that no problem.

There might be another alternative. On digikey, I found "PMIC - MOSFET,
Bridge Drivers, Internal/External Switch", which look like they do what I
want as well, with fewer discrete parts. It costs a bit more, and the
interface is "serial" rather than a simple 3-to-8 decoder I currently
have. Overall, I'm not sure which is the better approach. I think overall
I'd use a little less power when using the mosfet chip, which could be
useful if I make this a battery powered device.

What's the difference between internal and external for these "bridge
driver" chips? Also, what's the difference between High-Side and Low-Side
in these chips?

Thanks,
Daniel.

Usually there are more ways to skin a cat and I drew some of them below.
(View using fixed font like Courier)

-----+----+------------------+-----+--------------------------+----
| | | | |
| | .-. | |
| | 56R| | | |
| | | | | |
| | '-' | |
| | | |/ |
| | +---| BC635 |
| | | |> |
| | | | |
.-. | .-. | |
270R| | | 220R| | | |
| | | | | | |
'-' | '-' | |
|\ | | |\ | | |\ |
| \ | |/ | \ | | | \ ___ |<
| >--+--| BC635 | >---+ | | >---|___|---| BC369
| / |> | / | | / 270R |\
|/ | |/ | |/ |
+------ +------ +------
| | |
| | |
V V V
- LED - LED - LED
| | |
| | |
| | |
O___ O___ O___
O O O
controlled| | |
current | | |
limiter | | |
| | |
-----------+------------------------+--------------------------+----
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

The one to the left uses a 270R resistor to limit the current through your
driving circuit below 20mA when its output is low. As we do not know the
voltage on the anodes of the LEDs we cannot calculate what current is
driving the base of the BC635 when the output is high. That current may not
be enough to drive the BC635 into saturation as it has a limited gain. If so
you can look for a transistor that provides more gain (hard to find) or a
Darlington for example a TIP122. Pretty much a canon to swat a mosquito and
its saturation voltage may be to high.

A better solution seems to split up the current limiting resistor as in the
example in the middle. If you still have not enough base current you can
lower the 56R as long as you raise the 220R accordingly.

The example to the left uses a PNP-transistor. Note that this circuit lights
the LEDs whem your driving output is low. You only have to be sure that this
output when in high state is raised to at least 4.4V. If not, you may use a
pullup resistor to that output but that will increase the current to be
sinked through the output so you may have to raise the 270R.

petrus bitbyter

 

[George Herold]

On Apr 30, 1:31 pm, Tim Wescott <t...@seemywebsite.com> wrote:

On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:
I have a signal at 0 or 5v. The source of that signal can only handle
20mA of current.

I have another circuit which needs to have the same voltage level with
reference to ground, but it could have up to 500mA going through it. The
actual current is controlled by a current sink, and it may vary over
time.

In other words, when my signal is low, I want no current. When my signal
is high, I want to source as much current as possible for the sink to
take in.

I have some BJTs laying around, both PNP and NPN, both can support the
max Ic I need.  From what I figure, I'll need to hook my signal up to
the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v for the
resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually have 8
of these signals (although, only one is active at a time), so I was
hoping to avoid adding the 8 resistors and 8 transistors to my project,
though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then a PNP
with the emitter connected to +5V, a resistor to your 20mA driver, and
the collector connected to your load, is what you want.

Tim, stupid question, why is that better?

George H.

You need a PNP that has an HFE of at least 25 when it is saturated, which
may be hard to find if your junkbox has older parts in it.  This _may_
work at the 8kHz that you state elsewhere in the thread, particularly if
your driver can source to the base of the transistor; that'll help it
turn off.

A P-channel logic-level MOSFET would be perfect here, again assuming that
you can stand the inversion.  A gate resistor is still a good idea to
protect your source against current surges, but it's much easier to get
suitable PMOS parts than it is to find suitable high-HFE PNP parts.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Softwarehttp://www.wescottdesign.com- Hide quoted text -

- Show quoted text -

 

[Daniel Pitts]

On 4/30/13 2:48 PM, Tim Wescott wrote:

On Tue, 30 Apr 2013 13:17:28 -0700, George Herold wrote:

On Apr 30, 1:31 pm, Tim Wescott <t...@seemywebsite.com> wrote:
On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:
I have a signal at 0 or 5v. The source of that signal can only handle
20mA of current.

I have another circuit which needs to have the same voltage level
with reference to ground, but it could have up to 500mA going through
it. The actual current is controlled by a current sink, and it may
vary over time.

In other words, when my signal is low, I want no current. When my
signal is high, I want to source as much current as possible for the
sink to take in.

I have some BJTs laying around, both PNP and NPN, both can support
the max Ic I need. From what I figure, I'll need to hook my signal
up to the base, with at a minimum of 215ohms (Vbe = .7v, leaving 4.3v
for the resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually
have 8 of these signals (although, only one is active at a time), so
I was hoping to avoid adding the 8 resistors and 8 transistors to my
project, though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then a PNP
with the emitter connected to +5V, a resistor to your 20mA driver, and
the collector connected to your load, is what you want.

Tim, stupid question, why is that better?

I didn't qualify my statement, did I? Because the transistor will pull
up a lot closer to 5V -- with an emitter follower you have the B-E drop
in series with whatever drop your driving electronics has at the base
current. That not only means less voltage, but higher driver impedance
and hence more variation with current.
I don't care as much about the voltages at the various parts of the

transistor. I'm using a 74HC238 to select the row, though if necessary
I could swap that out for 74HC138 (active low instead of active high).
As I've said elsethread the 20ma is a maximum draw from the signal, not
a desired amount. Less would be better actually.
Datasheet for the 74HC238:
<http://www.nxp.com/documents/data_sheet/74HC_HCT238.pdf>

Come to think of it, maybe the OP is better off to use NPN emitter
followers and accept that the emitters will be somewhere between 4 and
4.3V (and look at the data sheet of whatever it is that delivers 20mA!!!).
Here's the current-sink (to actually drive the LEDs):

<http://www.ti.com/lit/ds/symlink/tlc5916.pdf>

BTW, I believe its a "conventional current sink", meaning an electron
source. Why does convention have to be backwards from physics!

So, I would either have a PNP with the emitter on the +5v line, or an
NPN with the collector on the +5. The opposite end would be on the LED
line, and that line may be somewhere between 5v and 0v depending on the
current sink on the other side of the LEDs. The end of the transistor
not tied to +5v would likely be between 2v and 5v. 5v means "all the
LEDs are off anyway", so I would expect no current. 2v means I'm
dropping 3v across the transistor at potentially 500mA. Hmm, 1.5 watts.
I wonder if my components can handle that. Unless (and likely) my
analysis is wrong.

So it's not a stupid question.
I'm glad he asked then ;-)

 

[Tim Wescott]

On Tue, 30 Apr 2013 15:43:37 -0700, Daniel Pitts wrote:

On 4/30/13 2:48 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 13:17:28 -0700, George Herold wrote:

On Apr 30, 1:31 pm, Tim Wescott <t...@seemywebsite.com> wrote:
On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:
I have a signal at 0 or 5v. The source of that signal can only
handle 20mA of current.

I have another circuit which needs to have the same voltage level
with reference to ground, but it could have up to 500mA going
through it. The actual current is controlled by a current sink, and
it may vary over time.

In other words, when my signal is low, I want no current. When my
signal is high, I want to source as much current as possible for the
sink to take in.

I have some BJTs laying around, both PNP and NPN, both can support
the max Ic I need. From what I figure, I'll need to hook my signal
up to the base, with at a minimum of 215ohms (Vbe = .7v, leaving
4.3v for the resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually
have 8 of these signals (although, only one is active at a time), so
I was hoping to avoid adding the 8 resistors and 8 transistors to my
project, though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then a
PNP with the emitter connected to +5V, a resistor to your 20mA
driver, and the collector connected to your load, is what you want.

Tim, stupid question, why is that better?

I didn't qualify my statement, did I? Because the transistor will pull
up a lot closer to 5V -- with an emitter follower you have the B-E drop
in series with whatever drop your driving electronics has at the base
current. That not only means less voltage, but higher driver impedance
and hence more variation with current.
I don't care as much about the voltages at the various parts of the
transistor. I'm using a 74HC238 to select the row, though if necessary
I could swap that out for 74HC138 (active low instead of active high).
As I've said elsethread the 20ma is a maximum draw from the signal, not
a desired amount. Less would be better actually. Datasheet for the
74HC238:
http://www.nxp.com/documents/data_sheet/74HC_HCT238.pdf

20mA is asking a lot from the part, and you want to be careful -- IIRC
not everyone specifies that much.

Come to think of it, maybe the OP is better off to use NPN emitter
followers and accept that the emitters will be somewhere between 4 and
4.3V (and look at the data sheet of whatever it is that delivers
20mA!!!).
Here's the current-sink (to actually drive the LEDs):
http://www.ti.com/lit/ds/symlink/tlc5916.pdf

BTW, I believe its a "conventional current sink", meaning an electron
source. Why does convention have to be backwards from physics!

Because the convention was established before people knew the polarity of
the electron. The coin came up tails...

So, I would either have a PNP with the emitter on the +5v line, or an
NPN with the collector on the +5. The opposite end would be on the LED
line, and that line may be somewhere between 5v and 0v depending on the
current sink on the other side of the LEDs. The end of the transistor
not tied to +5v would likely be between 2v and 5v. 5v means "all the
LEDs are off anyway", so I would expect no current. 2v means I'm
dropping 3v across the transistor at potentially 500mA. Hmm, 1.5 watts.
I wonder if my components can handle that. Unless (and likely) my
analysis is wrong.

If the transistor circuit is operating properly then it should drop 0.2V
or so in the PNP case, and 0.7V or so in the NPN case. It should not
vary by more than 0.2V in either direction, if that.

What transistor are you planning on using? Or maybe I should say -- what
candidates do you have in le box de junque?

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

 

[Daniel Pitts]

On 4/30/13 4:38 PM, Tim Wescott wrote:

On Tue, 30 Apr 2013 15:43:37 -0700, Daniel Pitts wrote:

On 4/30/13 2:48 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 13:17:28 -0700, George Herold wrote:

On Apr 30, 1:31 pm, Tim Wescott <t...@seemywebsite.com> wrote:
On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:
I have a signal at 0 or 5v. The source of that signal can only
handle 20mA of current.

I have another circuit which needs to have the same voltage level
with reference to ground, but it could have up to 500mA going
through it. The actual current is controlled by a current sink, and
it may vary over time.

In other words, when my signal is low, I want no current. When my
signal is high, I want to source as much current as possible for the
sink to take in.

I have some BJTs laying around, both PNP and NPN, both can support
the max Ic I need. From what I figure, I'll need to hook my signal
up to the base, with at a minimum of 215ohms (Vbe = .7v, leaving
4.3v for the resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually
have 8 of these signals (although, only one is active at a time), so
I was hoping to avoid adding the 8 resistors and 8 transistors to my
project, though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then a
PNP with the emitter connected to +5V, a resistor to your 20mA
driver, and the collector connected to your load, is what you want.

Tim, stupid question, why is that better?

I didn't qualify my statement, did I? Because the transistor will pull
up a lot closer to 5V -- with an emitter follower you have the B-E drop
in series with whatever drop your driving electronics has at the base
current. That not only means less voltage, but higher driver impedance
and hence more variation with current.
I don't care as much about the voltages at the various parts of the
transistor. I'm using a 74HC238 to select the row, though if necessary
I could swap that out for 74HC138 (active low instead of active high).
As I've said elsethread the 20ma is a maximum draw from the signal, not
a desired amount. Less would be better actually. Datasheet for the
74HC238:
http://www.nxp.com/documents/data_sheet/74HC_HCT238.pdf

20mA is asking a lot from the part, and you want to be careful -- IIRC
not everyone specifies that much.


Come to think of it, maybe the OP is better off to use NPN emitter
followers and accept that the emitters will be somewhere between 4 and
4.3V (and look at the data sheet of whatever it is that delivers
20mA!!!).
Here's the current-sink (to actually drive the LEDs):
http://www.ti.com/lit/ds/symlink/tlc5916.pdf

BTW, I believe its a "conventional current sink", meaning an electron
source. Why does convention have to be backwards from physics!

Because the convention was established before people knew the polarity of
the electron. The coin came up tails...
That was a lament, not a question ;-)

So, I would either have a PNP with the emitter on the +5v line, or an
NPN with the collector on the +5. The opposite end would be on the LED
line, and that line may be somewhere between 5v and 0v depending on the
current sink on the other side of the LEDs. The end of the transistor
not tied to +5v would likely be between 2v and 5v. 5v means "all the
LEDs are off anyway", so I would expect no current. 2v means I'm
dropping 3v across the transistor at potentially 500mA. Hmm, 1.5 watts.
I wonder if my components can handle that. Unless (and likely) my
analysis is wrong.

If the transistor circuit is operating properly then it should drop 0.2V
or so in the PNP case, and 0.7V or so in the NPN case. It should not
vary by more than 0.2V in either direction, if that.
Right, I realized this after I posted. Indeed, if I'm in saturation, the

voltage drop should be acceptable.

What transistor are you planning on using? Or maybe I should say -- what
candidates do you have in le box de junque?

I have plenty of SS8550DBU (PNP) and PN2222ATA (NPN). I was also
looking at some FETs on digikey last night.

I'm not opposed to ordering other parts. I know what I'm making isn't
novel, but if I can "do it cheaper", I might consider trying to sell my
end-product. In any case, I'm enjoying the learning experience.

 

[Tim Wescott]

On Wed, 01 May 2013 08:37:29 -0700, Daniel Pitts wrote:

On 4/30/13 4:38 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 15:43:37 -0700, Daniel Pitts wrote:

On 4/30/13 2:48 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 13:17:28 -0700, George Herold wrote:

On Apr 30, 1:31 pm, Tim Wescott <t...@seemywebsite.com> wrote:
On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:
I have a signal at 0 or 5v. The source of that signal can only
handle 20mA of current.

I have another circuit which needs to have the same voltage level
with reference to ground, but it could have up to 500mA going
through it. The actual current is controlled by a current sink,
and it may vary over time.

In other words, when my signal is low, I want no current. When my
signal is high, I want to source as much current as possible for
the sink to take in.

I have some BJTs laying around, both PNP and NPN, both can support
the max Ic I need. From what I figure, I'll need to hook my
signal up to the base, with at a minimum of 215ohms (Vbe = .7v,
leaving 4.3v for the resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually
have 8 of these signals (although, only one is active at a time),
so I was hoping to avoid adding the 8 resistors and 8 transistors
to my project, though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then a
PNP with the emitter connected to +5V, a resistor to your 20mA
driver, and the collector connected to your load, is what you want.

Tim, stupid question, why is that better?

I didn't qualify my statement, did I? Because the transistor will
pull up a lot closer to 5V -- with an emitter follower you have the
B-E drop in series with whatever drop your driving electronics has at
the base current. That not only means less voltage, but higher
driver impedance and hence more variation with current.
I don't care as much about the voltages at the various parts of the
transistor. I'm using a 74HC238 to select the row, though if
necessary I could swap that out for 74HC138 (active low instead of
active high). As I've said elsethread the 20ma is a maximum draw from
the signal, not a desired amount. Less would be better actually.
Datasheet for the 74HC238:
http://www.nxp.com/documents/data_sheet/74HC_HCT238.pdf

20mA is asking a lot from the part, and you want to be careful -- IIRC
not everyone specifies that much.


Come to think of it, maybe the OP is better off to use NPN emitter
followers and accept that the emitters will be somewhere between 4
and 4.3V (and look at the data sheet of whatever it is that delivers
20mA!!!).
Here's the current-sink (to actually drive the LEDs):
http://www.ti.com/lit/ds/symlink/tlc5916.pdf

BTW, I believe its a "conventional current sink", meaning an electron
source. Why does convention have to be backwards from physics!

Because the convention was established before people knew the polarity
of the electron. The coin came up tails...
That was a lament, not a question ;-)


So, I would either have a PNP with the emitter on the +5v line, or an
NPN with the collector on the +5. The opposite end would be on the LED
line, and that line may be somewhere between 5v and 0v depending on
the current sink on the other side of the LEDs. The end of the
transistor not tied to +5v would likely be between 2v and 5v. 5v means
"all the LEDs are off anyway", so I would expect no current. 2v means
I'm dropping 3v across the transistor at potentially 500mA. Hmm, 1.5
watts. I wonder if my components can handle that. Unless (and likely)
my analysis is wrong.

If the transistor circuit is operating properly then it should drop
0.2V or so in the PNP case, and 0.7V or so in the NPN case. It should
not vary by more than 0.2V in either direction, if that.
Right, I realized this after I posted. Indeed, if I'm in saturation, the
voltage drop should be acceptable.

What transistor are you planning on using? Or maybe I should say --
what candidates do you have in le box de junque?

I have plenty of SS8550DBU (PNP) and PN2222ATA (NPN). I was also
looking at some FETs on digikey last night.

I'm not opposed to ordering other parts. I know what I'm making isn't
novel, but if I can "do it cheaper", I might consider trying to sell my
end-product. In any case, I'm enjoying the learning experience.

I don't think you could trust a PN2222(whatever) to 500mA. You could use
one or two per LED with the base leads common, but that uses up a lot of
space.

Again, a P-channel logic-level MOSFET would be ideal for this. If you're
ordering from DigiKey anyway I think you'll find tons of candidate parts,
probably for pennies each.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

 

[petrus bitbyter]

"Daniel Pitts" <newsgroup.nospam@virtualinfinity.net> schreef in bericht
news:ZEagt.14010$VE.817@newsfe25.iad...

On 4/30/13 4:38 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 15:43:37 -0700, Daniel Pitts wrote:

On 4/30/13 2:48 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 13:17:28 -0700, George Herold wrote:

On Apr 30, 1:31 pm, Tim Wescott <t...@seemywebsite.com> wrote:
On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:
I have a signal at 0 or 5v. The source of that signal can only
handle 20mA of current.

I have another circuit which needs to have the same voltage level
with reference to ground, but it could have up to 500mA going
through it. The actual current is controlled by a current sink, and
it may vary over time.

In other words, when my signal is low, I want no current. When my
signal is high, I want to source as much current as possible for the
sink to take in.

I have some BJTs laying around, both PNP and NPN, both can support
the max Ic I need. From what I figure, I'll need to hook my signal
up to the base, with at a minimum of 215ohms (Vbe = .7v, leaving
4.3v for the resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually
have 8 of these signals (although, only one is active at a time), so
I was hoping to avoid adding the 8 resistors and 8 transistors to my
project, though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then a
PNP with the emitter connected to +5V, a resistor to your 20mA
driver, and the collector connected to your load, is what you want.

Tim, stupid question, why is that better?

I didn't qualify my statement, did I? Because the transistor will pull
up a lot closer to 5V -- with an emitter follower you have the B-E drop
in series with whatever drop your driving electronics has at the base
current. That not only means less voltage, but higher driver impedance
and hence more variation with current.
I don't care as much about the voltages at the various parts of the
transistor. I'm using a 74HC238 to select the row, though if necessary
I could swap that out for 74HC138 (active low instead of active high).
As I've said elsethread the 20ma is a maximum draw from the signal, not
a desired amount. Less would be better actually. Datasheet for the
74HC238:
http://www.nxp.com/documents/data_sheet/74HC_HCT238.pdf

20mA is asking a lot from the part, and you want to be careful -- IIRC
not everyone specifies that much.


Come to think of it, maybe the OP is better off to use NPN emitter
followers and accept that the emitters will be somewhere between 4 and
4.3V (and look at the data sheet of whatever it is that delivers
20mA!!!).
Here's the current-sink (to actually drive the LEDs):
http://www.ti.com/lit/ds/symlink/tlc5916.pdf

BTW, I believe its a "conventional current sink", meaning an electron
source. Why does convention have to be backwards from physics!

Because the convention was established before people knew the polarity of
the electron. The coin came up tails...
That was a lament, not a question ;-)


So, I would either have a PNP with the emitter on the +5v line, or an
NPN with the collector on the +5. The opposite end would be on the LED
line, and that line may be somewhere between 5v and 0v depending on the
current sink on the other side of the LEDs. The end of the transistor
not tied to +5v would likely be between 2v and 5v. 5v means "all the
LEDs are off anyway", so I would expect no current. 2v means I'm
dropping 3v across the transistor at potentially 500mA. Hmm, 1.5 watts.
I wonder if my components can handle that. Unless (and likely) my
analysis is wrong.

If the transistor circuit is operating properly then it should drop 0.2V
or so in the PNP case, and 0.7V or so in the NPN case. It should not
vary by more than 0.2V in either direction, if that.
Right, I realized this after I posted. Indeed, if I'm in saturation, the
voltage drop should be acceptable.

What transistor are you planning on using? Or maybe I should say -- what
candidates do you have in le box de junque?

I have plenty of SS8550DBU (PNP) and PN2222ATA (NPN). I was also looking
at some FETs on digikey last night.

I'm not opposed to ordering other parts. I know what I'm making isn't
novel, but if I can "do it cheaper", I might consider trying to sell my
end-product. In any case, I'm enjoying the learning experience.

You'll have to drive a PN2222A next to its limits and that's never a good
idea. If you want to give it a try, look at the datasheets of the
manufacturers. They have different specs for the same type of transistors.

That SS8550 suits better. A 15-18mA base current should be enough to turn
the transistor on. If the transistor is not driven into saturation, it may
become (too) hot. You have to account for that.

petrus bitbyter

 

[Daniel Pitts]

On 5/1/13 10:35 AM, Tim Wescott wrote:

On Wed, 01 May 2013 08:37:29 -0700, Daniel Pitts wrote:

On 4/30/13 4:38 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 15:43:37 -0700, Daniel Pitts wrote:

On 4/30/13 2:48 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 13:17:28 -0700, George Herold wrote:

On Apr 30, 1:31 pm, Tim Wescott <t...@seemywebsite.com> wrote:
On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:
I have a signal at 0 or 5v. The source of that signal can only
handle 20mA of current.

I have another circuit which needs to have the same voltage level
with reference to ground, but it could have up to 500mA going
through it. The actual current is controlled by a current sink,
and it may vary over time.

In other words, when my signal is low, I want no current. When my
signal is high, I want to source as much current as possible for
the sink to take in.

I have some BJTs laying around, both PNP and NPN, both can support
the max Ic I need. From what I figure, I'll need to hook my
signal up to the base, with at a minimum of 215ohms (Vbe = .7v,
leaving 4.3v for the resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I actually
have 8 of these signals (although, only one is active at a time),
so I was hoping to avoid adding the 8 resistors and 8 transistors
to my project, though that seems to be the cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then a
PNP with the emitter connected to +5V, a resistor to your 20mA
driver, and the collector connected to your load, is what you want.

Tim, stupid question, why is that better?

I didn't qualify my statement, did I? Because the transistor will
pull up a lot closer to 5V -- with an emitter follower you have the
B-E drop in series with whatever drop your driving electronics has at
the base current. That not only means less voltage, but higher
driver impedance and hence more variation with current.
I don't care as much about the voltages at the various parts of the
transistor. I'm using a 74HC238 to select the row, though if
necessary I could swap that out for 74HC138 (active low instead of
active high). As I've said elsethread the 20ma is a maximum draw from
the signal, not a desired amount. Less would be better actually.
Datasheet for the 74HC238:
http://www.nxp.com/documents/data_sheet/74HC_HCT238.pdf

20mA is asking a lot from the part, and you want to be careful -- IIRC
not everyone specifies that much.


Come to think of it, maybe the OP is better off to use NPN emitter
followers and accept that the emitters will be somewhere between 4
and 4.3V (and look at the data sheet of whatever it is that delivers
20mA!!!).
Here's the current-sink (to actually drive the LEDs):
http://www.ti.com/lit/ds/symlink/tlc5916.pdf

BTW, I believe its a "conventional current sink", meaning an electron
source. Why does convention have to be backwards from physics!

Because the convention was established before people knew the polarity
of the electron. The coin came up tails...
That was a lament, not a question ;-)


So, I would either have a PNP with the emitter on the +5v line, or an
NPN with the collector on the +5. The opposite end would be on the LED
line, and that line may be somewhere between 5v and 0v depending on
the current sink on the other side of the LEDs. The end of the
transistor not tied to +5v would likely be between 2v and 5v. 5v means
"all the LEDs are off anyway", so I would expect no current. 2v means
I'm dropping 3v across the transistor at potentially 500mA. Hmm, 1.5
watts. I wonder if my components can handle that. Unless (and likely)
my analysis is wrong.

If the transistor circuit is operating properly then it should drop
0.2V or so in the PNP case, and 0.7V or so in the NPN case. It should
not vary by more than 0.2V in either direction, if that.
Right, I realized this after I posted. Indeed, if I'm in saturation, the
voltage drop should be acceptable.

What transistor are you planning on using? Or maybe I should say --
what candidates do you have in le box de junque?

I have plenty of SS8550DBU (PNP) and PN2222ATA (NPN). I was also
looking at some FETs on digikey last night.

I'm not opposed to ordering other parts. I know what I'm making isn't
novel, but if I can "do it cheaper", I might consider trying to sell my
end-product. In any case, I'm enjoying the learning experience.

I don't think you could trust a PN2222(whatever) to 500mA. You could use
one or two per LED with the base leads common, but that uses up a lot of
space.
Here's the datasheet from my supplier:

<http://www.fairchildsemi.com/ds/PN/PN2222A.pdf>. 500ma is %5 above my
"worst case" (a pure white display) Looking at the datasheet though, the
current gain is 40 there, meaning I'd need at least 12.5ma to the base
to drive the thing to saturation. I could do that, but as you said that is

Again, a P-channel logic-level MOSFET would be ideal for this. If you're
ordering from DigiKey anyway I think you'll find tons of candidate parts,
probably for pennies each.
I was hoping for that, though I'm not seeing it. I'm seeing plenty of

N-channel mosfets for < $0.30 (although none that cheap can handle the
amperage I'm expecting), but the cheapest P-Channel is $0.42. That
would add $3.20 to my project.

I was also considering a Darlington, but that seems to be as expensive
as a similarly suitable P-channel MOSFET. It would be cheaper to hook up
two transistors myself.

<http://www.onsemi.com/pub/Collateral/2N6040-D.PDF> could be suitable. A
bit overkill (8A, 75W), but only $0.30 each. Comes in both PNP and NPN
flavors.


Actually, there is an octal array that might also be worthwhile...

<http://www.st.com/web/en/resource/technical/document/datasheet/CD00000179.pdf>

Still only 500ma max, but again, that is my worst-case scenario. This
array is only $0.27. It also says outputs could be parallel to increase
max current, so I could in theory add two to my project to support up to
one amp, and it would only add a half dollar to the cost.
The Toshiba PNP version:

<http://www.semicon.toshiba.co.jp/docs/datasheet/en/LinearIC/TD62783AFG_en_datasheet_091116.pdf>

Seems like one of these might be the way to go. Have an array of 8
simplifies a lot for me too, though it does add one big chip to the
project ;-)

Thanks again for all the advice.

 

[Tim Wescott]

On Wed, 01 May 2013 16:15:43 -0700, Daniel Pitts wrote:

On 5/1/13 10:35 AM, Tim Wescott wrote:
On Wed, 01 May 2013 08:37:29 -0700, Daniel Pitts wrote:

On 4/30/13 4:38 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 15:43:37 -0700, Daniel Pitts wrote:

On 4/30/13 2:48 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 13:17:28 -0700, George Herold wrote:

On Apr 30, 1:31 pm, Tim Wescott <t...@seemywebsite.com> wrote:
On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:
I have a signal at 0 or 5v. The source of that signal can only
handle 20mA of current.

I have another circuit which needs to have the same voltage
level with reference to ground, but it could have up to 500mA
going through it. The actual current is controlled by a current
sink, and it may vary over time.

In other words, when my signal is low, I want no current. When
my signal is high, I want to source as much current as possible
for the sink to take in.

I have some BJTs laying around, both PNP and NPN, both can
support the max Ic I need. From what I figure, I'll need to
hook my signal up to the base, with at a minimum of 215ohms (Vbe
= .7v, leaving 4.3v for the resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I
actually have 8 of these signals (although, only one is active
at a time), so I was hoping to avoid adding the 8 resistors and
8 transistors to my project, though that seems to be the
cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then
a PNP with the emitter connected to +5V, a resistor to your 20mA
driver, and the collector connected to your load, is what you
want.

Tim, stupid question, why is that better?

I didn't qualify my statement, did I? Because the transistor will
pull up a lot closer to 5V -- with an emitter follower you have the
B-E drop in series with whatever drop your driving electronics has
at the base current. That not only means less voltage, but higher
driver impedance and hence more variation with current.
I don't care as much about the voltages at the various parts of the
transistor. I'm using a 74HC238 to select the row, though if
necessary I could swap that out for 74HC138 (active low instead of
active high). As I've said elsethread the 20ma is a maximum draw
from the signal, not a desired amount. Less would be better
actually. Datasheet for the 74HC238:
http://www.nxp.com/documents/data_sheet/74HC_HCT238.pdf

20mA is asking a lot from the part, and you want to be careful --
IIRC not everyone specifies that much.


Come to think of it, maybe the OP is better off to use NPN emitter
followers and accept that the emitters will be somewhere between 4
and 4.3V (and look at the data sheet of whatever it is that
delivers 20mA!!!).
Here's the current-sink (to actually drive the LEDs):
http://www.ti.com/lit/ds/symlink/tlc5916.pdf

BTW, I believe its a "conventional current sink", meaning an
electron source. Why does convention have to be backwards from
physics!

Because the convention was established before people knew the
polarity of the electron. The coin came up tails...
That was a lament, not a question ;-)


So, I would either have a PNP with the emitter on the +5v line, or
an NPN with the collector on the +5. The opposite end would be on
the LED line, and that line may be somewhere between 5v and 0v
depending on the current sink on the other side of the LEDs. The end
of the transistor not tied to +5v would likely be between 2v and 5v.
5v means "all the LEDs are off anyway", so I would expect no
current. 2v means I'm dropping 3v across the transistor at
potentially 500mA. Hmm, 1.5 watts. I wonder if my components can
handle that. Unless (and likely)
my analysis is wrong.

If the transistor circuit is operating properly then it should drop
0.2V or so in the PNP case, and 0.7V or so in the NPN case. It
should not vary by more than 0.2V in either direction, if that.
Right, I realized this after I posted. Indeed, if I'm in saturation,
the voltage drop should be acceptable.

What transistor are you planning on using? Or maybe I should say --
what candidates do you have in le box de junque?

I have plenty of SS8550DBU (PNP) and PN2222ATA (NPN). I was also
looking at some FETs on digikey last night.

I'm not opposed to ordering other parts. I know what I'm making isn't
novel, but if I can "do it cheaper", I might consider trying to sell
my end-product. In any case, I'm enjoying the learning experience.

I don't think you could trust a PN2222(whatever) to 500mA. You could
use one or two per LED with the base leads common, but that uses up a
lot of space.
Here's the datasheet from my supplier:
http://www.fairchildsemi.com/ds/PN/PN2222A.pdf>. 500ma is %5 above my
"worst case" (a pure white display) Looking at the datasheet though, the
current gain is 40 there, meaning I'd need at least 12.5ma to the base
to drive the thing to saturation. I could do that, but as you said that
is

Again, a P-channel logic-level MOSFET would be ideal for this. If
you're ordering from DigiKey anyway I think you'll find tons of
candidate parts,
probably for pennies each.
I was hoping for that, though I'm not seeing it. I'm seeing plenty of
N-channel mosfets for < $0.30 (although none that cheap can handle the
amperage I'm expecting), but the cheapest P-Channel is $0.42. That
would add $3.20 to my project.

I was also considering a Darlington, but that seems to be as expensive
as a similarly suitable P-channel MOSFET. It would be cheaper to hook up
two transistors myself.

http://www.onsemi.com/pub/Collateral/2N6040-D.PDF> could be suitable. A
bit overkill (8A, 75W), but only $0.30 each. Comes in both PNP and NPN
flavors.


Actually, there is an octal array that might also be worthwhile...

http://www.st.com/web/en/resource/technical/document/datasheet/
CD00000179.pdf

Still only 500ma max, but again, that is my worst-case scenario. This
array is only $0.27. It also says outputs could be parallel to increase
max current, so I could in theory add two to my project to support up to
one amp, and it would only add a half dollar to the cost.
The Toshiba PNP version:

http://www.semicon.toshiba.co.jp/docs/datasheet/en/LinearIC/
TD62783AFG_en_datasheet_091116.pdf

Seems like one of these might be the way to go. Have an array of 8
simplifies a lot for me too, though it does add one big chip to the
project ;-)

Thanks again for all the advice.

If 500mA is only 5% of the maximum then you may be OK, because if I
understand what you're doing correctly, the transistor will only be on
for 1/8 of the time. That means that -- to the extent that the rating is
for thermal considerations -- you can push things.

Current gain will be a problem. Note that they're specifying the current
gain at 500mA and a Vce of 10V, and note how much the current gain drops
from 150mA and 10V vs. 150mA and 1V.

What about this PMOS: http://www.digikey.com/product-detail/en/
IRLML5103TRPBF/IRLML5103PBFCT-ND/812496. Or are you sticking with
through-hole?

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

 

[Daniel Pitts]

On 5/1/13 6:22 PM, Tim Wescott wrote:

On Wed, 01 May 2013 16:15:43 -0700, Daniel Pitts wrote:

On 5/1/13 10:35 AM, Tim Wescott wrote:
On Wed, 01 May 2013 08:37:29 -0700, Daniel Pitts wrote:

On 4/30/13 4:38 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 15:43:37 -0700, Daniel Pitts wrote:

On 4/30/13 2:48 PM, Tim Wescott wrote:
On Tue, 30 Apr 2013 13:17:28 -0700, George Herold wrote:

On Apr 30, 1:31 pm, Tim Wescott <t...@seemywebsite.com> wrote:
On Mon, 29 Apr 2013 21:58:15 -0700, Daniel Pitts wrote:
I have a signal at 0 or 5v. The source of that signal can only
handle 20mA of current.

I have another circuit which needs to have the same voltage
level with reference to ground, but it could have up to 500mA
going through it. The actual current is controlled by a current
sink, and it may vary over time.

In other words, when my signal is low, I want no current. When
my signal is high, I want to source as much current as possible
for the sink to take in.

I have some BJTs laying around, both PNP and NPN, both can
support the max Ic I need. From what I figure, I'll need to
hook my signal up to the base, with at a minimum of 215ohms (Vbe
= .7v, leaving 4.3v for the resistor. 4.3v/20ma=215ohms).

Will a circuit like that work? Is there an easier way? I
actually have 8 of these signals (although, only one is active
at a time), so I was hoping to avoid adding the 8 resistors and
8 transistors to my project, though that seems to be the
cheapest way.

Any advice is appreciated.
Thanks,
Daniel.

If you can stand having the sense of your signals reversed, then
a PNP with the emitter connected to +5V, a resistor to your 20mA
driver, and the collector connected to your load, is what you
want.

Tim, stupid question, why is that better?

I didn't qualify my statement, did I? Because the transistor will
pull up a lot closer to 5V -- with an emitter follower you have the
B-E drop in series with whatever drop your driving electronics has
at the base current. That not only means less voltage, but higher
driver impedance and hence more variation with current.
I don't care as much about the voltages at the various parts of the
transistor. I'm using a 74HC238 to select the row, though if
necessary I could swap that out for 74HC138 (active low instead of
active high). As I've said elsethread the 20ma is a maximum draw
from the signal, not a desired amount. Less would be better
actually. Datasheet for the 74HC238:
http://www.nxp.com/documents/data_sheet/74HC_HCT238.pdf

20mA is asking a lot from the part, and you want to be careful --
IIRC not everyone specifies that much.


Come to think of it, maybe the OP is better off to use NPN emitter
followers and accept that the emitters will be somewhere between 4
and 4.3V (and look at the data sheet of whatever it is that
delivers 20mA!!!).
Here's the current-sink (to actually drive the LEDs):
http://www.ti.com/lit/ds/symlink/tlc5916.pdf

BTW, I believe its a "conventional current sink", meaning an
electron source. Why does convention have to be backwards from
physics!

Because the convention was established before people knew the
polarity of the electron. The coin came up tails...
That was a lament, not a question ;-)


So, I would either have a PNP with the emitter on the +5v line, or
an NPN with the collector on the +5. The opposite end would be on
the LED line, and that line may be somewhere between 5v and 0v
depending on the current sink on the other side of the LEDs. The end
of the transistor not tied to +5v would likely be between 2v and 5v.
5v means "all the LEDs are off anyway", so I would expect no
current. 2v means I'm dropping 3v across the transistor at
potentially 500mA. Hmm, 1.5 watts. I wonder if my components can
handle that. Unless (and likely)
my analysis is wrong.

If the transistor circuit is operating properly then it should drop
0.2V or so in the PNP case, and 0.7V or so in the NPN case. It
should not vary by more than 0.2V in either direction, if that.
Right, I realized this after I posted. Indeed, if I'm in saturation,
the voltage drop should be acceptable.

What transistor are you planning on using? Or maybe I should say --
what candidates do you have in le box de junque?

I have plenty of SS8550DBU (PNP) and PN2222ATA (NPN). I was also
looking at some FETs on digikey last night.

I'm not opposed to ordering other parts. I know what I'm making isn't
novel, but if I can "do it cheaper", I might consider trying to sell
my end-product. In any case, I'm enjoying the learning experience.

I don't think you could trust a PN2222(whatever) to 500mA. You could
use one or two per LED with the base leads common, but that uses up a
lot of space.
Here's the datasheet from my supplier:
http://www.fairchildsemi.com/ds/PN/PN2222A.pdf>. 500ma is %5 above my
"worst case" (a pure white display) Looking at the datasheet though, the
current gain is 40 there, meaning I'd need at least 12.5ma to the base
to drive the thing to saturation. I could do that, but as you said that
is

Again, a P-channel logic-level MOSFET would be ideal for this. If
you're ordering from DigiKey anyway I think you'll find tons of
candidate parts,
probably for pennies each.
I was hoping for that, though I'm not seeing it. I'm seeing plenty of
N-channel mosfets for < $0.30 (although none that cheap can handle the
amperage I'm expecting), but the cheapest P-Channel is $0.42. That
would add $3.20 to my project.

I was also considering a Darlington, but that seems to be as expensive
as a similarly suitable P-channel MOSFET. It would be cheaper to hook up
two transistors myself.

http://www.onsemi.com/pub/Collateral/2N6040-D.PDF> could be suitable. A
bit overkill (8A, 75W), but only $0.30 each. Comes in both PNP and NPN
flavors.


Actually, there is an octal array that might also be worthwhile...

http://www.st.com/web/en/resource/technical/document/datasheet/
CD00000179.pdf

Still only 500ma max, but again, that is my worst-case scenario. This
array is only $0.27. It also says outputs could be parallel to increase
max current, so I could in theory add two to my project to support up to
one amp, and it would only add a half dollar to the cost.
The Toshiba PNP version:

http://www.semicon.toshiba.co.jp/docs/datasheet/en/LinearIC/
TD62783AFG_en_datasheet_091116.pdf

Seems like one of these might be the way to go. Have an array of 8
simplifies a lot for me too, though it does add one big chip to the
project ;-)

Thanks again for all the advice.

If 500mA is only 5% of the maximum then you may be OK, because if I
understand what you're doing correctly, the transistor will only be on
for 1/8 of the time. That means that -- to the extent that the rating is
for thermal considerations -- you can push things.
Yeah, I remember considering that at some point, but then it slipped my

mind again. Too much going on in my life outside of this hobby ;-)

So basically, if I have 480mA through one device with a duty-cycle of
12.5%, then as long as it can handle the "peak" without damage, the
average is about 60ma. That greatly improves my wattage as well. I
could also consider lowering my current by a little over a 3rd, and just
use the original IC I have now. This might be feasible since current
relates to LED brightness, and they may be bright enough at 6mA.
Especially for an indoor portable display, which was my original intent.

Something to ponder.

Current gain will be a problem. Note that they're specifying the current
gain at 500mA and a Vce of 10V, and note how much the current gain drops
from 150mA and 10V vs. 150mA and 1V.
On the NPN Darlington array I linked to, the current gain is 1000. Well

more than enough for my needs. I think this really is the way for me to go.

The Toshiba one I linked to (the TD62783APG) doesn't list current gain,
but the I-in(on) at with VIN = 3.85 V is a max of 260uA. Well below my 20mA.

What about this PMOS: http://www.digikey.com/product-detail/en/
IRLML5103TRPBF/IRLML5103PBFCT-ND/812496. Or are you sticking with
through-hole?

Yeah, pretty much through-hole only. This whole thing is on a
solderless breadboard. I've considered designing a PCB, but that's a
whole other experience. I have soldered SMT stuff before (had a kit I
bought at a workshop). It was interesting, but I don't think I'm ready
to design an actual circuit board from scratch.

Anyway, Thanks again for the feedback and advice. I really appreciate it.

 

[John Fields]

On Wed, 01 May 2013 23:30:04 -0700, Daniel Pitts
<newsgroup.nospam@virtualinfinity.net> wrote:

Yeah, pretty much through-hole only. This whole thing is on a
solderless breadboard. I've considered designing a PCB, but that's a
whole other experience. I have soldered SMT stuff before (had a kit I
bought at a workshop). It was interesting, but I don't think I'm ready
to design an actual circuit board from scratch.

Anyway, Thanks again for the feedback and advice. I really appreciate it.

---
I've read through most of the thread, and it seems to me that the best
bet for the high-side row driver would be a PNP or a PMOS FET which
you could run into saturation with your row logic driver.

That would give you the least power dissipation possible in the
high-side drivers and the most headroom possible for the low-side
sinks.

--
JF