# Solid State Switch

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1. Dec 21, 2014

### Daniel Baggett

My goal is to use a single signal to do two things at the same time...when DC signal is applied turn transistor1 off and turn transistor2 on. Then when signal is not applied transistor1 returns to on and transistor2 returns to off. Frequency will be no more than 60Hz.
What would this circuit look like and what components would I need? Thanks!

2. Dec 21, 2014

### Staff: Mentor

That depends on the rest of the circuit - what are the transistors supposed to drive?
Transistors of all types usually come in two sorts - one gets switched on with high input voltage and the other gets switched off. If some other constraints (voltage ranges, impedance, ...) match, you can simply connect your input to both transistor gates/bases.

3. Dec 21, 2014

### meBigGuy

As mfb said, without understanding what the transistors need to drive, there is no way to design a circuit. For example, do thye both need to turn on to ground, both need to turn on to power-supply, or one goes to ground and one goes to supply. Do they need to conduct 10Amps, or 1ma? Do they both need the same current capacity. What voltage drop can you tolerate across the transistors? What is the source current and voltage specification for the driving signal? It goes on and on.

Draw a diagram using "voltage controlled switches" (or relays) and indicate currents and voltages required and supplied. From that we can help you design a transistor based circuit to do what you want.

4. Dec 21, 2014

### Daniel Baggett

I'm trying to put together a diagram with Yenka (electronics app.) It might be a day or 2-3 (or more) before I can upload it.
Daniel

5. Dec 27, 2014

### Daniel Baggett

Regretably, I haven't had the time to draw the circuit. However, mfb has the answer I was looking for!
Ans: Transistors of all types usually come in two sorts - one (T1) gets switched ON with high input voltage and the other (T2) gets switched OFF. This appears to be what I'm looking for. Transistor1 only will be turned on to ground with respect to 12V source. Transistor1 will have 12V Source Voltage with no current flowing as the circuit will be switched at high enough frequency to prevent current from flowing...only voltage component will be utilized in Transistor1 circuit (Voltage preceeds Current in a resistive load). This circuit will charge Capacitor1.
Transistor2 (a completely separate circuit) will go high at exactly the moment Transistor1 goes low. Transistor2 will be releasing the stored energy in Capacitor1 through a LOAD and will be accessing the current and voltage stored in Capacitor1 when T1 goes low and T2 goes high. T1 and T2 both need to have the same signal applied at the same time to cause T1 to go low and T2 to go high at the same instant! T1 source voltage = 12V DC. T2 source voltage = 110V DC (C1). T2 will be "grounded" to the negative plates of C1. Frequency through T1 is between 300 and 500 Hz/1/60th sec. (using a 555 with potentiometer in a separate timing circuit). Frequency at T2 will be 60 Hz.I hope my description helps explain what I hope to accomplish. I think I can do the math to determine R and V in the circuits. The components are where I could use a little help.
Thanks,
Daniel

6. Dec 28, 2014

In purely resistive loads AC current and voltage goes in phase.So Voltage doesnt preceed current in resistive load like a resistor.
Any wire in room temperature is a small resistor with resistance increasing the longer it goes .

7. Dec 28, 2014

### Staff: Mentor

That is not a frequency no matter how I parse that expression.

A circuit diagram would be way easier to understand, and does not take much time to do (less time than thinking about a good description of the circuit).

8. Dec 28, 2014

### Daniel Baggett

What I meant by 300 and 500 Hz/1/60th sec is every 1/60 of a second there will be 300 to 500 pulses into C1. In other words, 18,000 to 30,000 pulses per second. And yes, my mistake... Current lags Voltage in an Inductor! The circuit is proprietary!