3 V signal (PWR-ENABLE) be enough to turn on this circuit? The collector of Q3 will be at 11 V before switching is to occur.
\$\begingroup\$
\$\endgroup\$
1
-
\$\begingroup\$ Note that the negative terminal of the 11.1V supply MUST be connected to GND. If that's obvious then it's obvious :-) - but some try to do this with the switched supply floating. \$\endgroup\$– Russell McMahon ♦Commented Jul 17 at 9:37
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
2
(Your current question title says 3.3 V while the text says 3 V, so I'll use the lower of the two.)
Driving PWR_ENABLE with 3 V will be plenty to switch the whole circuit on.
However, you need to lower R6 from 10 kOhm to below 9.14 kOhm. A reasonable value would be 6.8 kOhms.
Your transistor Q3 is a Darlington pair (see internal schematic below) and requires a higher voltage than a normal BJT. It also contains a pull-down resistance for its second transistor's base which acts as a series resistance until the second transistor's base begins to conduct.
Driving PWR_ENABLE with anything above something like 0.8 V can be enough to switch Q3 on (see graph below, at 125'C). Your Q1 MOSFET circuit needs very little current to start to switch on, under half a mA. So your driving circuit just needs to drive at, say 0.7 V max. for off, to guarantee that Q3 is firmly off.
To switch Q3 and the circuit ON at room temperature (see graph below), the input should be at least 1.4 V which allows for current through the 10K series resistor. Your 3 V is well above this.
R6(10 kOhm) will form a potential divider with the 4 kOhm internal resistance, along with dropping the Vbe of the first internal transistor. That Vbe is approx. 0.7 V which can be considered to leave 2.3 V across the potential divider. The voltage across the 4 kOhm must be above 0.7 V to switch the second transistor on, but it's only 0.65 V. Lowering R6 to 6.8 kOhm will increase the potential divider voltage to 0.85 V across the second base-emitter, which it will then clamp to about 0.7 V when it conducts.
The rest of the circuit should switch the 250 mA load fine. You don't give a switching rate or slew rate requirement, so I have to presume this is infrequent power switching.
Your R30/R29 potential divider is good, puts 6 V on the gate which is comfortably inside the +/-8 V max. for the MOSFET.
(diagrams from datasheet, text labels slightly rearranged)
-
\$\begingroup\$ @jwo, (on your proposed edit) Yes, I knew Q3 was a Darlington from the datasheet, I'd just typo'd and written '0.8V' instead of '1.4V'. Fixed now and answer expanded to emphasise Darlington. On your edit itself, it's best instead to describe the error to the answerer so they can decide if they need a correction (which I did, thank you). I appreciate you don't have enough Reputation to comment yet but a comment on an answer is the route to use when you can in the future. \$\endgroup\$– TonyMCommented Jul 15 at 12:27
-
\$\begingroup\$ @Tim, apologies, I didn't read the edit proposal properly. \$\endgroup\$– TonyMCommented Jul 15 at 12:27
\$\begingroup\$
\$\endgroup\$
1
Q3 is a darlington pair, meaning that you require \$V_{BE}\approx 2\times0.7V = 1.4V\$ to switch it on. A potential of +3.3V at PWR_ENABLE will be adequate in this application, because collector current will be small. To be sure, calculate base current:
$$ I_B = \frac{3.3V-V_{BE}}{R_6} = \frac{3.3-1.4}{10k} \approx 200\mu A $$
Given Q3's ridiculously high current gain \$h_{FE}>1000\$, this will be sufficient current for collector currents up to 200mA, which is far greater than R29 and R30 will ever pass.
Calculating Q3 collector current when it is saturated (fully on):
$$ I_C \approx \frac{11V}{10k\Omega + 10k\Omega} = 550\mu A $$
The current gain you require for that transistor will be at least:
$$ \beta = \frac{I_C}{I_B} = \frac{550\mu A}{200\mu A} \approx 3 $$
Typically you'd double that, to ensure saturation, \$\beta \ge 6 \$. You can see why a transistor with \$\beta > 1000\$ is absurd for this application.
You've also assumed that Q3 will have \$V_{CE}\approx 0V\$ when switched on, but this is not correct. A darlington pair will have \$V_{CE}\approx 0.7V\$ in saturation (for small collector currents), and this particular model is no exception. Therefore, your calculation of gate potential is slightly incorrect, but shouldn't affect operation here. This is just a warning for the future.
You don't need a darlington pair at all. You'd be fine with any regular, cheaper, low-gain NPN device, like the 2N3904 or 2N2222. In fact, while you'd have to be unlucky, with such a high gain any base current picked up from a strong radio source, a nearby node with fast-changing potential or large changing currents, could tend to switch it on.
Coupled with Q1's extremely low \$V_{GS(TH)}=0.4V\$, it would take very little to switch Q3 and Q1 on. I would strongly recommend using a regular transistor for Q3, not a darlington pair.
-
\$\begingroup\$ The Darlington transistor will work, but this answer adds a good observation that a Darlington transistor is not necessary. The solution suggested here will work and be less expensive to implement. \$\endgroup\$– JkingNHCommented Jul 20 at 4:25