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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?

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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:

Image

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:

Image

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:

Image

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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General comments (8 comments)
General comments
Circuit fantasist‭ wrote over 3 years ago · edited over 3 years ago

An interesting example... I like this circuit. As though, it is a voltage divider of three resistors in series. The middle "resistor" is variable and it controls the common current that, like an "electrical transmission", connects the voltage drops across R1 and R2 by Ohm's law - V1/R1 = V2/R2, so V1/V2 = R1/R2. It is like a "perfect voltage divider" with a simpler ratio. My idea was to make two non-linear loads (e.g., LEDs with slightly different VF) "help" the SPST switch when switching...

Circuit fantasist‭ wrote over 3 years ago · edited over 3 years ago

Thanks for the responce; it means a lot to me. My idea was to do the opposite - no equalizing resistors... and then the LEDs to act themselves as switches. Then the idea came to me to consider them together with the transistor as a "composite SPDT switch"... and I decided to ask this question... but I am still thinking about it...

(I slightly corrected the question to make it more clear - "switch loads" -> "switch between loads").

Circuit fantasist‭ wrote over 3 years ago

Thanks again for the efforts! I feel a little awkward about the time taken. Rather, I expected you to be a little affected by my naive question.

My idea was even simpler. An LED is supplied by a current and is lit. The SPST switch connects another LED with lower VF in parallel. It diverts the current of the first LED, turns it off and lights up. As though, the first LED acts as a NC SPST switch that complements the NO SPST switch... and the combination of them acts as a composite SPDT switch.

Olin Lathrop‭ wrote over 3 years ago · edited over 3 years ago

@Circuit: If you already had a specific idea you weren't sure of, then why didn't you ask about that? If you already had an answer, then what was the point of asking the question in the first place? Either way, it seems you were just wasting other people's time.

Circuit fantasist‭ wrote over 3 years ago · edited over 3 years ago

@Olin Lathrop‭, I am sorry if it is taken this way. I have already told how the idea for this conceptual question arose from something specific from the distant past. My idea was to get confirmation of my current hypothesis, so I asked it. I did not want to direct the answers in a certain direction in advance. For me, this is not a waste of time and I keep thinking about summarizing the problem... Thank you and sorry again.

Olin Lathrop‭ wrote over 3 years ago · edited over 3 years ago

@Circuit: But that's not what you did. If you have an idea in mind that you want validated, ask about it specifically. What you really did was grandstanding masquerading as a question. This will not be tolerated in the future. Consider yourself on thin ice. Future questions that aren't specific answerable questions will be closed.

Circuit fantasist‭ wrote over 3 years ago

@Olin Lathrop, OK, I see... I had two possibilities - to show my possible answer or to wait for others to show their solutions (as you)... and I chose the latter. Another time I chose the former... but result was the same - negative reactions. I thought it would be different here, so I was enthusiastic to do something more original and interesting... and here is the result. Everywhere the FORM dominates the CONTENT.

Circuit fantasist‭ wrote over 3 years ago

@Olin Lathrop, I asked a question in the meta section about this problem - https://electrical.codidact.com/questions/278579