Driving two transistors with same signal
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Posted by vic on May 8, 2008, 1:11 pm
Please log in for more thread options 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 |>
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| Display1 | | GND Is there a way to achieve this ? Thanks. | |||||||||||||||||||||||||
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Posted by ehsjr on May 8, 2008, 2:22 pm
Please log in for more thread options > Hi,
> > 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. Add diodes to isolate the bases, and resistors to bias the transistors off when the desired on signal is not present. Ed see below VCC + | +-------------+ | | [R] | | ___ |< o--------|<------+----|___|--| | |\ | VCC | | + | | | Display2 | | | | ___ |/ | Input---o--->|-+-|___|--| GND
[R] |
| |> | | | Display1 | | +----------+ | GND | |||||||||||||||||||||||||
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Posted by Dan Coby on May 8, 2008, 6:21 pm
Please log in for more thread options > vic wrote:
>> Hi,
I need to achieve
>> >> I have two common anode 7-segments displays, and only one wire to drive them. >> 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.
connecting their bases
>> >> I tried using a NPN transistor for display1 and a PNP for display2, >> 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. >
> Add diodes to isolate the bases, and resistors > to bias the transistors off when the desired > on signal is not present. > > Ed > > see below > > > > VCC > + > | > +-------------+ > | | > [R] | > | ___ |< > o--------|<------+----|___|--| > | |\ > | VCC | > | + | > | | Display2 > | | | > | ___ |/ | > Input---o--->|-+-|___|--| GND > | |> > [R] | > | | > | Display1 > | | > +----------+ > | > GND The revised circuit does not solve the problem. Both diodes will be conducting when the input is floating. As a result both displays will still be on. A simpler solution is to remove the diodes, change the locations of the added resistors, and move Display1. I.e.: VCC + | +---------+ | | [R1] | | |< o-------------[R2]---+-------| | |\ | VCC | | + | | | Display2 | Display1 | | | | | |/ | Input---o-----[R2]-+--------| GND | |>
[R1] |
| | +----------+ | GND The R1 / R2 resistor pairs need to be chosen so that there is only about 0.5 volts across the transistors' base-emitter junctions when the input is floating. When the input is floating, the various resistors will pull the input to VCC/2. The value of 0.5 volts was chosen to be low enough to keep the transistors from turning on when the input is floating but still allow the transistors to be turned on when the input is being driven to VCC or ground. | |||||||||||||||||||||||||
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Posted by Rich Grise on May 8, 2008, 7:49 pm
Please log in for more thread options On Thu, 08 May 2008 15:21:04 -0700, Dan Coby wrote:
> A simpler solution is to remove the diodes, change the
> locations of the added resistors, and move Display1. I.e.: > > VCC > + > | > +---------+ > | | > [R1] | > | |< > o-------------[R2]---+-------| > | |\ > | VCC | > | + | > | | Display2 > | Display1 | > | | | > | |/ | > Input---o-----[R2]-+--------| GND > | |> > [R1] | > | | > +----------+ > | > GND > > The R1 / R2 resistor pairs need to be chosen so that > there is only about 0.5 volts across the transistors' > base-emitter junctions when the input is floating. > When the input is floating, the various resistors > will pull the input to VCC/2. The value of 0.5 volts was > chosen to be low enough to keep the transistors from > turning on when the input is floating but still allow the > transistors to be turned on when the input is being driven > to VCC or ground. This looks a little iffy - when there's no drive, then both bases will still be forward biased unless you divide the drive down so dramatically that you don't^H^H^H^H^Hmight not get full saturation when it's supposed to be on. A much more elegant solution uses two NPN's (the same thing could be done with two PMP's, just turn the arrows around and the power supply upside down. ;-) ) Vcc Vcc | | | [Display 2] | | [Display 1] | | |/ +--[R]---+---| | | |>
|/ [R] |
Control --[R]--+--| | | | |> | |
[R] | | |
| | | | gnd gnd gnd gnd The resistors can be anything from around 4.7K to maybe 22K. Cheers! Rich | |||||||||||||||||||||||||
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Posted by petrus bitbyter on May 8, 2008, 8:13 pm
Please log in for more thread options
>> vic wrote:
>>> Hi,
>>> >>> 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. >>
>> Add diodes to isolate the bases, and resistors >> to bias the transistors off when the desired >> on signal is not present. >> >> Ed >> >> see below >> >> >> >> VCC >> + >> | >> +-------------+ >> | | >> [R] | >> | ___ |< >> o--------|<------+----|___|--| >> | |\ >> | VCC | >> | + | >> | | Display2 >> | | | >> | ___ |/ | >> Input---o--->|-+-|___|--| GND >> | |> >> [R] | >> | | >> | Display1 >> | | >> +----------+ >> | >> GND >
> The revised circuit does not solve the problem. Both diodes > will be conducting when the input is floating. As a result both > displays will still be on. > > A simpler solution is to remove the diodes, change the > locations of the added resistors, and move Display1. I.e.: > > VCC > + > | > +---------+ > | | > [R1] | > | |< > o-------------[R2]---+-------| > | |\ > | VCC | > | + | > | | Display2 > | Display1 | > | | | > | |/ | > Input---o-----[R2]-+--------| GND > | |> > [R1] | > | | > +----------+ > | > GND > > The R1 / R2 resistor pairs need to be chosen so that > there is only about 0.5 volts across the transistors' > base-emitter junctions when the input is floating. > When the input is floating, the various resistors > will pull the input to VCC/2. The value of 0.5 volts was > chosen to be low enough to keep the transistors from > turning on when the input is floating but still allow the > transistors to be turned on when the input is being driven > to VCC or ground. > This may work, (so there are some ifs.) Most important, the driving source must be able to provide the extra current (sink and source). The displays may need GND on one side. The OP did not said so, but his schematic suggests it. You need to do a little bit of serious calculation to find the values of the resistors. The voltages are important as you mentioned already, but the base currents need to be high enough to switch the transistors on when active. I leave it an excercise for the interested reader. petrus bitbyter | |||||||||||||||||||||||||
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