A .tar.bz2 is a BZip2 compressed tar file (tar is just a lump of the files), and indeed, widely used on UNIX/Linux.
I’ve [uploaded a .zip for you, no problem.
I made some custom libraries, so that’s why it’s archived.
Silly question.
I would like to drive the PNP from the 12V source on the board, as follows:
http://img131.imageshack.us/img131/1360/pnpfl8.png
Where VCC is 5V, and the load connects between the two open nets.
Basically all of the PNP transistor datasheets I’ve seen specify a max Vebo (Emitter-Base Voltage) as 5.0Volts (or -5.0Volts, depends on your perspective). Is this a problem (12V - 5V = 7V which is > 5V), or am I thinking this specification is something else?
Silly question I know, I may have known the answer at one point in time, when I first read my dad’s electronics books, but I tend to forget stuff when you don’t use it for a while.
-Nate
reklipz:
Silly question.I would like to drive the PNP from the 12V source on the board, as follows:
…
Where VCC is 5V, and the load connects between the two open nets.
Greetings Nate,
So when the PNP transistor is conducting (on) the load is connected
between Vcc (+5V) and ground?
The output of the PNP would be the collector, the input would
be the Emitter tied to the Vcc rail, and the base of the PNP
would be either Vcc (off) or Vcc - Vbe (on).
Your control signal needs to be lower than Vcc - Vbe (4.3V) to
activate the transistor and could return to the 12V rail. Or Vcc
rail when off, such as a 5V CMOS logic signal or a uC port.
If your signal rests at 12V you’ll need a protection diode on the
PNP base and emitter to reduce the reverse voltage on
the base (as you noted in your question).
What current in the load? What transistor type? (Need to
know Hfe at Ic = load current).
Also, is the input signal 0V to 12V? Or 0V to 5V? Or ??
Comments Welcome!
Hmm, after reading your post, and reviewing my post, I’m not quite sure why I decided to upload that image. I had another one laid out as you described, but the one in the picture is how it would fit into my schematic (Vcc rail is the output, input would be between the break). Sorry about the confusion.
The control signal swings between 12V and 0V. This would mean I would need to drop it by at least 2V in order to meet the spec. I could just use 3 diodes, but that seems like a bad idea for some reason. If it is a bad idea, I wonder if this may be better; The 12V source feeds a switching regulator, which outputs 3.3V and 6.8V. Would using the 6.8V source as the control signal be a bad idea? (I’m wondering about whether or not this is ok because it is a switching regulator, and takes a small amount of time to settle)
Sorry for the delay, and that I can’t seem to get my facts straight before posting, which only provides confusion.
-Nate
reklipz:
The control signal swings between 12V and 0V. This would mean I would need to drop it by at least 2V in order to meet the spec.
Nate,
The spec of 7V is a max rating for the transistor under
reverse bias. It would be unwise to stress the device to
the spec limit.
http://www.stonard.com/SFE/High_side1.jpg
This is how you might switch a 5V load from a 12V control
signal. The transistor conducts when the control signal is
near ground, and when the control signal is above 5V (i.e.
at 12V) the diode protects the transistor. The transistor
inverts the signal. If you want the output to be at 5V when
the control is at 12V, a second invertion is needed.
Comments Welcome!
Thanks for the drawing.
I feel as though I’m just a hassle now, as I always seem to find something negative to come back with, unfortunately.
I tried simulating the design in multisim, and cant get the transistor to ever turn off with the 1K resistor. When I increased the value to 10K (this effectively lowers the base current, allowing the transistor to stay in its conducting state, correct?)
I’ve also got another question for you. When I first viewed the circuit, I thought to myself, would the diode and resistor not raise the 5V rail to 12V? As it turns out, it doesn’t. Is this because the transistor sinks all of the current when the input goes high, thus not affecting the 5V rail?
If you feel that I’m not fully understanding how this works, and wouldn’t mind giving a bit more detail, I would truly appreciate it. I find it a bit, embarrassing, that I can put together that whole schematic / board presented earlier, yet trivial things such as this completely stump me (well, not completely, as I was on the right track, but hit a road block).
Thanks very much for the solution! I think it’s the last thing I need, and it’s off to fabrication (after a full review of course!).
-Nate
I’m a hobbyist, not an engineer, so take this for what it’s worth, but to switch on a 5v rail from a 12v signal, I’d just use a pair of mosfets - small signal N-channel for the signal sense which in turn switches an appropriately rated P-channel device to switch the 5v rail. Most small signal mosfets have a Vgs(max) of 20 volts and so are easily man enough to take a 12 volt signal. (Actually, when I’m switching a load from a signal, I try and arrange it so that the gnd line is the one that needs switching, then all you need is a single N-channel mosfet). I like mosfets - for the things I do they almost always result in a simpler circuit with higher impedance inputs.
Greetings Nate,reklipz:
I tried simulating the design in multisim, and cant get the transistor to ever turn off with the 1K resistor.
There are several different ways to do what you want,
Winston suggested a P-Ch FET. My circuit can be analyzed
intuitively, or with pSpice if you’d rather have a full simulation.
It works!
http://www.stonard.com/SFE/High_side1_sim1.jpg
http://www.stonard.com/SFE/High_side1_sim_results1.jpg
reklipz:
I’ve also got another question for you. When I first viewed the circuit, I thought to myself, would the diode and resistor not raise the 5V rail to 12V? As it turns out, it doesn’t. Is this because the transistor sinks all of the current when the input goes high, thus not affecting the 5V rail?
Remember that current takes the path of least resistance,
so the current from the 12V signal defined by R1 does flow
into the 5V rail via the diode and R2. The 5V rail is very
low impedance, say one ohm, so the 5V rail will be raised
by 1/3900 or 0.025% - it would be hard to measure with a
good DMM. Under these conditions the transistor is off, the
base-emitter is reverse-biased, and only device leakage
current flows into the base.
reklipz:
When I increased the value to 10K (this effectively lowers the base current, allowing the transistor to stay in its conducting state, correct?)
No. The value of R3 shunts the current from emitter to
base. In a PNP device the base current flows out of the
base when the device is conducting. Increasing the value
or R3 increases the current in the base, bacause less is
shunted around the device. The function of R3 is to remove
the base charge when the input goes positive (to 12V in this
case) so that the transistor switches off quickly, and leakage
current from the transistor doesn’t turn it back on (unlikely
in this case). R3 could be removed, many circuits do not
place a resistor from emitter to base, but doing so costs
one resistor and defines that the circuit will work reliably.
I’m sure there are a lot of people lurking on this thread
and learning from your mistakes. This is an open forum
for discussion, anything here can be questioned, others
will likely have different opinions or ways to approach the
problems presented.
Comments Welcome!
OK, I think I’ve found what I did wrong.
In my simulation, I placed a switch inline with the 12V source, and added a 10K resistor from the base of the transistor to ground. The reason I added a switch instead of just setting the voltage to 0V is because the only pathway back to ground in the actual circuit would be the transformer, the smoothing cap, or the switching regulator. I assume I can count out the switching regulator as a pathway; relying on the capacitor in reverse bias is just a bad idea; and the transformer would be really high impedance, so that pathway is out too, right? I lowered the value of the resistor to 1K, and it seems to work;
Is my reasoning above correct, or can I do away with the resistor?](ImageShack - Best place for all of your image hosting and image sharing needs)
reklipz:
In my simulation, I placed a switch inline with the 12V source, and added a 10K resistor from the base of the transistor to ground.
Greetings Nate,
That’s a fancy simulator…
For the simulation to be accurate the switch should be in
parallel with V1, so that when the switchis open the input
is 12V and when the switch is closed the input is zero volts.
That’s a bit rough on the V1 source, so a better method is
to change the value of V1 and run the simulation twice.
reklipz:
The reason I added a switch instead of just setting the voltage to 0V is because the only pathway back to ground in the actual circuit would be the transformer, the smoothing cap, or the switching regulator.
Actually, the real world path impedance is very low
(only a few ohms).
reklipz:
Is my reasoning above correct, or can I do away with the resistor?
Please run your simulation with R4 removed and V1= 12V,
then again with V1 =0V. That is what I did (scroll up and
check the table data).
Voltage sources should be treated as ‘ideal’ with no series
impedance. We only need the DC stimulus for this circuit,
so capacitors are open and inductors shorted (if any are
used).
Does your simulator spit out numerical data?
Comments Welcome!
I’m sure Multisim (my simulator) does output those goods, but it’s so involved I’ve not yet figured out how to do it.
As for switching the 12V to 0V, it does exactly the same as yours, I was just worried that this would not be the real world result, but it seems I was wrong.