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I want to get 24 V from the MCU instead of 5 V so I decided to use a TIP141 transistor with current rated up to 10 A. My circuit is the MCU 5 V output power the transistor base basing through 10 kΩ resistor then the 24 V connected to the transistor collector and get the power from emitter.

When I open the transistor and use the voltmeter to measure the output from the emitter I found the power is just 4 V not 24 V as expected.

enter image description here

Can anyone explain that and tell me what should I do?

Edit: To explain my question more let me share the application which I need to do.

It’s a multi systems in a bus and my ECU controls them all with 24V . All the systems, even my ECU, have a common ground which is the bus body, so each system needs only 24V to be powered:

enter image description here

If I use the normal transistor connection:

enter image description here

I will have to connect two wires to my ECU and in buses it is a long distance and we already have a common ground, so I need it to be like this:

enter image description here

That way there is only one wire from the system to my ECU.

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  • \$\begingroup\$ FYI. "opening a transistor" happens when a transistor is destroyed and no longer conducts. It is "open" like an open switch, or an open circuit. One "turns on" a transistor to make it conduct. \$\endgroup\$ Feb 8 at 14:50
  • \$\begingroup\$ You're missing one little detail. Is the 24V input to SYS1 high-impedance or low-impedance? If low-impedance, then AnalogKid's answer will suffice. If it's high-impedance, then your "normal transistor" diagram will work if you use a pullup resistor and only route one wire from the collector to SYS1. \$\endgroup\$
    – Ste Kulov
    Feb 8 at 16:47

3 Answers 3

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TIP141 is an old Darlington pair with some internal resistors. For an output current of 5A it is characterized at 10mA base current, so your base resistor should be much less than 10kΩ, more like 300Ω.

Being a Darlington, the voltage drop will be relatively high, 2V at 5A, which is 10W of dissipation, requiring a rather large heat sink or fan. In 2024 we would generally prefer to use a logic-level MOSFET which would not require a heatsink, or perhaps just a very small one.

In any case, you must ground the emitter and put the load on the high side, as so:

schematic

simulate this circuit – Schematic created using CircuitLab

If your load is inductive (like a motor or a solenoid) you will also need a suitably rated diode across the load as shown or the TIP141 may be damaged the first time it tries to turn off.

Also, keep in mind that with a circuit involving 24V, wiring misconnections or even an open connection (say the emitter) can result in your MCU and perhaps other things which may be connected to it being destroyed.


Since your edit suggests you must have a high-side driver, you can accomplish that with two transistors, or (preferably, in my opinion) if the current requirements are less than a few A, you can use a protected high side driver IC such as NCV8460A which is a convenient (for SMT) SOIC-8 part designed for automotive applications:

enter image description here

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  • \$\begingroup\$ Darlingtons aren't very suitable at all for driving motors or solenoids, even if you provide flyback diodes. When I started with electronics some 20 years ago, Darlingtons had already been long since phased out and replaced with MOSFET drivers. The reason being that Darlingtons popped like popcorns in an industrial setting. \$\endgroup\$
    – Lundin
    Feb 8 at 15:18
  • \$\begingroup\$ Thanks @spehro Perfhany for ur awesome answer. I edited the question to be more clear to u. Can u please answer it to me please. \$\endgroup\$ Feb 8 at 15:25
  • \$\begingroup\$ See above edit. \$\endgroup\$ Feb 8 at 17:10
  • \$\begingroup\$ @Lundin Yes, you have to pay close attention to the SOA (Safe Operating Area) limits or the transistor will die. As to using MOSFETs, that was not an option in the early days, and many TIP121s were used to drive fuel injectors, a particularly nasty inductive load because you have to allow a large voltage peak to get them to shut off quickly. \$\endgroup\$ Feb 8 at 17:12
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It sounds like you have created an emitter follower. Try putting your load between the 24V supply and the collector while grounding the emitter. Also make sure the MCU and 24V supply share a ground.

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  • \$\begingroup\$ Thanks @vir for ur answer. I edited the question to be more clear to u. Can u please answer it to me please. \$\endgroup\$ Feb 8 at 15:27
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Because of your grounding and distance situations, it will take two transistors to switch the +24 V power using a +5 V signal. One NPN small-signal transistor and one PNP power transistor. In general, this type of circuit is called a high-side switch.

The small-signal part can be just about anything - 2N4401, 2N3904, 2N2222, etc.

Since you are familiar with the TIP components, the power transistor can be a TIP145 -146 -147. Here is an intergoogle image grab. In your case, Q3 is the TIP part. Increase R2 to 4.7K and R3 to 10K.

schematic

Image source: Copied from question "PNP and NPN circuit" by Ahmed Shady on Electrical Engineering Stack Exchange, originally licensed under CC BY-SA 3.0

With this circuit, the output voltage will be around 23 V (depending on the output current). A variation of this circuit lets you use the TIP141 you already have, but the output voltage will be around 22 V or less, and the power dissipation in the power transistor will be greater.

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