J
John Fields
- Jan 1, 1970
- 0
When the 7406 is off, can I treat it as a 0.7 Voltage load with 40 mA
running through it to ground?
---
Depends on what you mean by "off".
It might be simpler to look at the 7406 as a single transistor with
its base pulled up to Vcc and nothing connected to its collector.
(View in Courier)
Vcc
|
[10K]
| C
| <--O--+-----B
| E
O S1 |
| |
GND---+---------------+
Normally open switch S1 is also connected to the base and is used to
turn the transistor on and off. In the open position there will be
charge flowing from base to emitter and then to ground, and that
current will turn the transistor on, making the collector-to-emitter
resistance drop to a low value. When the switch is made, the base
current will be diverted to ground, through the switch, and the
transistor will turn off, causing the base-to-emitter resistance to
rise to a very high value. Notice that whether there is current
into the base or not, nothing comes _out_ of the collector.
Now let's change the circuit a little and see what happens.
30V
|
|R2
[750]
5V |
| +--->OUT
[10K] |
| C
| <--O--+-----B NPN
| E
O S1 |
| |
GND---+---------------+
Now, with S1 open the transistor will be on because charge will be
flowing through the base-emitter junction to ground. This will then
allow charge to flow from Vcc through R2 and through the
collector-to-emitter junction of the transistor to ground. If
there's enough current into the base to cause the transistor to be
saturated, then the maximum voltage dropped across the
collector-to-emitter junction will be 0.7 volts with 40 mA of
collector current (7406, remember?). So, since we have 0.7V dropped
across the transistor and we have a 30V collector supply the
collector current must be:
E 30V - 0.7V
I = --- = ------------ = 39mA
R 750R
Now let's close the switch and see what happens:
30V
|
|R2
[750]
5V |
| +--->OUT
[10K] |
| C
|<--O--+-----B NPN
| E
OS1 |
| |
GND----+--------------+
Since the base current is now diverted to ground the transistor will
be turned off and collector current will cease flowing since the
path to ground will no longer exist. When that happens, the voltage
on the collector will rise to the supply voltage, 30 volts, which is
what the 7406 must stand off when it's turned off.
So, to get back to your question, "When the 7406 is off, can I treat
it as a 0.7 Voltage load with 40 mA running through it to ground?",
the answer is "no".
When the 7406 is off you should treat it like a reverse biased diode
in series with your load, and when it's on you should treat it like
a forward biased diode in series with your load.
Now, to address your original post (I may be repeating someone else
since I haven't read through the whole thing, but anyway...
5V
|
[Rs]
|
|A
7406 [LED]
| \ |
MCU I/O>---| >O---+
| /
Assuming that you want to drive a 20 mA LED with a Vf of 2.2V,
Looking at the 7406 data sheet shows that when it's sinking 16mA
worst case Vol will be 0.4V and when it's sinking 40mA worst case
Vol will be 0.7V. If we assume the change is linear, then we have a
24mA change in current causing a 300mV change in voltage, which is
12.5mV/mA.
So, for 20mA, worst case Vol would be 0.45V.
For Rs,
Vcc - (Vled + Vol) 5V - 2.65V
R = -------------------- = ------------- = 117.5 ohms
Iled 0.02A
The closest standard 5% value is 120 ohms, which would be fine, and
the power it would need to dissipate with 20mA through it would be:
P = IE = 0.02A * 2.35V = 0.047 watts.
So since there's a little less than 20 mA through it, a standard 1/4
watt resistor would be fine.
So, here's what you need:
5V
|
[120]
|
|A
7406 [LED]
| \ |
MCU I/O>---| >O---+
| /
And the LED will light when the MCU I/O goes high or goes high-Z.