Vce,sat vs Rds,on

[Jasen Betts]

On 2017-05-24, jurb6006@gmail.com <jurb6006@gmail.com> wrote:

"You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.) "

Yes, but it still acts more like a diode than a resistor.

only by the colours of smoke available.




build this.


C --[1K]--- 10VAC

+5vdc ---[10k]--- B <== 2n3904 or whatever jellybean.

E

|
|
--+-- 0V


scope the voltage on the collector.


that's a pretty lousy diode.



--
This email has not been checked by half-arsed antivirus software

 

[rickman]

John Larkin wrote:

On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

At Ib=0, Vout is zero.

Can someone explain to me how this is an inverted switdch? From what
I've read an inverted switch is one that takes advantage of the lower
Vce sat when the collector is used as the emitter and vice versa. Do I
have that wrong? This circuit is just an emitter follower fed by a
current source with a higher voltage than the V+ rail.

As Ib increases, Vout gets close to V+, like a regular saturated
transistor. But more Ib makes Vout keep increasing, to exactly V+ or
eventually a bit above V+.

I built a bunch of 16-bit DACS when I was a kid, using this switching
mode into an R-2R ladder network of wirewound resistors, with the Ib's
of the first few MSBs trimmed for exactly zero saturation.

But this is not an inverted switch. It's just a transistor base driven
from a high voltage which will drive the output rather than current from
the power rail.

--

Rick C

 

[rickman]

Jasen Betts wrote:

On 2017-05-24, jurb6006@gmail.com <jurb6006@gmail.com> wrote:
"You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.) "

Yes, but it still acts more like a diode than a resistor.

only by the colours of smoke available.




build this.


C --[1K]--- 10VAC

+5vdc ---[10k]--- B <== 2n3904 or whatever jellybean.

E

|
|
--+-- 0V


scope the voltage on the collector.


that's a pretty lousy diode.

It's a much worse resistor.


Version 4
SHEET 1 880 680
WIRE 320 -48 256 -48
WIRE 400 -48 320 -48
WIRE 256 0 256 -48
WIRE 400 0 400 -48
WIRE 256 96 256 80
WIRE 256 128 256 96
WIRE 400 128 400 80
WIRE -48 176 -80 176
WIRE 16 176 -48 176
WIRE 144 176 96 176
WIRE 192 176 144 176
WIRE -80 208 -80 176
WIRE 256 272 256 224
WIRE -80 320 -80 288
FLAG 256 272 0
FLAG -80 320 0
FLAG -48 176 Vbb
FLAG 144 176 Vb
FLAG 256 96 Vc
FLAG 320 -48 Vcc
FLAG 400 128 0
SYMBOL npn 192 128 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL res 112 160 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL res 272 96 R180
WINDOW 0 36 76 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL voltage -80 192 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 5
SYMBOL voltage 400 -16 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value SINE(0 10 1000)
TEXT -114 344 Left 2 !.tran 10ms


--

Rick C

 

[George Herold]

On Wednesday, May 24, 2017 at 12:33:24 AM UTC-4, John Larkin wrote:

On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

Huh, OK. I got that to work.
But I had to make Ib>Ie, like twice as much.
It was kinda weird, 'cause the extra current going from b-c
was feeding voltage into my triple power supply.

George H.

At Ib=0, Vout is zero.

As Ib increases, Vout gets close to V+, like a regular saturated
transistor. But more Ib makes Vout keep increasing, to exactly V+ or
eventually a bit above V+.

I built a bunch of 16-bit DACS when I was a kid, using this switching
mode into an R-2R ladder network of wirewound resistors, with the Ib's
of the first few MSBs trimmed for exactly zero saturation.


--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[George Herold]

On Wednesday, May 24, 2017 at 1:28:06 AM UTC-4, rickman wrote:

John Larkin wrote:
On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

At Ib=0, Vout is zero.

Can someone explain to me how this is an inverted switdch? From what
I've read an inverted switch is one that takes advantage of the lower
Vce sat when the collector is used as the emitter and vice versa. Do I
have that wrong? This circuit is just an emitter follower fed by a
current source with a higher voltage than the V+ rail.


As Ib increases, Vout gets close to V+, like a regular saturated
transistor. But more Ib makes Vout keep increasing, to exactly V+ or
eventually a bit above V+.

I built a bunch of 16-bit DACS when I was a kid, using this switching
mode into an R-2R ladder network of wirewound resistors, with the Ib's
of the first few MSBs trimmed for exactly zero saturation.

But this is not an inverted switch. It's just a transistor base driven
from a high voltage which will drive the output rather than current from
the power rail.

Ya know, you can call things whatever you want.
I guess it's 'inverted' 'cause I'd usually think of the
npn being on the low side.

George H.

--

Rick C

 

[George Herold]

On Wednesday, May 24, 2017 at 2:37:39 AM UTC-4, rickman wrote:

Jasen Betts wrote:
On 2017-05-24, jurb6006@gmail.com <jurb6006@gmail.com> wrote:
"You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.) "

Yes, but it still acts more like a diode than a resistor.

only by the colours of smoke available.




build this.


C --[1K]--- 10VAC

+5vdc ---[10k]--- B <== 2n3904 or whatever jellybean.

E

|
|
--+-- 0V


scope the voltage on the collector.


that's a pretty lousy diode.

It's a much worse resistor.


Version 4
SHEET 1 880 680
WIRE 320 -48 256 -48
WIRE 400 -48 320 -48
WIRE 256 0 256 -48
WIRE 400 0 400 -48
WIRE 256 96 256 80
WIRE 256 128 256 96
WIRE 400 128 400 80
WIRE -48 176 -80 176
WIRE 16 176 -48 176
WIRE 144 176 96 176
WIRE 192 176 144 176
WIRE -80 208 -80 176
WIRE 256 272 256 224
WIRE -80 320 -80 288
FLAG 256 272 0
FLAG -80 320 0
FLAG -48 176 Vbb
FLAG 144 176 Vb
FLAG 256 96 Vc
FLAG 320 -48 Vcc
FLAG 400 128 0
SYMBOL npn 192 128 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL res 112 160 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL res 272 96 R180
WINDOW 0 36 76 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL voltage -80 192 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 5
SYMBOL voltage 400 -16 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value SINE(0 10 1000)
TEXT -114 344 Left 2 !.tran 10ms


--

Rick C

Add a ~10V DC offset to the 1kHz Sine wave.
and goose R1 down to 2k ohms.
Looks like an OK resistor to me.
(but not worth getting your knickers in a knot over it.)

George H.

 

[rickman]

George Herold wrote:

On Wednesday, May 24, 2017 at 1:28:06 AM UTC-4, rickman wrote:
John Larkin wrote:
On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

At Ib=0, Vout is zero.

Can someone explain to me how this is an inverted switdch? From what
I've read an inverted switch is one that takes advantage of the lower
Vce sat when the collector is used as the emitter and vice versa. Do I
have that wrong? This circuit is just an emitter follower fed by a
current source with a higher voltage than the V+ rail.


As Ib increases, Vout gets close to V+, like a regular saturated
transistor. But more Ib makes Vout keep increasing, to exactly V+ or
eventually a bit above V+.

I built a bunch of 16-bit DACS when I was a kid, using this switching
mode into an R-2R ladder network of wirewound resistors, with the Ib's
of the first few MSBs trimmed for exactly zero saturation.

But this is not an inverted switch. It's just a transistor base driven
from a high voltage which will drive the output rather than current from
the power rail.
Ya know, you can call things whatever you want.
I guess it's 'inverted' 'cause I'd usually think of the
npn being on the low side.

You're not familiar with an emitter follower? The other name for that
is common collector.

The "inverted" aspect has not to do with high side or low side
switching, it refers to swapping the role of the collector and emitter
to use the transistor in a different way that gives it very different
properties including a very low saturation voltage.

--

Rick C

 

[rickman]

George Herold wrote:

On Wednesday, May 24, 2017 at 2:37:39 AM UTC-4, rickman wrote:
Jasen Betts wrote:
On 2017-05-24, jurb6006@gmail.com <jurb6006@gmail.com> wrote:
"You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.) "

Yes, but it still acts more like a diode than a resistor.

only by the colours of smoke available.




build this.


C --[1K]--- 10VAC

+5vdc ---[10k]--- B <== 2n3904 or whatever jellybean.

E

|
|
--+-- 0V


scope the voltage on the collector.


that's a pretty lousy diode.

It's a much worse resistor.


Version 4
SHEET 1 880 680
WIRE 320 -48 256 -48
WIRE 400 -48 320 -48
WIRE 256 0 256 -48
WIRE 400 0 400 -48
WIRE 256 96 256 80
WIRE 256 128 256 96
WIRE 400 128 400 80
WIRE -48 176 -80 176
WIRE 16 176 -48 176
WIRE 144 176 96 176
WIRE 192 176 144 176
WIRE -80 208 -80 176
WIRE 256 272 256 224
WIRE -80 320 -80 288
FLAG 256 272 0
FLAG -80 320 0
FLAG -48 176 Vbb
FLAG 144 176 Vb
FLAG 256 96 Vc
FLAG 320 -48 Vcc
FLAG 400 128 0
SYMBOL npn 192 128 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL res 112 160 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL res 272 96 R180
WINDOW 0 36 76 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL voltage -80 192 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 5
SYMBOL voltage 400 -16 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value SINE(0 10 1000)
TEXT -114 344 Left 2 !.tran 10ms


--

Rick C

Add a ~10V DC offset to the 1kHz Sine wave.
and goose R1 down to 2k ohms.
Looks like an OK resistor to me.
(but not worth getting your knickers in a knot over it.)

I believe a diode will look very similar in that test. If you don't
look at both polarities it's not much of a diode test.

Try measuring the resistance. Plot V(vc)/Ic(Q1). You'll see it is
anything *but* a resistor.


--

Rick C

 

[John Larkin]

On Wed, 24 May 2017 07:11:31 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Wednesday, May 24, 2017 at 1:28:06 AM UTC-4, rickman wrote:
John Larkin wrote:
On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

At Ib=0, Vout is zero.

Can someone explain to me how this is an inverted switdch? From what
I've read an inverted switch is one that takes advantage of the lower
Vce sat when the collector is used as the emitter and vice versa. Do I
have that wrong? This circuit is just an emitter follower fed by a
current source with a higher voltage than the V+ rail.


As Ib increases, Vout gets close to V+, like a regular saturated
transistor. But more Ib makes Vout keep increasing, to exactly V+ or
eventually a bit above V+.

I built a bunch of 16-bit DACS when I was a kid, using this switching
mode into an R-2R ladder network of wirewound resistors, with the Ib's
of the first few MSBs trimmed for exactly zero saturation.

But this is not an inverted switch. It's just a transistor base driven
from a high voltage which will drive the output rather than current from
the power rail.
Ya know, you can call things whatever you want.
I guess it's 'inverted' 'cause I'd usually think of the
npn being on the low side.

George H.



--

Rick C

Ignore rickman. He's nasty and not especially bright.


--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[John Larkin]

On Wed, 24 May 2017 07:08:28 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Wednesday, May 24, 2017 at 12:33:24 AM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

Huh, OK. I got that to work.
But I had to make Ib>Ie, like twice as much.

Depends on the transistor and the currents, but that's about right.

It was kinda weird, 'cause the extra current going from b-c
was feeding voltage into my triple power supply.

Yup!


--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[rickman]

George Herold wrote:

On Wednesday, May 24, 2017 at 12:33:24 AM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

Huh, OK. I got that to work.
But I had to make Ib>Ie, like twice as much.
It was kinda weird, 'cause the extra current going from b-c
was feeding voltage into my triple power supply.

Sure, because the base voltage was higher than either the collector or
the emitter so was pumping current into *both*. It's an odd design but
will bring the load up to the voltage of the positive rail. But then so
would a two diode circuit. Essentially this circuit is using the the
NPN as two diodes rather than a transistor.

JL can wave his hands in the air, but try it, you'll see the two diode
circuit works exactly the same as the "current source" brings Vb near
the positive rail or is turned off dragging Vout to ground. The only
difference is in the middle ground where the transistor is actually
acting as a transistor.

--

Rick C

 

[George Herold]

On Wednesday, May 24, 2017 at 11:43:12 AM UTC-4, rickman wrote:

George Herold wrote:
On Wednesday, May 24, 2017 at 2:37:39 AM UTC-4, rickman wrote:
Jasen Betts wrote:
On 2017-05-24, jurb6006@gmail.com <jurb6006@gmail.com> wrote:
"You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.) "

Yes, but it still acts more like a diode than a resistor.

only by the colours of smoke available.




build this.


C --[1K]--- 10VAC

+5vdc ---[10k]--- B <== 2n3904 or whatever jellybean.

E

|
|
--+-- 0V


scope the voltage on the collector.


that's a pretty lousy diode.

It's a much worse resistor.


Version 4
SHEET 1 880 680
WIRE 320 -48 256 -48
WIRE 400 -48 320 -48
WIRE 256 0 256 -48
WIRE 400 0 400 -48
WIRE 256 96 256 80
WIRE 256 128 256 96
WIRE 400 128 400 80
WIRE -48 176 -80 176
WIRE 16 176 -48 176
WIRE 144 176 96 176
WIRE 192 176 144 176
WIRE -80 208 -80 176
WIRE 256 272 256 224
WIRE -80 320 -80 288
FLAG 256 272 0
FLAG -80 320 0
FLAG -48 176 Vbb
FLAG 144 176 Vb
FLAG 256 96 Vc
FLAG 320 -48 Vcc
FLAG 400 128 0
SYMBOL npn 192 128 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL res 112 160 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL res 272 96 R180
WINDOW 0 36 76 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL voltage -80 192 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 5
SYMBOL voltage 400 -16 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value SINE(0 10 1000)
TEXT -114 344 Left 2 !.tran 10ms


--

Rick C

Add a ~10V DC offset to the 1kHz Sine wave.
and goose R1 down to 2k ohms.
Looks like an OK resistor to me.
(but not worth getting your knickers in a knot over it.)

I believe a diode will look very similar in that test. If you don't
look at both polarities it's not much of a diode test.

OK maybe we think about diodes differently. To me a diode has
an exponential I-V relation.

Try measuring the resistance. Plot V(vc)/Ic(Q1). You'll see it is
anything *but* a resistor.

So a quick plot... looks like ~6mV of DC offset and then about a 1.5 ohm resistor.
(Curve flattens out at higher currents some, opposite from what I'd
expect for a diode.)

George H.

--

Rick C

 

[George Herold]

On Wednesday, May 24, 2017 at 12:44:31 PM UTC-4, John Larkin wrote:

On Wed, 24 May 2017 07:08:28 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Wednesday, May 24, 2017 at 12:33:24 AM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

Huh, OK. I got that to work.
But I had to make Ib>Ie, like twice as much.

Depends on the transistor and the currents, but that's about right.
2n4401, Ic ~6mA

GH

It was kinda weird, 'cause the extra current going from b-c
was feeding voltage into my triple power supply.

Yup!


--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[George Herold]

On Wednesday, May 24, 2017 at 1:12:50 PM UTC-4, rickman wrote:

George Herold wrote:
On Wednesday, May 24, 2017 at 12:33:24 AM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

Huh, OK. I got that to work.
But I had to make Ib>Ie, like twice as much.
It was kinda weird, 'cause the extra current going from b-c
was feeding voltage into my triple power supply.

Sure, because the base voltage was higher than either the collector or
the emitter so was pumping current into *both*. It's an odd design but
will bring the load up to the voltage of the positive rail. But then so
would a two diode circuit. Essentially this circuit is using the the
NPN as two diodes rather than a transistor.

JL can wave his hands in the air, but try it, you'll see the two diode
circuit works exactly the same as the "current source" brings Vb near
the positive rail or is turned off dragging Vout to ground. The only
difference is in the middle ground where the transistor is actually
acting as a transistor.

--

Rick C

No no! (Rick, It's very hard to have a conversation with you.)
It wouldn't back feed the power supply when it was working 'properly'.
But is was something I had to watch out for as I twiddled knobs on my
bench supply.

JL (in his youth) made an ideal switch from a transistor.
That's kinda fun, I played with it for ~15 min this morning.

Maybe you can make it work in spice?

George H.

 

[John Larkin]

On Wed, 24 May 2017 11:21:13 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Wednesday, May 24, 2017 at 1:12:50 PM UTC-4, rickman wrote:
George Herold wrote:
On Wednesday, May 24, 2017 at 12:33:24 AM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

Huh, OK. I got that to work.
But I had to make Ib>Ie, like twice as much.
It was kinda weird, 'cause the extra current going from b-c
was feeding voltage into my triple power supply.

Sure, because the base voltage was higher than either the collector or
the emitter so was pumping current into *both*. It's an odd design but
will bring the load up to the voltage of the positive rail. But then so
would a two diode circuit. Essentially this circuit is using the the
NPN as two diodes rather than a transistor.

JL can wave his hands in the air, but try it, you'll see the two diode
circuit works exactly the same as the "current source" brings Vb near
the positive rail or is turned off dragging Vout to ground. The only
difference is in the middle ground where the transistor is actually
acting as a transistor.

--

Rick C

No no! (Rick, It's very hard to have a conversation with you.)
It wouldn't back feed the power supply when it was working 'properly'.
But is was something I had to watch out for as I twiddled knobs on my
bench supply.

JL (in his youth) made an ideal switch from a transistor.
That's kinda fun, I played with it for ~15 min this morning.

Maybe you can make it work in spice?

George H.

The transistor is truly bi-directional ohmic, at least until it runs
out of reverse beta. The upside-down config lets you tune Ib to make
the residual DC offset go to zero. Two diodes do NOT make a
transistor! [1]

Ricky gets confrontational when he doesn't understand things. Ignore
him.

[1] Lots of us, in our childhoods, tried to make two diodes have
transistor action.


--

John Larkin Highland Technology, Inc
picosecond timing precision measurement

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[rickman]

George Herold wrote:

On Wednesday, May 24, 2017 at 1:12:50 PM UTC-4, rickman wrote:
George Herold wrote:
On Wednesday, May 24, 2017 at 12:33:24 AM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

Huh, OK. I got that to work.
But I had to make Ib>Ie, like twice as much.
It was kinda weird, 'cause the extra current going from b-c
was feeding voltage into my triple power supply.

Sure, because the base voltage was higher than either the collector or
the emitter so was pumping current into *both*. It's an odd design but
will bring the load up to the voltage of the positive rail. But then so
would a two diode circuit. Essentially this circuit is using the the
NPN as two diodes rather than a transistor.

JL can wave his hands in the air, but try it, you'll see the two diode
circuit works exactly the same as the "current source" brings Vb near
the positive rail or is turned off dragging Vout to ground. The only
difference is in the middle ground where the transistor is actually
acting as a transistor.

--

Rick C

No no! (Rick, It's very hard to have a conversation with you.)

What is hard about it? You have reported the facts and I'm just
addressing what you reported.


> It wouldn't back feed the power supply when it was working 'properly'.

You need to define "properly". The only way this circuit will get the
emitter voltage up to the rail is if the base voltage is about 0.7 volts
higher than the rails, no? At that point the BC junction will also be
forward biased and conduct as well as the BE junction. What is wrong
with that analysis?

Isn't that the goal, to bring the load up to the voltage of the rail as
closely as possible?

Heck, you said yourself the base current was twice the current in the
emitter. Where else would the excess current be flowing but out the
collector to the power rail?

But is was something I had to watch out for as I twiddled knobs on my
bench supply.

JL (in his youth) made an ideal switch from a transistor.
That's kinda fun, I played with it for ~15 min this morning.

Maybe you can make it work in spice?

Why simulate it when you have already built it and shown the base
current feeding into the power rail?

--

Rick C

 

[rickman]

George Herold wrote:

On Wednesday, May 24, 2017 at 11:43:12 AM UTC-4, rickman wrote:
George Herold wrote:
On Wednesday, May 24, 2017 at 2:37:39 AM UTC-4, rickman wrote:
Jasen Betts wrote:
On 2017-05-24, jurb6006@gmail.com <jurb6006@gmail.com> wrote:
"You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.) "

Yes, but it still acts more like a diode than a resistor.

only by the colours of smoke available.




build this.


C --[1K]--- 10VAC

+5vdc ---[10k]--- B <== 2n3904 or whatever jellybean.

E

|
|
--+-- 0V


scope the voltage on the collector.


that's a pretty lousy diode.

It's a much worse resistor.


Version 4
SHEET 1 880 680
WIRE 320 -48 256 -48
WIRE 400 -48 320 -48
WIRE 256 0 256 -48
WIRE 400 0 400 -48
WIRE 256 96 256 80
WIRE 256 128 256 96
WIRE 400 128 400 80
WIRE -48 176 -80 176
WIRE 16 176 -48 176
WIRE 144 176 96 176
WIRE 192 176 144 176
WIRE -80 208 -80 176
WIRE 256 272 256 224
WIRE -80 320 -80 288
FLAG 256 272 0
FLAG -80 320 0
FLAG -48 176 Vbb
FLAG 144 176 Vb
FLAG 256 96 Vc
FLAG 320 -48 Vcc
FLAG 400 128 0
SYMBOL npn 192 128 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL res 112 160 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL res 272 96 R180
WINDOW 0 36 76 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL voltage -80 192 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 5
SYMBOL voltage 400 -16 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value SINE(0 10 1000)
TEXT -114 344 Left 2 !.tran 10ms


--

Rick C

Add a ~10V DC offset to the 1kHz Sine wave.
and goose R1 down to 2k ohms.
Looks like an OK resistor to me.
(but not worth getting your knickers in a knot over it.)

I believe a diode will look very similar in that test. If you don't
look at both polarities it's not much of a diode test.
OK maybe we think about diodes differently. To me a diode has
an exponential I-V relation.

I think we do see them very differently. To me the first order I-V
curve is to conduct in one polarity and to not conduct in the other.
The second order effect is an exponential I-V curve with positive
voltage at low currents and a linear curve at higher currents.

Try measuring the resistance. Plot V(vc)/Ic(Q1). You'll see it is
anything *but* a resistor.
So a quick plot... looks like ~6mV of DC offset and then about a 1.5 ohm resistor.
(Curve flattens out at higher currents some, opposite from what I'd
expect for a diode.)

I think you need to check it again. Change the collector supply to a
ramp and you can see Vc/Ic in a plot range over very wide values and
nothing like a linear ramp!

But if you really want to consider this characteristic, go back to
having a negative voltage input.

--

Rick C

 

[rickman]

George Herold wrote:

On Wednesday, May 24, 2017 at 1:12:50 PM UTC-4, rickman wrote:
George Herold wrote:
On Wednesday, May 24, 2017 at 12:33:24 AM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 12:20:30 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Tuesday, May 23, 2017 at 2:21:55 PM UTC-4, John Larkin wrote:
On Tue, 23 May 2017 06:47:14 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Thursday, May 18, 2017 at 9:33:58 PM UTC-4, jurb...@gmail.com wrote:
"But a forward biased diode might have .6 volts of drop or more. And
it's exponential, not ohmic. "

Pretty much like a saturated bipolar, eh ?

You do know that if you drive an npn on hard (saturate) you can
get Vce to be less than ~0.1 V (Depending on all sorts of other things.)

Millivolts. And in the inverted switch configuration, microvolts or
even negative.
Grin... I had to build it on my bench supply to be sure.
It went down below 50 mV... but I was being conservative.

By inverted do you mean switching C and E?


Not exactly.

https://dl.dropboxusercontent.com/u/53724080/Parts/Transistors/Inverted_NPN.JPG

Huh, OK. I got that to work.
But I had to make Ib>Ie, like twice as much.
It was kinda weird, 'cause the extra current going from b-c
was feeding voltage into my triple power supply.

Sure, because the base voltage was higher than either the collector or
the emitter so was pumping current into *both*. It's an odd design but
will bring the load up to the voltage of the positive rail. But then so
would a two diode circuit. Essentially this circuit is using the the
NPN as two diodes rather than a transistor.

JL can wave his hands in the air, but try it, you'll see the two diode
circuit works exactly the same as the "current source" brings Vb near
the positive rail or is turned off dragging Vout to ground. The only
difference is in the middle ground where the transistor is actually
acting as a transistor.

--

Rick C

No no! (Rick, It's very hard to have a conversation with you.)
It wouldn't back feed the power supply when it was working 'properly'.
But is was something I had to watch out for as I twiddled knobs on my
bench supply.

JL (in his youth) made an ideal switch from a transistor.
That's kinda fun, I played with it for ~15 min this morning.

Maybe you can make it work in spice?

Ok, here is a two diode circuit that does pretty much the same thing as
John's NPN circuit. Notice the current in the two diodes is equal when
turned on. In this case they will be exact because the diodes are
exact. Use a single NPN transistor and they won't be exactly equal
because the junctions are not identical.

In JL's circuit the transistor is not acting as an amplifier when
switched on fully. It's not even acting as a switch. The current
source is the switch and the transistor is acting as two diodes to give
equal voltage drops so the output voltage will be equal to the rail.

For a transistor to operate as a gain device the BC junction has to be
reverse biased. In this circuit the BC junction is forward biased.


Version 4
SHEET 1 880 680
WIRE -32 80 -32 48
WIRE -32 208 -32 160
WIRE 32 208 -32 208
WIRE 80 208 32 208
WIRE 112 208 80 208
WIRE 240 208 176 208
WIRE 272 208 240 208
WIRE 272 240 272 208
WIRE 272 352 272 320
WIRE 80 416 80 208
WIRE 112 416 80 416
WIRE 208 416 176 416
WIRE 240 416 208 416
WIRE 288 416 240 416
WIRE 208 448 208 416
WIRE 208 560 208 528
FLAG -32 48 0
FLAG 208 560 0
FLAG 272 352 0
FLAG 32 208 Base
FLAG 240 208 Collector
FLAG 240 416 Emitter
SYMBOL diode 112 432 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D1
SYMATTR Value 1N914
SYMBOL diode 112 224 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D2
SYMATTR Value 1N914
SYMBOL res 192 432 R0
SYMATTR InstName R1
SYMATTR Value 200
SYMBOL voltage 272 224 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 5
SYMBOL current -32 80 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName I1
SYMATTR Value PULSE(0 50mA 5ms 1ns 1ns 5ms 10ms)
TEXT -66 584 Left 2 !.tran 10ms



--

Rick C