I have a basic circuit for converting the 3.3V signal of a microcontroller board to a 24V constant voltage needed to control a device switched by voltage.

enter image description here

When I measure voltage at the output while the transistor is on, it's 0V, and when it's off I get 24V, as expected. The problem is, whenever I connect the output to the device while the transistor is off, the voltage drops to 14.8V. The voltage I'm getting seems to be dependent on the load resistor, smaller resistor gives less voltage drop, but increases current when the transistor is closed.

So the question - what causes the voltage drop and how can I avoid it? In my eyes, the resistor appears to be causing the voltage drop, but why is it not the case while the external circuit is disconnected? Is the only way to avoid this using less resistance at the emitter? Thank you.

  • \$\begingroup\$ So are you basically asking why in a voltage divider set for X volts, when you change one of the resistors, it changes these X volts? \$\endgroup\$
    – PlasmaHH
    Jan 27, 2015 at 14:54
  • \$\begingroup\$ The only objective here is to get constant 24V at the output, I am asking how the voltage divider is formed and how to avoid the division. \$\endgroup\$ Jan 27, 2015 at 14:59
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    \$\begingroup\$ Decrease both the resistances (e.g., as in the figure above) to reduce the voltage drop across the collector resistor... Or, if it does not help, connect another PNP transistor with its emitter connected to +24 V, the base controlled by the current 1 k collector resistor, and the collector - as an output. Add also (a few kom) resistor between the base and emitter. The voltage divider is formed by the 1 k collector resistor and the next load resistance. \$\endgroup\$ Jan 27, 2015 at 15:04
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    \$\begingroup\$ The output impedance of your common emitter is going to be around 4.7 kOhm because of your collector resistance. As a rule of thumb uou'll need your external circuit to present an input impedance of at least 10x as large to avoid your voltage dropping when you connect it. When the external circuit is disconnected, you don't have a voltage divider...so the voltage is not dropping. If you don't want to change Rc, you can insert a buffer (something with very high input impedance and very low output impedance) after your common emitter. \$\endgroup\$ Jan 27, 2015 at 15:05
  • \$\begingroup\$ Show the schematic you are actually using. No, it's not acceptable to show one schematic and then say in the question how your actual values are different. -1 for the laziness. While you're at it, add proper component designators so that we can talk about the circuit easily. That's how its done. \$\endgroup\$ Jan 27, 2015 at 15:15

3 Answers 3


The problem is that R2 (oh, right, you don't have component designators, you'll just have to figure out from context which one that is) is the impedance of your signal when high. This together with the input ipedance of whatever is being connected to your output form a voltage divider.

For example, using the circuit you show, let's say the output is 7 V when connected to the device. That means R2 is dropping 5 V, which means 5 mA is flowing thru it, which also means that is how much current the input of your device is sinking when held at 7 V. To compute the input resistance of the device at this operating point, use Ohm's law. (7 V)/(5 mA) = 1.4 kΩ.

To get a higher output voltage, lower R2. However, keep in mind that this comes at the cost of higher current when the output is low. Also keep the dissipation of R2 in mind. In your example, the current thru R2 when the output is low is (12 V)/(1 kΩ) = 12 mA, and the dissipation is (12 V)2(1 kΩ) = 144 mW. Note that the power increases with the square of the supply voltage. Doubling that while keeping R2 the same will cause 4x the dissipation, or 576 mW. Even a ½ W resistor would be too small in that case.


Here is one way to fix your circuit without using a lot of current when the output is low:

R1, Q1, and R2 are basically what you have now. Q2 is a emitter follower that provides much more current output than the bare R2. The output will be about 700 mV lower than the bottom end of R2, but I expect that small amount is irrelevant for a digital signal that swings to 24 V.

If you can arrange the output from the microcontroller to be the opposite polarity, then you can use this circuit:

This circuit is non-inverting, unlike your original and my example above. This has both higher output current capability and lower output voltage drop compared to the circuit above.

  • \$\begingroup\$ This is perfect. I will try the circuits out, thank you. \$\endgroup\$ Jan 28, 2015 at 11:49

The circuit you are using is for a common-positive-supply-based load i.e. if the load is in parallel with the resistor, turning the transistor on will apply 24 volts across the load (within constraints of what current the transistor can supply).

It sounds like you need a circuit for a common-ground-connected load and this would be achieved by using a high-side bipolar transistor circuit like this: -

enter image description here

Here's also a little article on high side and low side switching.

  • \$\begingroup\$ Could you please clarify? My goal here is provide 24V to a voltage switched device, why is the first circuit wrong and the second one right for such an application? \$\endgroup\$ Jan 27, 2015 at 15:58
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    \$\begingroup\$ The first circuit provides 24 volt to a load but via a 4k7 resistor and as soon as you begin drawing current the 24 volt drops. If your device could be put in parallel with the 4k7 then no problem but, if it needs to be tied to 0V then you have to use a circuit like #2 \$\endgroup\$
    – Andy aka
    Jan 27, 2015 at 16:05
  • \$\begingroup\$ I see, in that case it's exactly what I need, thank you! \$\endgroup\$ Jan 28, 2015 at 11:37

@I have no idea what I'm doing, the role of the resistor that should be connected between the base and the emitter (across the base-emitter junction) of a BJT is crucial to turn off the transistor... but it's not so easy to explain why. Well, let's try... it's a great challenge:)

The problem is that you can surely turn off the PNP BJT by lowering the base-emitter voltage (below approximately 0.6 V); you can't do it only by decreasing the base current (by cuting off the previous NPN transistor). Here you can see another voltage divider composed by the PNP collector-emiter junction (the lower leg) and the resistor in question (the upper leg). When the PNP transistor is cut-off, the ratio of this divider is low enough so that the base-emitter voltage of the NPN transistor is low enough as well.

You can see also this resistor in the @Andy aka's solution where it is absolutely necessary to drive (not only to turn off but to turn on it as well) the gate-source junction of the "voltage-driven" FET. You can think of this resistor as of a current-to-voltage converter.

As I can see, @Olin Lathrop has already added this (NPN + PNP) BJT circuit solution.


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