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In the circuit below, V1+D1 controls when Q1 conducts: when supplying a negative voltage, Q1 conducts. When the output of V1 is changed to 0V, Q1 does not conduct.

How should I change it, so that instead of being controlled by a negative voltage source, it can be controlled by a positive voltage in the 5 - 12 V range?


Thanks for the replies. I made a correction in the circuit and added a current limiting resistor. Also I created the circuit in CircuitLab so you can see it in action.

I tried the suggestion about removing R3 and to make V1 a positive voltage source (without D1), but that didn't work.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Check which terminal of each voltage source you want connected to ground. Didn't "unipolar sources" use to be supported? \$\endgroup\$
    – greybeard
    Jan 27 at 0:29
  • \$\begingroup\$ what does the circuit do? ... there is no output \$\endgroup\$
    – jsotola
    Jan 27 at 3:54

5 Answers 5

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Perhaps you mean that you want a potential above +5V to switch on the transistor, and a potential below +5V to switch it off:

schematic

simulate this circuit – Schematic created using CircuitLab

It works by using zener diode D1 to "offset" the input by −4.7V, meaning that only when \$V_{IN}\$ exceeds +4.7V does base potential rise above zero. You could replace D1 with a model of any reverse breakdown voltage you like, to set the transition threshold anywhere you like.

Sweeping \$V_{IN}\$ from 0V to 12V yields this \$V_{OUT}\$:

enter image description here

There's some chance that reverse leakage current through D1 could partially switch on Q1, so it's best to include a resistance to sink that current to ground, as I've done with R3 below. Note that R3 must be much larger than R1, since they form a potential divider which will reduce base potential somewhat.

Also, if you envisage \$V_{IN}\$ becoming negative, then you should probably take measures to avoid reverse-biasing the the transistor's base-emitter junction, with a another diode, D2 here:

schematic

simulate this circuit

It's fairly trivial to employ the same trick to obtain a negative switching threshold. In fact, this is more like your original circuit, in which R3's role was to keep Q1 switched on by default:

schematic

simulate this circuit

With the zener diode flipped, and R3 acting to raise base potential (switch Q1 on) in the absence of an input to pull it low, the threshold is around \$V_{IN}\approx -4.5V\$. Here I sweep \$V_{IN}\$ between −12V and +6V:

enter image description here


In the comments accompanying my answer, you have cleared up a whole bunch of ambiguity about the circuit behaviour you desire. You said:

What's needed is that, 0V yields 0V at Out (like now) +5V, like from an Arduino board, yields 12V at Out

You really should put that in your original question, because that's really not made clear.

Anyway, what you require, then, is called a "level translator". Everything I described above, while they produce the correct output potentials, they also invert the signal, so that high in --> low out, and vice versa. You are requesting a circuit that does not invert:

schematic

simulate this circuit

The version on the right is employed by the various "level shifter" boards that you can buy from places like SparkFun and Adafruit. On the left I'm using a BJT instead of a FET, which performs similarly, but not quite as emphatically. Here are \$V_{OUT1}\$ (blue) and \$V_{OUT2}\$ (orange), responding to input \$V_{IN}\$ sweeping from 0V to +12V:

enter image description here

The BJT version switches when the input is little too close to +5V for comfort, so you may want to lower that threshold a little:

schematic

simulate this circuit

enter image description here

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  • \$\begingroup\$ What's needed is that, 0V yields 0V at Out (like now) +5V, like from an Arduino board, yields 12V at Out The circuits you posted don't do that. Thanks for trying anyway. \$\endgroup\$
    – primare
    Jan 28 at 8:45
  • \$\begingroup\$ @primare Now you've made it clearer what you want, I've added a potential solution to my answer. \$\endgroup\$ Jan 28 at 10:17
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what about this circuit? I made it with falstad.com's simulator...

falstad.com

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  • \$\begingroup\$ With such a low base current, there is no guarantee that the transistor will go into saturation as its beta changes with temperature, aging, and lot-to-lot production tolerances. \$\endgroup\$
    – AnalogKid
    Jan 27 at 4:43
  • \$\begingroup\$ @AnalogKid ok... i thought, that Ib·β is always far above 131µA... even at 5V... \$\endgroup\$
    – RRIDDICC
    Jan 28 at 10:40
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You have 2 options to make Q6 turn ON, thus the collector will drop to almost zero volts.

A) Remove v1 and v2 from the circuit. R4 will provide enough current to turn on Q6.

B) Flip v1 and v2 so the base of Q6 is driven by a positive voltage. Limit the current to a few milliamps by inserting a 10 kohm resistor in series with v2.

NOTE: You should avoid negative voltage on the base of Q6 at all cost. Look up how to properly bias a NPN transistor.

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In this circuit, V2 controls when Q6 conducts: when supplying a negative voltage, Q6 conducts.

Before we get to an answer, let's start with your initial statement, which is incorrect.

When the base of an NPN transistor is negative with respect to its emitter, the transistor does not conduct - normally. The is such a thing as reverse breakdown of the base-collector junction that can lead to current conduction, but this is far from normal, especially when there is no current-limiting resistance in the base circuit.

Reverse the polarity of V2 and insert a resistance in series with the base. Because the output impedance of a voltage source in LTSpice is zero ohms, you do not need R4. Once you have that circuit, it can be adjusted for just about any input voltage range.

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How should I change it, so that instead of being controlled by a negative voltage source, it can be controlled by a positive voltage in the 5 - 12 V range?

Invert V2. You have inserted it upside down.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1.

  • (a) Normally orientation of 5 V voltage source gives 5 V.
  • (b) "Upside-down" orientation of 5 V voltage source gives - 5 V (negative).
  • (c) "Upside down" orientation of - 5 V voltage source gives +5 V (positive) output!

enter image description here

Figure 2. OP's circuit.

Note that you have no current limiting resistor between V2 and the base of Q6. When V2 exceeds the (positive) forward voltage of D1 (0.7 V typical) + the forward voltage of Q6 base-emitter (0.7 V typical) - so 1.4 V total - a very high current will flow and the components will be pushed beyond their rated limits.

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  • \$\begingroup\$ @greybeard: Good man! Fixed. \$\endgroup\$
    – Transistor
    Jan 27 at 9:50

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