Comments on Is it possible for one transistor to switch between two loads?
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Is it possible for one transistor to switch between two loads?
The output collector-emitter part of a transistor can be thought of as a 2-terminal SPST switch controlled by the input base-emitter voltage or base current. So this transistor switch can control only a single collector load.
Is it possible to make the transistor act as a 3-terminal SPDT switch? If so, one transistor will be able to control (switch between) two loads.
In general, can a SPST switch act as a SPDT switch?
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It's not clear what you are really asking, but here is something that might fit your requirements:
First, R1 and R2 can be considered separate loads. The transistor therefore switches two loads.
This circuit can also be used to produce two signals, one inverted from the other. This used to be more common in old tube amplifiers, and was called a "splitter". SIG+ and SIG- would then go off to control the drivers at different ends of the center tapped push-pull output stage, for example.
With bipolar transistors, the currents thru R1 and R2 aren't exactly equal, but can be close enough for many purposes. The discrepancy is 1/gain of the transistor. When this matters, use a Darlington pair, which acts like a single transistor with gain2. Or, nowadays, you'd usually just use a FET.
My idea was to make two non-linear loads (e.g., LEDs with slightly different VF) "help" the SPST switch when switching.
We do engineering here. That includes describing specifications properly so that others can understand them. I've mention before that your descriptions are vague. This needs to be fixed.
I can't even guess what "help the SPST switch when switching" is supposed to mean. SPST switches don't generally need help. If you're talking about a relay, then you either energize the coil or not. No "help", whatever that means, is otherwise required.
A single transistor can certainly drive multiple LEDs with different forward voltages. Just give each their own series resistor to get the current you want thru that LED:
In this example, R2 and R3 can be adjusted so that the current thru each LED is the same, even if they have different forward voltages. For this to work, the V+ voltage must be known and reasonably constant.
My idea was ... the LEDs to act themselves as switches ... consider them together with the transistor as a "composite SPDT switch"
Now it's even less clear what you are asking than before. However, one thing can't be changed, which is that a SPST switch has two output states. No matter how many LEDs you have, there will ultimately be only two different configurations of on/off for the LEDs that you get to select between.
Define what states you want each LED to have for the switch being on and off, and it will be straight forward to design a circuit to achieve that. Put another way, give us a truth table. This must have only two rows, one for switch off and the other for on. It can have as many columns as you like, one for each LED that is to be controlled.
Enough with the hand waving already!
Is it possible for one transistor to switch between two loads?
It seems we finally have a somewhat coherent question. Yes, a single SPST switch can switch between loads. Here is an example where one or the other LED is always lit, depending on the state of a SPST switch:
In this example, the forward voltage of the LEDs is 2.1 V with 10 mA thru them.
When the switch is open, current flows thru R1 and D2. Since D2 drops 2.1 V, the voltage across R1 is (3.3 V - 2.1 V) = 1.2 V. The current thru R1 is therefore 10 mA. This current flows thru D2, so D2 is lit. The voltage across D1 and R2 is only 1.2 V, which is insufficient to get any meaningful current thru D1, so D1 is unlit.
When the switch is closed, D2 has 0 V across it, so is off. The bottom end of D1 will be at ground, so its top end will be at 2.1 V. Using the same math as above, that means 10 mA flows thru R2, which also flows thru D1, so D1 is lit.
Therefore, D2 is on and D1 off when the switch is open, and the opposite when the switch is closed.
However, there is a drawback to this scheme, which is the wasted current thru R1 when the switch is closed. Since the full supply voltage will be applied to R1 in that case, the current thru it will be (3.3 V)/(120 Ω) = 27.5 mA, which dissipates 91 mW as heat.
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