Bipolar transistor bias question

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I have a switching application where a single bipolar transistor might be  
used. The AC input is about 300mV and the transistor needs about 700 mV to  
switch on. My idea is to construct a voltage divider using a diode which  
will produce about a 600mV drop across the diode and a voltage divider of 2  
equal resistors to set the transistor base voltage at about 300 mV so the  
transistor will switch on with an additional 300mV. I understand the  
temperature problems, but it seems to be minimal. What am I missing?






Re: Bipolar transistor bias question
On Monday, 19 December 2016 00:41:55 UTC, Bill Bowden  wrote:
Quoted text here. Click to load it

a circuit diagram.


NT

Re: Bipolar transistor bias question
On Sun, 18 Dec 2016 16:41:42 -0800, "billbowden"

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More detail would help. What is the drive waveform like? What is the
load? Got a schematic?

A PNP emitter follower can be a neat way to drive an NPN transistor.
It moves the effective threshold near zero volts and does some
temperature compensation.

Or just add an IC, an opamp or a comparator.



--  

John Larkin         Highland Technology, Inc

lunatic fringe electronics  


Re: Bipolar transistor bias question

Quoted text here. Click to load it

All I want to do is detect a positive half cycle going sine wave at about  
20Hz and 300mV peak and produce a negative rectified DC change of about 3  
volts. I figure I can do this with one transistor. I'm not interested in  
adding more parts. I can easily do it with more parts. All I want to know is  
the circuit stability of biasing a transistor near it's conduction point so  
it works under all conditions of temperature and supply voltage. It needs to  
work between a supply voltage of 2 to 3.2 volts. Here's a typical circuit  
using LTSpice.
.
Version 4
SHEET 1 880 680
WIRE 208 0 32 0
WIRE 336 0 208 0
WIRE 32 32 32 0
WIRE 128 64 96 64
WIRE 208 128 208 80
WIRE 32 176 32 112
WIRE 128 176 128 144
WIRE 128 176 112 176
WIRE 144 176 128 176
WIRE 128 208 128 176
WIRE 32 224 32 176
WIRE 128 288 128 272
FLAG 336 80 0
FLAG 208 224 0
FLAG 128 368 0
FLAG 32 288 0
FLAG 96 64 0
SYMBOL npn 144 128 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL res 192 -16 R0
WINDOW 0 -400 416 Left 0
WINDOW 3 39 52 Left 0
SYMATTR InstName R1
SYMATTR Value 2k
SYMBOL voltage 336 -16 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 2
SYMBOL res 128 160 R90
WINDOW 0 257 402 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R2
SYMATTR Value 15k
SYMBOL res 48 128 R180
WINDOW 0 269 -273 Left 0
WINDOW 3 39 61 Left 0
SYMATTR InstName R3
SYMATTR Value 5.1k
SYMBOL diode 16 224 R0
WINDOW 0 -100 172 Left 0
WINDOW 3 -103 34 Left 0
SYMATTR InstName D1
SYMATTR Value 1N4148
SYMBOL cap 112 208 R0
WINDOW 0 -238 190 Left 0
WINDOW 3 -22 55 Left 0
SYMATTR InstName C1

SYMBOL voltage 128 272 R0
WINDOW 0 36 87 Left 0
WINDOW 3 94 53 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 94 81 Left 0
SYMATTR InstName V2
SYMATTR Value SINE(0 .3 20 0 0 0 15)
SYMATTR SpiceLine Rser=1k
SYMBOL res 112 48 R0
WINDOW 0 -281 351 Left 0
WINDOW 3 -45 65 Left 0
SYMATTR InstName R4
SYMATTR Value 51k
TEXT -32 396 Left 0 !.tran 1.5




Re: Bipolar transistor bias question
On 12/19/2016 8:41 PM, billbowden wrote:
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(snip)

Okay. Add the following to your schematic using the LTSpice directives:

.step v1 list 2 3.2
.step temp list -40 25 70

This will give you V1 steps of 2 to 3.2
And temperature in 3 steps from -40 to 70C

Check the help section to refine your desired range.


Re: Bipolar transistor bias question

Quoted text here. Click to load it

Yes, that worked well and all looked good. The base voltage varies from 360  
to 430 to 534mV with temp changes and in all cases the transistor is off.  
That's what I wanted to know. Thanks.







Re: Bipolar transistor bias question
On Monday, December 19, 2016 at 9:41:11 PM UTC-5, Bill Bowden wrote:
Quoted text here. Click to load it
[snip LTSpice]

  +2V          +2V
  -+-          -+-
   |            |
  .-. R1       .-. RL
  | | 5.6k     | | 2k
  '-'          '-'
   |            |
   |  R2      |/
   +--15k--+--| Q1
   |       |  |>.
D1 V   C1 ---   |
  ---  2uF---  ===
   |       |
  ===      ^
          Vin
Six parts.


This has almost zero offset, is drift-compensated,
and uses two fewer parts:
        +2V          +2V
        -+-          -+-
         |            |
        .-. R1       .-. RL
        | | 5.6k     | | 2k
        '-'          '-'
         |            |
         |          |/
         +----------|    Q1
         |          |>. 2n3904
       |<             |
 Vin>--|   Q2        ===
       |\ 2n3906
         |
        ===



Version 4
SHEET 1 880 680
WIRE 208 0 32 0
WIRE 336 0 208 0
WIRE 208 16 208 0
WIRE 32 32 32 0
WIRE 336 32 336 0
WIRE 336 128 336 112
WIRE 32 144 32 112
WIRE 32 144 0 144
WIRE 144 144 32 144
WIRE 0 176 0 144
WIRE -64 224 -112 224
WIRE 208 224 208 192
WIRE -112 240 -112 224
WIRE 0 304 0 272
WIRE -112 336 -112 320
FLAG 336 128 0
FLAG 208 224 0
FLAG -112 336 0
FLAG 0 304 0
SYMBOL npn 144 96 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL res 192 0 R0
WINDOW 0 -42 44 Left 2
WINDOW 3 -37 73 Left 2
SYMATTR InstName R1
SYMATTR Value 2k
SYMBOL voltage 336 16 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 2
SYMBOL res 48 128 R180
WINDOW 0 46 90 Left 2
WINDOW 3 39 61 Left 2
SYMATTR InstName R3
SYMATTR Value 5.1k
SYMBOL voltage -112 224 R0
WINDOW 0 36 87 Left 2
WINDOW 3 34 115 Left 2
WINDOW 123 0 0 Left 2
WINDOW 39 37 140 Left 2
SYMATTR InstName V2
SYMATTR Value SINE(0 .3 20 0 0 0 15)
SYMATTR SpiceLine Rser=1k
SYMBOL pnp -64 272 M180
SYMATTR InstName Q2
SYMATTR Value 2N3906
TEXT -32 400 Left 2 !.tran 1.5

Cheers,
James Arthur

Re: Bipolar transistor bias question
On Mon, 19 Dec 2016 21:15:45 -0800 (PST), snipped-for-privacy@yahoo.com
wrote:

Quoted text here. Click to load it

Remove D1 and R2: four parts.


--  

John Larkin         Highland Technology, Inc

lunatic fringe electronics  


Re: Bipolar transistor bias question
On Tuesday, December 20, 2016 at 12:22:22 AM UTC-5, John Larkin wrote:
Quoted text here. Click to load it


That might be better for his application -- C1 could store a
negative voltage to keep Q1 fully 'off' most of the time.

Bill showed me a gadget earlier this month. I assume this question
is related.

I think he wants to produce 5-to-10% duty pulses at about 20Hz, and
save battery the rest of the time.  A darlington might be worth it
to save the R1 idle-current.

Cheers,
James Arthur

Re: Bipolar transistor bias question
On 12/20/2016 12:58 AM, snipped-for-privacy@yahoo.com wrote:
Quoted text here. Click to load it

How exactly will that happen?  R1 and the BE junction will set the DC  
voltage at the base.  The capacitor can't influence that other than the  
AC signal that is passed.  The DC set point will remain the same, no?

I don't think any of these circuits do what the OP wanted, or did he  
change that?  This thread has gotten pretty long.  He wanted something  
that would detect when the AC input got above 300 mV peak, no?  The  
above schematic cuts off when the input goes negative and is on the rest  
of the time.  Removing the diode and R2 just turn it into an AC coupled  
small signal amplifier.

--  

Rick C

Re: Bipolar transistor bias question
On Tuesday, December 20, 2016 at 4:23:52 PM UTC-5, rickman wrote:
Quoted text here. Click to load it

C1 charges every time you use it to drive Q1(b). When the drive drops back
to zero, the C1-Q1(b) node drops to (Vbe - drive voltage).

Quoted text here. Click to load it

Cheers,
James Arthur

Re: Bipolar transistor bias question
On 12/21/2016 10:00 AM, snipped-for-privacy@yahoo.com wrote:
Quoted text here. Click to load it

Actually, I don't think that is correct - unless we are saying the same  
thing.  I'm not sure the electrical definition of "use it".

The base node will be driven above and below the DC set point no matter  
the DC content of the input signal.  The only effect of the capacitor is  
blocking the DC content of the input signal.  So what ever the set point  
is (as determined by R1 and the BE characteristics), if the input signal  
is applied and the voltage on the capacitor is allowed to settle, the  
bias on the base is going to be determined by that set point.  Q1 will  
*not* be off most of the time.  Just the opposite.  It is on at the set  
point and will only be turned off by significantly negative portions of  
the AC component on the input.

--  

Rick C

Re: Bipolar transistor bias question
On Wednesday, December 21, 2016 at 10:36:38 AM UTC-5, rickman wrote:
Quoted text here. Click to load it

The capacitor is charged rapidly when driving Q1, and discharges slowly,
creating a d.c. offset across C1 that depends on the time constants.

If you don't understand it, LTSpice might clear things up.

Cheers,
James Arthur

Re: Bipolar transistor bias question
On Wed, 21 Dec 2016 11:32:30 -0800 (PST), snipped-for-privacy@yahoo.com
wrote:

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[snip]

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My main advantage in engineering is due to education and design BEFORE
simulators.  Most of the respondents to Bowden's post really have no
"feel" for how circuits work... totally missing the forward conduction
of the B-E junction.
        
                                        ...Jim Thompson
--  
| James E.Thompson                                 |    mens     |
| Analog Innovations                               |     et      |
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Re: Bipolar transistor bias question
On 12/21/2016 2:32 PM, snipped-for-privacy@yahoo.com wrote:
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This makes assumptions about the waveform I believe.  What happens if  
the input waveform is only positive 1% of the time?

--  

Rick C

Re: Bipolar transistor bias question
On Wednesday, December 21, 2016 at 2:58:50 PM UTC-5, rickman wrote:
Quoted text here. Click to load it

LTSpice is your friend.

Cheers,
James Arthur

Re: Bipolar transistor bias question
On Wed, 21 Dec 2016 07:00:51 -0800 (PST), snipped-for-privacy@yahoo.com
wrote:

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Problem is that a coupling capacitor will get charged by base current
on the positive stroke, but then will discharge slowly via the bias,
creating a threshold movement.

So a simple transistor stage can't maintain a constant threshold.
        
                                        ...Jim Thompson
--  
| James E.Thompson                                 |    mens     |
| Analog Innovations                               |     et      |
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Re: Bipolar transistor bias question
On Wednesday, 21 December 2016 15:59:24 UTC, Jim Thompson  wrote:
Quoted text here. Click to load it

Either you guys are missing soemthing or I am. This looks like a 0.3v threshold detector:

.        _________+
.       |     |
.       R     R
.       |    _|
.  --R--+---(_)
.             |
.             |
.  ___________| 0v


Keep P_diss right down for Vbe stability.


NT

Re: Bipolar transistor bias question
On Wed, 21 Dec 2016 11:07:30 -0800 (PST), snipped-for-privacy@gmail.com wrote:

Quoted text here. Click to load it

Maybe I'm missing something, but suppose there are amplitudes higher
than 300mV... is the threshold supposed to stay put?
        
                                        ...Jim Thompson
--  
| James E.Thompson                                 |    mens     |
| Analog Innovations                               |     et      |
We've slightly trimmed the long signature. Click to see the full one.
Re: Bipolar transistor bias question
On Wednesday, December 21, 2016 at 2:30:05 PM UTC-5, Jim Thompson wrote:
Quoted text here. Click to load it
  
Quoted text here. Click to load it
         
We're all left guessing here, but I think the answer is that Bill has
a 300mV or greater a.c. signal, and wants that to drive a switch.

I forgot a resistor--R3--when drawing his original LTSpice circuit:

  +2V              +2V
  -+-              -+-
   |                |
  .-. R1           .-. RL
  | | 5.6k         | | 2k
  '-'              '-'
   |                |
   |  R2           |/
   +--15k--+----+--| Q1
   |       |    |  |>.
D1 V   C1 ---  .-.   |
  --- 2uF ---  | |  ===
   |       |   '-' R3
  ===      |    |  51k
           |   ===
           ^
          Vin


If D1 were schottky, the result is a quasi-temperature stable
~.3V threshold, R3 not needed.

  +2V             +2V
  -+-           -+-
   |             |
  .-. R1        .-. RL
  | | 10k       | | 2k
  '-'           '-'
   |             |
   |  R2        |/
   +--15k--+----| Q1
   |       |    |>.
D1 V   C1 ---     |
  ,-' 2uF ---    ===
   |       |
  ===      |
           ^
          Vin

Cheers,
James Arthur

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