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I got a few 2SD1048's, as I wanted to add a switch to an LCD project I have, so that I could turn the LCD screen on and off via a microcontroller pin.

The LCD screen is this and pulls ~100mA (rounded up). I'm operating it at 3.3v and the 2SD1048's were selected for their very low Vce(sat), since I can't afford to drop much voltage across the switch into the screen. I'm still pretty new to reading transistor datasheets, but it looks like the Vce(sat) for the 2SD1048 is 80mV at max, so probably 3.22V should reach the LCD screen. Marginal, but it should power on and work.

However, what I'm seeing is instead 3.3V at the collector, but 2.60V at the emitter, which isn't enough to power the LCD enough to function.

What am I doing wrong here, or what am I misinterpreting in the datasheet? Here is my circuit, voltages measured at Vin and Vout.

enter image description here

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  • \$\begingroup\$ Ah, let me see. So you expect Vce(sat) = 80mV but now Vce is 3.3 - 2.6 = 0.7V. Let us read the datasheet first. \$\endgroup\$ – tlfong01 Jul 22 at 5:20
  • \$\begingroup\$ That's Vbe, not Vcesat. Use the transistor as a low side switch, not an emitter follower. Or a PNP high side switch. \$\endgroup\$ – Brian Drummond Jul 22 at 9:18
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You should use a PNP eg. 2SB815 and invert the drive. Emitter to +3.3, and collector to the load.

schematic

simulate this circuit – Schematic created using CircuitLab

The way you are doing it will not saturate the transistor, because you would need more than 3.3V at the base resistor- it’s an emitter-follower which will always have at least one diode drop.

Something like this would give you the datasheet performance for the 2SD1048, however it's less convenient because you probably don't have a 6.6V supply or a way to level shift the control signal. There's also a potential issue if the load is removed the 3.3V supply could rise in value and damage something.

schematic

simulate this circuit

P.S. If by "screen" you mean only the backlight, you can simply use the D1048 as a low-side switch, but that's not a great way to switch the entire display power.

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  • \$\begingroup\$ The high side configuration with the 2SB815 worked like a charm. For clarity, I'm interested in switching the entire LCD module off (this is for when the controller goes into low power mode), but I'm curious why the low-side configuration would be appropriate for the backlight but not the power for the whole module? I'm not very clear on the difference. \$\endgroup\$ – justynnuff Jul 22 at 7:25
  • \$\begingroup\$ If you switch it off with a high side switch the outputs should all be high or high Z to prevent sneak currents through the inputs and the IC isolation junction and protection diodes. With a low side switch they should be low or high Z, which is more natural. \$\endgroup\$ – Spehro Pefhany Jul 22 at 7:33
  • \$\begingroup\$ Now that I know a little bit more about the Vbe threshold, wouldn't the base voltage only have to be 3.3v + 0.6v = 3.9v? I'm confused on where the 6.6v base voltage figure comes from. \$\endgroup\$ – justynnuff Jul 25 at 5:19
  • \$\begingroup\$ @justynnuff yes, you are correct, the base voltage would approximately be that, however in order to get current to flow you need voltage across that base resistor. In the case of a 220 ohm resistor you get (6.6-0.7-3.3)/220= 11.8mA of base current. Of course you would adjust the base resistor value if you had 6V or 10V available but in any case it has to be substantially more than 3.3V to get a well-controlled base current. 5V nominal would probably be about the lowest practically feasible, and there remains the issue that the 3.3V rail could be affected. \$\endgroup\$ – Spehro Pefhany Jul 25 at 6:24
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Question

NPN BJT = 2SD1048

LCD Screen = 100mA

Expect Vce(Sat) = 80mV

But now Vce = 3.3 - 2.6 = 700mV

Why?


Answer

I think the LCD should be on the high side, as shown below. You might like to refer to the Electronics Tutorials on Using Transistor as a Switch (Ref 2).

2sd1048


The datasheet specifies the following.

(1) At Ic = 100mA, Vce(sat) should be around 40mV

(2) At Ic = 100mA, hFE should be around 400.


Let us use the NPN BJT switching configuration recommended by the Electronics Tutorials and do the circuit analysis.

2sd1048 2


Circuit Analysis

(1) If Vin = 3V, Ib = (3V - 0.6V) / 220R = 2.4V / 220R ~= 2400 / 220 ~= 10mA

(2) Ic = Ib * hFE = 10mA x 400 = 4000mA max > 100mA (LCD loading)

(3) At 100mA, Ice(sat) should be around 35mV, therefore in saturation region.

(4) Therefore Vcc at LCD = 3V3 - 35mV ~= 3V, should be OK to drive LCD.

Errata - Equation (4) above should read 3V3 - 35mV = 3.26V. Many thanks to @justynnuff for pointing out my careless calculation mistake.

ad1024 sat region


Notes:

(1) My always dodgy calculation has not been proofread.

(2) I am just a friendly hobbyist. No guarantee no nothing won't melt down or blow up.


Discussion, Conclusion, and Recommendation

Discussion

There are a couple of things to discuss, such as

(1) Should we use a PNP BJT or NPN BJT, why?

(2) Should we place the LCD load place on high or low side of switch, why?

(3) Should we use a power MOSFET with a very very low on resistance, instead of BJT, which a a relatively big Vce(sat)?

(4) power BJT/MOSFET are usually used to switch current (eg relay, solenoid) and not a power supply. Power supplies are usually enabled/disabled by a logic signal. Switching off a current always creates a back EMF which might damage the power supply or other components.


Conclusion

The NPN BJT suggested should solve the OP's problem. But a better method is to enable/disable a power supply (Appendix B)


Recommendation

(1) NPN BJT switch with loading at high side is preferred over PNP BJT switch, for easier biasing and signal control.

(2) Instead of switching current to the LCD loading, enabling/disabling switching on/off power supply with over current protection etc, should be used, to avoid back EMF. (Appendix B)

Update 2020jul23

(3) Many thanks to @justynnuff for pointing out one important thing that many newbies or even old timers don't know that they don't know, that is using MOSFET as switches.

MOSFETs has very low "On resistance" which effectively means that the voltage drop (corresponding to BJT's Vce(sat) ) is typically 10 times smaller, therefore wasting much less energy and typically needs no heat sinks.

See Appendix C for a summary of Electronics Tutorials on using MOSFET as a switch.

/ to continue, ...


References

(1) 2SD1048 NPN BJT -15V -700mA Low Vce(sat) - On Semi 2013nov

(2) Transistor as a Switch - Electronics Tutorials


Appendices

Appendix A - 2SD1048 Datasheet Summary

2sd1048


Appendix B - Manual/Auto Switching On/off Enabling/Disabling Power Supply Unit

switch psu


Appendix C - MOSFET Switches - Electronics Tutorials

MOSFET Switches - Electronics Tutorials

MOSFETS can also be used as switches. The advantage of MOSFET over BJT is that the "On Resistance" is very small, in the order of 100mΩ. That means if the current being switched is big, say, 1A, the voltage drop is 100mΩ* 1A = 100mV = 0.1V. For a corresponding power NPN BJT, the voltage drop is Vce(sat) at 1A is of the order of 1V, or 10 times bigger.

mosfet switch


End of answer

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  • 1
    \$\begingroup\$ Excellent supplemental material. Small typo in the circuit analysis (4): 3.3V - 0.035V is 3.26V. I just had some SOT-23 BJBs on hand, but I think a MOSFET would be a better option for switching the LCD module. Switching off the power supply itself isn't an option, as the 3.3V power supply controls the microcontroller, too, so it needs to remain on. \$\endgroup\$ – justynnuff Jul 22 at 16:09
  • \$\begingroup\$ @justynnuff, thanks a lot for pointing out my calculation mistake, and one important thing missing in my answer, using MOSFET as a switch. So I have updated my answer. Please feel free to make more suggestions to improve our answer. Cheers. \$\endgroup\$ – tlfong01 Jul 23 at 7:51

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