Re: Driving two transistors with same signal
- From: "BobW" <nimby_NEEDSPAM@xxxxxxxxxxxxxx>
- Date: Fri, 9 May 2008 20:40:27 -0700
"vic" <news@xxxxxxxxxxxxx> wrote in message
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BobW wrote:
"BobW" <nimby_NEEDSPAM@xxxxxxxxxxxxxx> wrote in message
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"vic" <news@xxxxxxxxxxxxx> wrote in message
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Hi,Try this:
I have two common anode 7-segments displays, and only one wire to drive
them. I need to achieve the following : when the control signal is +5V,
display1 is ON and display2 is OFF. When the signal is 0V, display1 is
OFF and display2 is ON. When the signal is not connected (high
impedance), both displays are OFF.
I tried using a NPN transistor for display1 and a PNP for display2,
connecting their bases together. It works when the driving signal is
present, but when the signal is floating current flows from the base of
the PNP to the base of the NPN and both transistors turn each other on,
resulting in both displays being ON.
The circuit that didn't work :
VCC
+
|
|
___ |<
o---------------------|___|--|
| |\
| VCC |
| + |
| | Display2
| | |
| ___ |/ |
Input---o---|___|--| GND
|>
|
|
Display1
|
|
GND
Is there a way to achieve this ?
Thanks.
5V
| 5V
R5 |
| |<---
|/-------------| Q3
----| Q1 |\--------- (to display 1 and then to GND)
| |>-- 5V
| | |
| | R3
| | |
---R1-----R2----------
| | |
| | R4
| |<--- |
|--| Q2 GND
|\
| |/--------- (to display 2 and then to 5V)
-------------| Q4
| |>----
R6 |
| GND
GND
Q1 and Q2 form a comparator so that when the input is floating they will
both be off. R3 and R4 set the input threshold.
When the input is high (5V) then Q1 will be on and that will turn on Q3.
When the input is low (GND) then Q2 will be on and that will turn on Q4.
You can figure out the resistor values. They shouldn't be too critical,
but R3 and R4 need to be small enough to ensure enough drive for the
four transistors.
I hope I got the ascii art right as I had to compose it in a separate
word processor.
Bob
--
== NOTE: I automatically delete all Google Group posts due to
uncontrolled SPAM ==
I see, now, that your displays are both common anode. You'll need to add
another PNP (common emitter mode) driven by Q4.
As Monica Lewinsky used to say, "Close, but no cigar."
Bob
*gasp* 5 transistors needed to do what seemed simple at first glance ...
I don't quite understand what R5 and R6 are for, when Q1 and Q2 do not
conduct, the base current of Q3 and Q4 would be zero so the ressitors do
not seem necessary ?
Well I guess I could just try it and see if it works :)
There is always some leakage from collector to base, so R5 is there to
insure that Q3 doesn't conduct when Q1 is off. Same goes for R6/Q4.
Yeah, it's a lot of parts, but your requirements are kinda tough.
If you're only building one of these, and your supply is tightly regulated,
and your temperature range is limited, then the method (I forget who
proposed it) that keeps Vbe at about 0.4V when the drive is high impedance
may work okay.
If it were my project and the thing had to ALWAYS work (in production
quantities, over a varying range of Vcc, temperature, and Voh/Vol), then I'd
use this scheme.
If you have trouble with the resistor values then give a yell. R3 and R4
should be equal if your drive is from a CMOS output (i.e. the output swings
from supply to supply). If it's from an old TTL output, then the voltage at
the R3/R4 connection should be set to about ((2.4V-0.4V)/2)+0.4V = 1.4V
(i.e. the middle of the output swing range for worst-case TTL).
Have fun with it.
Bob
--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==
.
- References:
- Driving two transistors with same signal
- From: vic
- Re: Driving two transistors with same signal
- From: BobW
- Re: Driving two transistors with same signal
- From: BobW
- Re: Driving two transistors with same signal
- From: vic
- Driving two transistors with same signal
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