Uncategorized

bjt switch vce sat problem

High Vce(sat) of 2SD1048

Ask QuestionAsked todayActive todayViewed 30 times1

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

transistorsbjtshareedit  follow  flag asked 3 hours agojustynnuff14344 bronze badges

  • 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. – tlfong01 2 hours ago   

add a comment

2 Answers

ActiveOldestVotes3

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.shareedit  follow  flagedited 2 hours agoanswered 2 hours agoSpehro Pefhany255k99 gold badges210210 silver badges529529 bronze badges

  • 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. – justynnuff 49 mins ago
  • 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. – Spehro Pefhany 41 mins ago 

add a comment1

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.

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)

/ 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 – Manual/Auto Switching On/off Enabling/Disabling Power Supply Unit

switch psu

shareeditdeleteflagedited 3 mins agoanswered 2 hours agotlfong0127411 silver badge55 bronze badgesadd a comment

.END

Categories: Uncategorized

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.