Bipolar transistor bias question

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?

Reply to
billbowden
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a circuit diagram.

NT

Reply to
tabbypurr

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
Reply to
John Larkin

Why not use a Schottky diode with it's lower voltage drop eliminating the need for the divider resistors?

I think the schematic would be like this. Size the resistor to give the voltage drop required when the input is at 0.3 volts.

Vcc ___ | \ / +---- Out \ | | |/ -->>---S

Reply to
rickman

Fundamentals. ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| STV, Queen Creek, AZ 85142    Skype: skypeanalog |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     |
Reply to
Jim Thompson

You may be pushing what can be done (reliably) with a single transistor. When you provide us with more information (as others have suggested) be sure to include the collector circuit - or at least a more solid definition of what "on" and "off" are.

Reply to
Frank Miles

The fact that you should just build a friggin drive circuit for it.

You use a diode to up a 300 mV drive to positive 700 mV to turn the transis tor on, first of all it might not be enough to saturate it. And then when i t drops off it is probably not going to turn it off very fast.

If you are dealing with an inductive load this is poison. The Vce rises so fast it results in so much power dissipation it is almost not worth using a switcher.

In an SMPS with a bipolar, usually there is some drive enhancements, some o f which drive the base reverse by a few volts to get the thing to turn off really fast, if the load is highly inductive, but you did not say it is. Ho wever a switcher is hard to use without an inductor, at least effectively w hen it comes to efficiency.

So just take your 300 mV and amplify it, that'll work. One simple little tr ansistor and a few resistors should do it, if it is some sort of self oscil lating nightmare, then it takes two transistors.

Are there space constraints ? If it is cost tell the boss "You want to issu e refunds to ALL your customers ?". Another workaround in some applications is just to use a transistor that costs ten times as much, and they won't l ike that either.

Any solutions, they will depend on the scale. How much voltage and current ?

Reply to
jurb6006

You are not providing enough detail. What kind of current is involved, what is the impedance of the source providing the 300mV, etc.

Transistors are switched mainly by current, not voltage. Your source will need to provide the driving current of the transistor, and it cannot be told from "it is 300mV" whether it will be able to do that.

Reply to
Rob

(Pardon my snarky response last night... I fell asleep watching a movie on TV and didn't read your post carefully :-) ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| STV, Queen Creek, AZ 85142    Skype: skypeanalog |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     |
Reply to
Jim Thompson

If the input can be AC coupled, this might work:

+5 whatever | | | | | | R | | | | c in------C---------+---------b e | gnd

It's automagically self-biasing.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

When you say the AC input is 300mV is that pk-pk or rms? If the latter then the 840mV pk-pk may be enough without pre-biasing. Capacitor couple the input and use diode anti-parallel to base-emitter to serve as voltage doubler.

piglet

Reply to
piglet

If you want to really square it up....

...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| STV, Queen Creek, AZ 85142    Skype: skypeanalog |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     |
Reply to
Jim Thompson

and a voltage divider of 2

if you want the transistor to switch there is no reason to REDUCE the input level swing keep the swing as large as possible and figure out how to shift that level as needed. m

Reply to
makolber

For an ideal transistor, Ib (and Ic) increase a decade per 60 mV of base voltage. So 300 mv p-p can swing Ic over 5 decades of current.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin

Easiest way to bias it, is with a second matched transistor, in current-mirror fashion; add in base resistors and a capacitive coupling to the active transistor's base.

The thing still isn't an ideal switch, though. For one thing, the base current into the active transistor can do some bias damage, by charging up the coupling capacitor during saturation.

Reply to
whit3rd

Didn't I see something like that in:

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Reply to
bitrex

"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

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

Reply to
billbowden

(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.

Reply to
John S

[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

Reply to
dagmargoodboat

Remove D1 and R2: four parts.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

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