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I'm new to circuits in general but have a basic understanding of it. However, it seems I can't wrap my head around the following issue:

(I'm using Tinkercad Circuits for trying out different approaches. All values come from the software's "multimeter".)

I have a very simple circuit, consisting of a power source, a transistor and a consumer. I want the transistor to output the same voltage as the input is. However, I don't know how.

The transistor logically consumes power itself. So having 12V in, the transistor has an output of about 10.8V.

Using a nMOS or pMOS didn't help either - at least in my approaches that is. I feel like I'm missing something very basic. Maybe something like a pull-up resistor?

How to make the output of a transistor the same as the input?

schematic

simulate this circuit – Schematic created using CircuitLab

Thanks!

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    \$\begingroup\$ Your text says the transistor has "12 V in", but your circuit diagram onky shows a 5 V source. Please edit one or the other to make your question clear. \$\endgroup\$ – The Photon Apr 15 '18 at 17:49
  • \$\begingroup\$ Try a common emitter configuration (see google) \$\endgroup\$ – pjc50 Apr 15 '18 at 17:53
  • \$\begingroup\$ Where is the node where you measure "output" ? \$\endgroup\$ – Mitu Raj Apr 15 '18 at 17:59
  • \$\begingroup\$ @The Photon you are right, I used 5V in the diagram because in Tinkercad I used a 12 V Lightbulb as the consumer and a LED in the diagram. So 5V was more "realistic" :) \$\endgroup\$ – NitricWare Apr 15 '18 at 18:40
  • \$\begingroup\$ @MITU RAJ i connect the multimeter + side to between the resistor and the transistor and the minus between the diode and the battery. Is this correct? \$\endgroup\$ – NitricWare Apr 15 '18 at 18:46
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It's not clear to me where you are headed, but your statement:

I want the transistor to output the same voltage as the input is.

Suggests a thought to me. If it is totally NOT what you want, no harm done. At worst, we're talking at cross-purposes and I'll remove by answer.

schematic

simulate this circuit – Schematic created using CircuitLab

The output will follow the input pretty closely, but within some reasonable limits with respect to the power supply. (This topology will have a very familiar look to some audio amplifier aficionados where it is one part of a so-called diamond buffer.)

In general, \$V^-\lt V_\text{IN}\lt \left(V^+-1\:\text{V}\right)\$ to avoid saturation. But the output load, in combination with the values of \$R_1\$ and \$R_2\$, matters. (Not shown in the diagram.) So the maximum input voltage will be:

$$V_\text{IN}\le \frac{\beta_1\:R_1\:R_\text{LOAD}\left(V^+-V_\text{BE}\right) + R_2\:V^-\left(R_1+R_\text{LOAD}\right)}{R_1\:R_2 + R_\text{LOAD}\left(\beta_1\:R_1+R_2\right)}$$

For a single rail supply, where also \$R=R_1=R_2\$, this becomes:

$$\begin{align*}V_\text{IN}&\le \frac{\beta_1\:R_\text{LOAD}\left(V^+-V_\text{BE}\right)}{R + R_\text{LOAD}\left(\beta_1+1\right)}\\\\&=\frac{V^+-V_\text{BE}}{\frac{\beta_1+1}{\beta_1}+\frac{R}{\beta_1\:R_\text{LOAD}}}\end{align*}$$

And from this, you can see that you want \$R\ll \beta_1\:R_\text{LOAD}\$ in order to support input voltages that are within a diode drop of the positive voltage rail.


See following simulation:

enter image description here

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  • \$\begingroup\$ I tried to rebuild your circuit in Tinkercad. It measured 4.3V at the OUT while feeding it 5V so i guess this is as close as it gets, right? Also, there is no way, to make the Emitter Voltage indepentent from the Base Voltage, right? \$\endgroup\$ – NitricWare Apr 16 '18 at 12:24
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    \$\begingroup\$ @NitricWare Make sure you use -10 V and +10 V as the rail supplies. It should be very close. Q2 would emitter follow the input, feeding Q1, which will emitter follow in the other direction. So it should be way closer. If not, show me your work, if not. \$\endgroup\$ – jonk Apr 16 '18 at 12:33
  • \$\begingroup\$ Could you maybe check my Tinkercad Circuit and check if I wired it all up correctly? The circuit in the upper left corner is yours. The one in the lower right is what I wanted to accomplish in the first place... tinkercad.com/things/94Lv5jDQpsQ-smooth-bombul-amur/… \$\endgroup\$ – NitricWare Apr 16 '18 at 17:18
  • \$\begingroup\$ @NitricWare Your link requires a sign-in (or appears to, to me.) I can't examine it. \$\endgroup\$ – jonk Apr 17 '18 at 8:06
  • \$\begingroup\$ oh! I tried to set it to public - maybe it works now? If not - no worries :) \$\endgroup\$ – NitricWare Apr 17 '18 at 17:15
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It sounds like you want to build a voltage follower circuit. Here's a link to a great article about it: Voltage Follower. From what I understand though, the output will differ from the input by about 0.7V( which is one diode drop, due to the nature of transistors). If you want your output to follow the input more closely, I think you will need to add an op-amp, which seems to be out of the realm of your question.

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  • \$\begingroup\$ Thanks for the link! But as you stated, and the article showed, there still is a voltage loss. So... Basically, like @MITURAJ noted and also as The Photon showed a little voltage drop is always there, right? \$\endgroup\$ – NitricWare Apr 15 '18 at 19:57
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So having 12V in, the transistor has an output of about 10.8V.

In forward active mode, an NPN BJT has a \$V_{be}\$ of somewhere above 0.6 V. For low currents, 0.6 or 0.7 V is common. If your device was dropping 1.2 V, it must have been sourcing a pretty large current (on the order of amps, but certainly not the 10-20 mA that would be drawn by the circuit you showed in your schematic).

Using a nMOS or pMOS didn't help either - at least in my approaches that is.

If you can push the NMOS gate above the supply voltage by a couple volts (\$V_{th}\$ plus a bit), you should be able to get very close to zero voltage (maybe a few millivolts) drop across the channel. To do this you will need a gate driver circuit, which of course adds complexity and cost, and consumes power of its own.

A more common approach is to use a low-side switch:

schematic

simulate this circuit – Schematic created using CircuitLab

But why try to minimize loss in the transistor, if you're already dumping the majority of your power in a resistor anyway?

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  • \$\begingroup\$ Thanks for your answer!! So basically, as @MITURAJ said, the voltage drop is inevitable, right? \$\endgroup\$ – NitricWare Apr 15 '18 at 19:52
  • \$\begingroup\$ @NitricWare, unless you have another power source available at higher voltage, like in Jonk's answer. Or, again, with FETs and a gate driver circuit you can get the output within millivolts of the input. \$\endgroup\$ – The Photon Apr 15 '18 at 21:04
  • \$\begingroup\$ @NitricWare, what are you trying to actually achieve? Why not just connect the power source to the load directly? \$\endgroup\$ – The Photon Apr 15 '18 at 21:05
  • \$\begingroup\$ Here's what I want to do: I want the light to turn on when sensor (a photoresistor in this case) "reads" light. With an arduino that would be easy. As soon as the Analog In reads some specific value, my digital out goes to high. But I'd like to achieve this without a Controller :) \$\endgroup\$ – NitricWare Apr 16 '18 at 9:41

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