Simple (I think) transistor question

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hondgm 14 years ago

I'm using part of a circuit that looks like this:

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R3 is a thermistor that is intended to be self heated so as to detect moving air. Accuracy isn't important (it's for an electronic candle). The output from the collector of Q1 is going to the A/D on a

5V microcontroller. I have Q1 in there to keep >5V away from the micro, and was hoping to also amplify the voltage change at the junction of R3/R4. But it's not. For example, a 1.15V change at R3/ R4 results in about a 0.7V change at the output.

I can use op-amps and make a microcontroller play music. But I know jack about transistors other than using them as on/off drivers. I could use an op-amp here, but it seems to be overkill.

Any clever ideas to make this better, specifically to at least cause the output voltage change to equal or exceed the voltage change at R3/ R4? I'd prefer to not add more transistors.

BTW I tried simulating this on LTspice, as it even had a model for a

2n4401. Not surprisingly, the results are far off from reality.

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hondgm 14 years ago

One more thing I forgot. I did try different R2 and R1 values to try and maximize the output deflection. R1 should actually be about 1.5K and R2 330K or 390K.

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Jamie 14 years ago

this:

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remove the 12 volt reg, use 5 volts for every thing.

Lower the value of R2 down to around 10k or so.

The idea is to get the voltage at R3 and R4 close or slightly higher than 0.8volts or more, this is where the transistor starts to turn on.

And by selecting R2 with a low value, a slight change at R3,R4 will make a big change in the Q1

You will need to adjust your values from there.

To do it correctly with that circuit, you should have a pot for the zero set and one for the gain set.. Pot being the R4 and gain the R2. But, if you're not picky, I guess anything close will do.

Jamie

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Richard Henry 14 years ago

Since this is feeding a micro A/D input, I would just set up a voltage divider between the thermistor and a fixed resistor the micro's supply voltage, with the divider junction connected to A/D input. Protection (series resistor, shuntcapacitor, diodes, etc) can be added as you think necessary.

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Jan Panteltje 14 years ago

On a sunny day (Tue, 13 Sep 2011 18:19:05 -0700 (PDT)) it happened snipped-for-privacy@yahoo.com wrote in :

In that case you actually only want AC gain:

+5 | R3 |-------------> ADC +5 - R2- | | | | R1 + | |/ |---------------|||-------| NPN si thermistor C1 |\/ | | /// ///

C1 should then be a few uF (say 100 uF) electrolytic. R3 should be that for this transistor, without any signal, Vce is about 2.5 V. If R3 is 1k, and beta 250, then the voltage across R2 is about 2.5 - 0.7 = 1.8V. You need 2.5 mA to drop 2,5 V over R2, so Ib is 2.5 / 250 = 10 uA. This makes R2 1.8 / 10.10^-6 = 180 k (for a beta of 250). C1 sets the lowest frequency, in combination with R1 parallel with the thermistor and the input impedance seen at the base. Beta depends on temperature too... I think this gives you maximum AC gain with full output swing with one transistor.

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hondgm 14 years ago

.5 V.

=3D 1.8V.

ermistor

ansistor.

Jan I think this is worth trying. Jamie I think you missed the part about the heated thermistor. That's why I have a 12v supply. The thermistor has to get noticeably warm. Richard I'm not quite following you. You may have also overlooked the heated thermistor requirement.

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John Fields 14 years ago

Try this: (View with a fixed-pitch font) . +12 +5 . | RT1 | R3 . [RTH] [1k] . | | . | +---> Vout . | | . |R1 R2 C Q1 . [500]

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meg 14 years ago

Save yourself a whole bunch of parts, power the thermistor off the same supply voltage as the a/d and have the a/d sense the voltage directly.

If you=92re hell bent on screwing with the tranny than research single transistor amps. Sometimes all the fun is in the journey.

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John Fields 14 years ago

this:

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Did you miss the part where he's using the 12V to self-heat the thermistor?

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meg 14 years ago

no i didn't. read what i said.

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John Fields 14 years ago

this:

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I did, but it doesn't make any sense to me. Can you post a schematic to show what you mean?

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hondgm 14 years ago

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I'm not understanding either. If I power the thermistor from 5V, it will not dissipate enough power. That's why I have to use 12V. Sure I can get a different thermistor, but that's more hassle that it's worth for one or two units. And I can't power the A/D from 12V if that's what you meant, it's in a 5V PIC micro.

By the way John, you are awesome. I tried the pot, adjusted for 500mV and it's very responsive now! I need to go back to my undergrad electronics texts I guess.

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John Fields 14 years ago

this:

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I'm glad it worked out, and thanks for the compliment. :-) BTW, just for fun, here's an LTspice circuit list if you want to fool around with the circuit: Version 4 SHEET 1 880 680 WIRE 304 -64 -256 -64 WIRE 0 -32 -144 -32 WIRE 0 0 0 -32 WIRE 304 0 304 -64 WIRE 0 112 0 80 WIRE 304 128 304 80 WIRE 336 128 304 128 WIRE 304 208 304 128 WIRE 0 256 0 192 WIRE 112 256 0 256 WIRE 240 256 192 256 WIRE -256 304 -256 -64 WIRE -144 304 -144 -32 WIRE 0 304 0 256 WIRE -256 416 -256 384 WIRE -144 416 -144 384 WIRE -144 416 -256 416 WIRE 0 416 0 384 WIRE 0 416 -144 416 WIRE 304 416 304 304 WIRE 304 416 0 416 WIRE -256 464 -256 416 FLAG -256 464 0 FLAG 336 128 Vout SYMBOL npn 240 208 R0 SYMATTR InstName Q1 SYMATTR Value 2N4401 SYMBOL res 208 240 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 10k SYMBOL res 288 -16 R0 SYMATTR InstName R2 SYMATTR Value 1000 SYMBOL res -16 -16 R0 SYMATTR InstName R3 SYMATTR Value {R1} SYMBOL res -16 288 R0 SYMATTR InstName R4 SYMATTR Value 90 SYMBOL voltage -144 288 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value 12 SYMBOL voltage -256 288 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V2 SYMATTR Value 5 SYMBOL res -16 96 R0 SYMATTR InstName R5 SYMATTR Value 410 TEXT 16 448 Left 0 !.step param R1 584 750 TEXT 16 472 Left 0 !.op I assumed the thermistor changed from 750 ohms at 25C to 584 ohms at 50C, and if you run the sim and probe the collector of Q1 you'll see the change in thermistor resistance (X) causing a change in collector voltage (Y). You can change the ratio of R5 to R4 (carefully) ;) while keeping the sum at 500 ohms (which is what you do when you adjust the pot) and change the slope of the output.

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Tim Wescott 14 years ago

Transistor circuits can get amazingly complicated amazingly fast, sometimes. When DC parameters mean something, an OP-amp is often the way to go.

www.wescottdesign.com

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meg 14 years ago

ect

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R3/

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I probably should have been explicit

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meg 14 years ago

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I should have been more explicit, I haven=92t looked for a while, but digikey use to have a great selection of thermistors.

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John S 14 years ago

Good work, John.

Add this to your LTSpice schematic: .step temp list 10 25 40

It will not include the thermistor. What do you think?

John S

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John Fields 14 years ago

this:

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And that justifies your disrespectful: "no i didn't. read what i said." indefensible cocky answer to: "Did you miss the part where he's using the 12V to self-heat the thermistor?"?

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John S 14 years ago

On second thought, add this: .step temp list 25 50 And, change R4 to 75 ohms.

Looks great! Again, good work.

John S

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John S 14 years ago

Oops! got the temps backwards: .step temp list 50 25

Sorry.

John S

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Simple (I think) transistor question

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