Do I need a voltage doubler for this?

metron9

I want to trigger a pin on the BS2 from
a source that measures about 0.6 volts DC
the voltge is 16 milliseconds in in total length
and has a square wave that alternates 32 times
on and off in that length of time.

I know this because of my handy occilloscope I just got last Friday.

First, from the information above, what is a better way to describe the voltage I am dealing with here.
Second, would I need to amplify the signal (i.e. voltage)
What is the minimum time a pin needs to go from low to high.
I found this voltage doubler circuit[noparse]:http:[/noparse]//www.reconnsworld.com/power_voltdoubler.html

Beau Schwabe

metron9,

With 0.6V, you should be able to "direct drive" the base on a transistor.

If you are sure that the voltage is only 0.6V, try something like this:

           2n3904            4.7K 
GND >--------EBC--------o----/\/\---->+5V 
              |         |
0.6V >--------o         o------------>To Scope

Note:
The signal will be inverted.



BTW)

You have 16mS with 16 periods (1 period = 1 rise and 1 fall) that’s 1mS per period or a frequency 1KHz at 0.6V

"Time per Period" = "Time in Seconds" / "Number of Periods"

"Frequency in Hertz" = "Number of Periods" / "Time in Seconds"

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Beau Schwabe

IC Layout Engineer
Parallax, Inc.

Post Edited (Beau Schwabe (Parallax)) : 8/29/2005 3:50:02 AM GMT

metron9

I have Radio Shack part 276-1617 and on the transistor is: NPN-Type Switching Transistor

F912
MPS
A13

I am hooking up as you say, I am getting this: Picture below

I guess it's not the right part, I thought at first thought it was a signal but when I turned the pulse off and even unplugged the wire connecting it the scope stayed the same or did funnny stuff as I moved the probe wire.

I put my finger on the +5 and touched the BASE indeed the scope measured the voltage. I assume the transistor I have here must be raited for higher current but I dont know how to read the info on the back.
Typical hfe = 200 ?
Max Rating 30 V (got that)
Ic 800 ma (it can handle power for 40 LEDs at 20ma each for example)?
Power dissipation 1.8W (thought I knew this one but gotta re read the book on watts again something to do with heat)

I understand a small current can be applied to the base to allow a larger current to flow from the collector to the emmitter thus pulling the pin low (connecting the scope to a ttl pin)

I am guessing there is not enough current for this transistor to do its job.

Anything above I stated wrong please correct me.

Thanks Beau. I will look at RS Monday to see if they have the 2n3904

Others I have are
2n3906


OOPS found one, 2n394 in the pile however,
I do get a inverted signal only it's strength is about the same and the signal is dirty not a clean line.
What's the next step, and is the assumption the current is not enough on the base?

Beau Schwabe

Sorry about that....

Try biasing the transistor input like this by setting up a voltage divider.

            2n3904            
 GND >--------EBC--------o------------> Output (To Scope)
               |         |    4.7K
               |         o----/\/\--o
               |                    |
          10K  |              100K  |
Input >--/\/\--o--------------/\/\--o-->+5V


Input:    Output:
0V        4.97V
0.6V      72mV or 0.072V

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Beau Schwabe

IC Layout Engineer
Parallax, Inc.

metron9

You da man Beau. Here is the output, I cranked up the freq on the input so my camera shutter could pick up the image I am at 2ms time sweep and 1 volt resolution. Looks like about 1.7 volt strong signal.

I bet if I want a signal closer to +5 I could use two 2n3904s by feeding the output to the second transistors base and re-invert the signal at the same time?

Now lets get into the calculations and specs on that transistor so I understand and in the future I can select the proper transistor for the job at hand.

Is there a web site you use for specs on transistors or do you just google the part number? Or you have an electronics book that has lists of specs. I see on some transistors the EBC is reversed from the small flat area on the component.

I use findchips.com for chips so I am wondering is there is a similar web page for other components.

Beau Schwabe

metron9,

Did you move the horizontal position to "fit" the 1 volt resolution on the scope?
·
I will assume you did, and that the top trace is about 5V and the bottom trace is·about 3.3V· (5V - 1.7V)
·
That said, what you are looking at is the voltage across the C-E transistor junction, which means that the
voltage across the 4.7K resistor is 1.7V ... current across the 4.7K resistor is about 362uA. (1.7V / 4.7K)
·
The hfe or "DC gain" can vary·between transistors and degrades sharply depending on the amount of current.
So in this case it might be better to calculate the hfe based on real measurements.
·
We know that the voltage drop across the B-E junction will be about 0.6V, so from 5V we can apply 4.4V across
the 100K resistor to come up with about 44uA··(4.4V / 100K)
·
So now we know current going in (44uA) and current going out (362uA).· From here we can calculate the hfe to
be about 8.2 (362uA / 44uA)
·
Why is this important?· Well from here we can calculate the amount of current necessary at the transistors base
to bring the voltage down from 3.3V to something closer to 0V. (in actuallity 0.6V)
·
Assume we keep the 4.7K resistor as a pull-up resistor, this time we want to drop close to 5V across it.· I say
close, because we are limited to 4.4V because of the 0.6V drop across the C-E transistor junction.· So 4.4V across
a 4.7K resistor is about 936uA.
·
Knowing now what the hfe is of 8.2 we can determine the amount of Base current necessary by dividing 936uA by 8.2
to get 114uA.
·
Now we need to determine the resistor value to limit the Base current to 114uA.· Since the voltage will be constant (4.4V),
the resistor value·should be about 38.6K ... The closest standard value to this is 39K so we will use that.
·
Finally, we need to determine an appropriate resistor value that will turn off the transistor.· This is the voltage applied
across the B-E junction·( I usually use 450mV to start with).· Remember that we already have a current established across
here from the 39K resistor of about 113uA, so to determine a resistor value, simply divide 450mV by 113uA to get a
resistor value of about 4K ... The closest standard value here is·3.9K so we will use that.
·
The final circuit should look like this...·

            2n3904            
 GND >--------EBC--------o------------> Output (To Scope)
               |         |    4.7K
               |         o----/\/\--o
               |                    |
         3.9K  |              39K   |
Input >--/\/\--o--------------/\/\--o-->+5V

As far as looking up datasheets...

Google:
2n3904 datasheet













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Beau Schwabe

IC Layout Engineer
Parallax, Inc.

Post Edited (Beau Schwabe (Parallax)) : 8/29/2005 7:07:59 PM GMT

metron9

Right on the money a good signal over 4 volts. (No picture)

I intended to go over the math when I got home but couldnt stop playing around with the stuff

so I popped in a PNP transistor, guessing at resistors as it was getting late again I was able to flash an LED thru the PNP using the output of your circuit. The pictures show the slow drop back to 0 volts when the LED and Bias resistor not connected (on my PNP circuit) when I saw that slow drop back to 0 I thought connecting the base thru a 56k resistor would allow it to pull down faster and it did.

I will try and go over the math and post a PNP circuit that drives an LED like this with how I came up with the values but there is only so much time in the day.

That way I can be sure I understand the math that goes into it.

This is like having a private tutor and you don't know how much I appreciate your time, thank you very much and I hope other newcommers to electronics are able to learn from this exchange.