Transistor question? PNP

judoGTI · 07-25-2005, 01:02 PM

I am curious how I can determine the amount of voltage I need to supply a PNP transistor to keep the collector-emitter gate closed?

Is this listed as a spec on the transistor packaging? Is't it mathmatically determined?

I tried Google with no luck, but Im extremely new at this. Thx.

rsd212 · 07-25-2005, 01:36 PM

I could definately answer this better if I had my notes, but honestly I would just play with it. Solving a BJT circuit can be time consuming. Since its a current controlled device you pretty much have to guess what set of equations to use, solve it, check your outcome, then try again until you get something that works.

I recommend putting a large resistor to start on the base. 100K-ish. Crank up the voltage until you see the output you expect. Adjust the resistor from there (potentiometer?) until it trips at the voltage you want.

whiskeytech · 07-25-2005, 10:00 PM

I am probably wrong, but I thought it was constant at .7 volts for silicon (most
common) and like .3-.4 for germanium ones. Wow that class was a long time
ago.

rsd212 · 07-25-2005, 10:10 PM

Any way you can use a FET? They're so much easier

whiskeytech · 07-25-2005, 10:13 PM

oh .7 and all that crap is for base-emitter I think. lemme see if I can find my notes.

judoGTI · 07-26-2005, 09:51 AM

Quote: Originally Posted by whiskeytech

I am probably wrong, but I thought it was constant at .7 volts for silicon (most
common) and like .3-.4 for germanium ones. Wow that class was a long time
ago.

So you think if I go below .7 the gate opens and above .7 the gate closes?

What is a FET? I only know about PNP transistors from what I have read on the internet. Basically teaching myself as I go.

rsd212 · 07-26-2005, 09:56 AM

FET (field-effect transistor) is another type of transistor which tends to be easier to use. Bipolar Junction Transistors (like the PNP you're using) are kinda funny in that your base is controlled by current running through the device. Basically this voltage you apply to control it will leak. In a FET the gate (similar to the BJT's base) is isolated from the rest of the transistor. With a MOSFET or JFET, the only thing that will matter to determine if its 'open' or not is the voltage at the gate.

Also, NPN transistors may be slightly easier to use in your case. In an NPN, 0 volts between the base and the emitter will mean its closed. Then just arbitrarily bias the base (say with 5V and a 1K-10K resistor) and it should open just fine, allowing current to flow through the other two terminals.

zPilott · 07-26-2005, 10:40 AM

I am not an expert in this either, so feel free to correct me if I am wrong, but the .7v is FULLY open, and lower voltage values will let some current through. Lower voltages will let less current through.

rsd212 · 07-26-2005, 12:25 PM

0.7V represents the diode drop across the B-E junction. However, putting a greater than 0.7V source across this junction can cause it to burn out. This is why I recommend a FET, or if you insist on BJT start with a high resistance going to the base and lower it until it switches. Much easier to figure out what you need to know by experiment rather than by trying to solve the bias circuit.

judoGTI · 07-26-2005, 01:59 PM

Yeah after doing some reading it looks like I could use a p-channel JFET for the same thing. It looks like they operate on and off around 1v. I'll play around and see what I can get to work.

rsd212 · 07-26-2005, 02:10 PM

Just remember that the 1V needed to switch it is in reference to the other pins. If your drain pin is held at 5V, that means you'd need 6V at the gate to trigger it. With a FET you should be able to use arbitrarily high voltages at the gate to switch it, just read the gate breakdown voltages on your datasheet and dont exceed them. Most of them are pretty high though, I think the ones I used in lab were around 30V.

User Name		Remember Me?
Password