# Would a transistor be the correct device?

I am going to create the following circuits. However I need to be able to have a switch that completes the lightbulb circuit when VM1 input > 4.3v

I think a transistor may be the answer from what I have read. Please could someone give me some help. I have thought about using a relay, but this just seems to mess up the voltage to VM1 Any help is greatly appreciated.

simulate this circuit – Schematic created using CircuitLab

I believe this is the solution I was looking for. Please read this carefully before testing. POT's R1 & R2 are attached to the same wiper, so what happens to one must happen to the other. The switch SW1 is a push to break switch. Please could you tell me if the resistors I have selected are compatible with the zener diodes I have selected. Thank you.

simulate this circuit

• If you need to be precise, a comparator (to detect the trip voltage) and a transistor for driving the necessary current for the lightbulb would be my starting point On mobile so no schematic; there may be enough information here to get you started. – Peter Smith Dec 26 '15 at 17:03
• To be clear, you want LAMP1 to turn on when VM1 node is >4.3V? – Krunal Desai Dec 26 '15 at 17:07
• I have heard about the comparitor, but not sure how to implement one. And yes when VM1 >4.3v the lamp would turn on. – James Shill Shillinglaw Dec 26 '15 at 17:10
• Q1) What's so special about 4.3 V. Q2) What is providing the 4.3 V. Q3) At what voltage do you want to turn off the lamp. e.g., 4.3 V or something a bit less to give some hysteresis? Update your question rather than post the additional information in comments. – Transistor Dec 26 '15 at 17:13
• if you read the post correctly Q2 and Q3 are already answered. The answer to Q1 is that it represents a threshold for another set of events, regulated by VM1 – James Shill Shillinglaw Dec 26 '15 at 17:36

Transistor is the correct answer but to get good control over this 4.3V threshold you may want something more than a simple transistor. Something like this:

simulate this circuit – Schematic created using CircuitLab

The LM393 in the middle is a comparator chip with open-collector output, comparing your input voltage against the reference voltage generated from the Zener diode D1 1N4731A (a 4.3V Zener diode, with R1 providing its 58mA current from a 5V supply) and shifting the signal level of 5V to 12V as required by the lamp with the pull-up resistor R3. The comparator outputs logic low when voltage on the "-" input is higher than "+" input, and high impedance (or logic high on a normal comparator, since LM393 in particular have an open-collector output) otherwise. The MOSFET M1 (any P-channel MOSFET will work here, IRF4905 is just my personal favorite) switches the lamp from the high side when it sees a low input at its gate, which is tied to the output of the comparator.

• Firstly thank you for your input. I am guessing that VIN is the voltage input to VM1 on my circuit? I understand the part where the zener diode drops the 5v to 4.3v for the comparator to compare the VM1 input voltage with. The MOSFET is a switch that closes the circuit when it receives the signal from comparator saying "-" > "+" thus lighting the bulb? I am quickly going to try simulate this and see if the results are as I require. – James Shill Shillinglaw Dec 26 '15 at 18:01
• Did you mean fro R1 to be 12 k? Right now it is just 12, which seems excessive. – mkeith Dec 26 '15 at 18:01
• It may be too complicated for the OP, but this design could be improved greatly if a shunt reference is used instead of an ordinary Zener. For example TL431. – mkeith Dec 26 '15 at 18:03
• Yes complicated is best avoided as I am not an electrician lol I am hoping to use this on an automotive application if that is any help – James Shill Shillinglaw Dec 26 '15 at 18:09
• @mkeith If you punch the calculator, $\frac {(5 - 4.3)V} {58mA} = 12Ω$. – Maxthon Chan Dec 26 '15 at 18:11

You can use a comparator to pull this off in the most flexible way. One method is to connect a known voltage reference (say from a precision Zener reference) to the minus input of the comparator, and then a voltage-divided version of VM1 to the positive input. You would select the divider values such that VM1 * divider ratio allows a voltage of 4.3V at VM1 to become a voltage of reference + a few mV. When that condition is true, the comparator will output a '1' that can be used to directly drive the gate of a transistor or some other transistor / relay circuit.

If your 5V rail is fairly stable, you could use that as your reference input by dividing it down via divider and feeding it to the minus input as well, saving the external reference.

The comparator can likely be powered off your existing 5V rail as well. Common models include the LM311. You may see a difference between open-drain and push-pull models; the open-drain types will only ever pull down actively; otherwise the output value floats to whatever you pull it too. A push-pull will actively drive high and low.

Finally, you can leave a resistor option between the output and - input to allow you to tune hysteresis in the future.

Option two (to avoid the comparator) is to use a precision analog MOSFET (from ALD, for example) with a very tight threshold voltage. Fed with a divider, this can then trigger other devices to turn on the lamp.

• Thank you for your input. The 5v is a reliable voltage, and I was hoping to use this as a reference if it were needed, which I assumed it would. I am going to try simulate based on your help, see what I can come up with. – James Shill Shillinglaw Dec 26 '15 at 18:07

The "4.3V threshold with 5V supply" is a bit trippy - as it enables a tricky yet simple solution here, leveraging the fact that BJT start to conduct reliably when there is 0.7V between base and emitter.

simulate this circuit – Schematic created using CircuitLab

This circuit leverages this fact: when VIN is lower than 4.3V, Q1 start to conduct and pull the gate voltage of M1 up. Then it would start conduct and ground the gate of M4, shutting the lamp off. When VIN is higher than 4.3V Q1 cuts off, and in turn M1 cuts off, and in turn M4 switches on, allowing the lamp to be on.

What you see here is in fact a open-drain dual-supply level-shifting logic inverter with a threshold voltage of about 0.7V below low-side supply. The lamp have one of its ends tied high so it lights up when the circuit outputs low.

• Thank you for this help, I am most grateful. I will simulate the first answer with my circuit and see if it works as I desire. I will report back with the results. I will then try this one and see which is most suitable. – James Shill Shillinglaw Dec 26 '15 at 19:09