![]() I want a single input control two transistors of different typeĪre sure that's what you need, or is that just a particular solution you have in mind, but not the actual need? If so, that woul be a case of an XY Problem. Had you asked about your real application, you'd have gotten relevant replies immediately, without having to go back-and-forth with comments and dosing the information in little drips. You have asked a question and you will get answers exactly to that question you wrote, not to whatever you know but don't tell us. See, it is really important to actually say what you need, not to imply things with very little experience. For that load, you will need either logic level mosfets, or two stages of transistors. When having a specific schematic and no text to indicate otherwise, the schematic is all we got to work with to answer your question - it speaks in the words you didn't.įrom your scattered comments I gather that you're driving a much heavier load than you're not telling us about. A schematic is a graphical language to convey information. Your circuit implies a lot about the load current and load behavior. The schematic was the simplest way to explain that I want a single input control two transistors of different type.Ībsolutely not. With more reasonable transistor gains, it performs even better, letting you use larger base resistors.Īdding LEDs back into the circuit decreases the currents and lowers the saturation voltage even further. You can simulate this circuit, click on SW1 to select it, then use space to flip back and forth between logic states on the input. With LEDs added, the saturation voltage is even lower, since only 30-50% of the current flows compared to when the LEDs were shorted out. ![]() They still saturate just fine with LEDs removed and 10k base resistors. None of the modern transistors you buy would be that bad at the currents in your circuit. Higher collector current will be drawn by removing the LEDs, and even then lousiest of transistors I got in my junk box get saturated in this application. So you're either misunderstanding what saturation is, or have some truly atrocious transistors, or are making a measurement mistake, or you're not telling us something - possibly some combination of those. The transistors would be considered saturated at V(CE) of about 0.1-0.2V, and I can guarantee that that's the case in your circuit as long as it's what you're showing. See why providing too little information wastes your and our time? After all, that's what you show in the question, even though I have an inkling that's not the case. I don't know what transistors you're using, but I don't believe for a second that they are not saturated at the couple mA of current you're flowing through them. In my working voltages with base resistors higher than 2K none of the transistors go into saturation mode that I need. Well, that doesn't work very well at 3.3 V, but it would be OK for 5 V logic. With either 3.3 V or 0 V applied, the 10k resistor provides about 100 uA to the base, which should be enough to to drive the LEDs with up to 10 mA with a beta of 100. R1, R7, R8, and R9 bias the B-E junctions of the transistors to 0.53 V when there is no input, so they are both off. (edit) Here is a circuit that lights each LED with about 6 mA, and with no input, there is only about 10 uA through the LEDs. (edit) If the emitters are separated to drive D1 and D2 individually (I won't update the schematic), the circuit works properly, as illustrated here (off LED current is 33-43 uA): And perhaps green, blue, or white LEDs might stay off.Īdding 1N4148 diodes in series with the LEDs reduces the no connection current to about 70 uA, but even with 100 ohm resistors for R2 and R3, ON current is only about 3 mA. With the input disconnected, there is still about 600 uA flowing in the LEDs, so they will both be on, but dim. Unfortunately, there is a flaw in this circuit. Simulate this circuit – Schematic created using CircuitLab If they need to be brighter, try 100 ohms. ![]() I changed the LED resistors to 220 ohms to get 3 mA through the LEDs. ![]() Here is another way to perform the function you require. ![]()
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