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I am looking for a kind of "voltage controlled open-closed switch", "tri state switch" or "solid state relay", implemented with transistors.

enter image description here

Consider a \$200\Omega\$ resistive load, and the \$GND=0\$ to \$VCC=5V\$ range for both a control and an input signal ("inputs to the gate").

A single PNP | NPN transistor, could be the simplest trial, switching from \$OC\$ to \$0V\$ | \$5V\$ if the load is connected to \$GND\$ | \$VCC\$, and only within some ranges of \$V_{E}\$, between \$1-5V | 0-4V\$ respectively.

The PNP with the transistor in the cut zone works well as open circuit: No current, independent of the voltage input and the voltage at the load end (Figure 1 ).

The problem is the PNP as closed circuit (Figure 2). For the given voltage ranges, the transistor is in the active|sat zone, and \$I_C\$ mean valuechanges "smoothly" on the voltage at the load end, instead of being independent of it, hence not achieving the purpose.

  • Green is the voltage input,
  • Red is the voltage output,
  • Blue is the voltage at the load end (GND),
  • Black is \$I_C\$.

What kind of transistor circuit should I use for implement a more real voltage controlled open-close switch?

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    \$\begingroup\$ If I am understanding the diagrams and plots correctly, then you are not obeying your own rules. In your diagram you state that V2 must be less than 4V, but in the plot you show V2 varying from 1V-5V. What happens if you lower V2 to 0V-4V? I think your circuit would work then. \$\endgroup\$ – Mr. Snrub Jul 4 '19 at 5:47
  • \$\begingroup\$ I corrected the images. If i use a single transistor, the \$V_E\$ should keep the 1V gap for working. But if i have a digital signal as V1 or V2 breaking that gap the transistor enter into cut zone and then i lost the closed circuit. Hence this do not work at all actually. \$\endgroup\$ – Brethlosze Jul 4 '19 at 5:59
  • \$\begingroup\$ The purpose is to connect/disconnect a load. A single transistor is an example, which works only when properly polarized. A single DC relay is a non-transistor solution. The context is not the design of a FET/MOSFET, hence drain is not understood under that meaning. An alternative replacement for "open circuit" switch could be "tri-state", "high impedance" or others... \$\endgroup\$ – Brethlosze Jul 4 '19 at 7:03
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    \$\begingroup\$ You can't get tri-state from a single switching element. You can only get two states: on / off. Tri-state is applicable when the output has both a high-side and a low-side switch. Your schemes can only switch one way and the load pulls the other way. \$\endgroup\$ – Transistor Jul 4 '19 at 7:38
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    \$\begingroup\$ For a real tri-state circuit you need two transistors in series, not only one. \$\endgroup\$ – Uwe Jul 8 '19 at 10:21
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(This is more of a comment than an answer, but the URL has more characters than a comment will allow.)

The second circuit works just fine if V2 is limited to 0V - 4V. See this CircuitJS simulation here. If you set Vctrl to +5V then Vout is equal to V2. If you set Vctrl to 0V then Vout stays stuck at +5V. I'm not sure why your second plot doesn't agree with this. What were your resistor values?

One minor detail: note that the output is never truly "open circuit" since it's always connected through that resistor to +5V.

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  • \$\begingroup\$ Thanks for your answer comment. Yes, if the signal is limited I could say the circuit works. But eventually one would require to have 0-5V or something closer to that. If \$V_{ctrl}\$ is 5V, we see there is no current into the load, independently to the voltage at the other voltage end, thus i could say it is an open circuit. \$\endgroup\$ – Brethlosze Jul 4 '19 at 14:24

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