Why I am I unable to turn on a load when Wheatstone bridge is connected in series with the load

Question

I am trying to switch on a load by using a Wheatstone bridge and planning to provide the differential voltage to the transistor.

Before doing that, I am trying to use two batteries (second one substituting the differential voltage from transistor).

However, I am getting a 0 V across the load and all of 9 V across the Wheatstone bridge.

^{simulate this circuit – Schematic created using CircuitLab}

When I remove the transistor, I get 8 V across the bridge, but 0 across the load.

^{simulate this circuit}

Either way, it's not turning on the light bulb or even a LED.

Final objective

Trying to switch on the load when the room gets darker (as resistance of LDR increased).

At ambient room light, I have set the Wheatstone bridge to be balanced or output a negative voltage.

As the darkness increases, resistance of the LDR increases, making the Wheatstone bridge un-balanced.

In our case, the voltage divider on right side (VRout), will decrease as R4 will remain constant whereas LDR1 will keep increasing.

At the end, the differential voltage between, VLout-VRout >=0.7V .

By connecting VLout to the base of an NPN transistor and VRout to ground, I am trying to switch the transistor to active state, in turn switching on the light bulb.

^{simulate this circuit}

Answer 1

There are too many reasons why your circuit fails.

1. Lack of low impedance current loop.
2. Lack of differential bias threshold for each Vbe = 0.7V (consider 0.6 your bridge reference +/-0.1 for gradual ON/OFF)

This design I made solves both but is still weak due to the relatively low resistance of the bulb compared to the battery ESR for a Panasonic Alkaline which all have a wide tolerance, based on the short circuit transient current.

The Threshold is around 1.4V and the resistance of the Rce of the driver NPN for a PN2222A is around 1 Ohm depending on saturation current. I used a 3.3W 9V bulb to match your bulb R which rises from 1.2 ohm to around 11x this value near 2/3 A.

Adjust R ratios for night Vbe=1.4V and 1.2V to minimize battery drain in daylight.

Corrected design

In order to turn on 12 to 14 ohm lamp with 10 k pullup to base, LDR must be much higher and the current gain must be much greater than this R Ratio of almost 1k. so two transistors in series as a Darlington, as shown above. These hFE's current gains multiply to enable saturation (low Vce) where the current gain is reduced to nearly 10% of rated hFE. Otherwise Vce rises until the gain is achieved.

In your case, the emitter resistance was far too high to conduct and bulb current to ground and back thru the batter to the bulb (even if Vbe was sufficient) So no bias voltage and no low R current loop.

Final remarks

A good alkaline battery can store 5 watt-hours [Wh]. A 9V bulb with these resistances cold and hot is a 3 W bulb. The battery will have a very short life and LED's are far more efficient.

There is no better design result than this one, if you restrict it as requested with BJT's, 12.3 ohm tungsten lamp at 9V with a battery and have no other requirements.

Answer 2

1. Your transistor is an emitter follower.

   A BJT is not controlled by the base terminal. It is controlled by the base terminal and emitter terminal relative to each other.

   BAT2 is referenced to GND and therefore applies voltage to the base relative to ground. However, the base relative to the emitter is what the transistor cares about. It does not know or care what you have defined ground in your circuit to be. It cannot see that ground because it has no terminals connected to it. It only cares about what's between the base and emitter.

   So imagine your transistor does conduct. Current flows through your wheatstone bridge causing a voltage drop across it which makes the voltage of the emitter node relative to ground rise. If you are applying a ground referenced voltage to the base, the the base voltage relative to ground remains the same while the emitter voltage relative to ground rises. The end result is that the difference between base and emitter gets reduced, potentially to the point where the transistor can no longer conduct enough to do its job. Negative feedback. The more it conducts the more it resists conducting further.

But that point is moot because you have more fundamental issues with your circuit:

2. You have a short-circuit across your lamp. So your lamp will never turn on.

3. Your "wheatstone bridge" makes no sense in a bunch of ways. The voltmeter put across isn't what you do for a wheatstone bridge, but even bigger is that you said you are trying to use your wheatstone bridge to switch a lamp. Wheastone bridges don't switch anything. They measure. Transistors switch things, but it was already established your transistor was wired up incorrectly, but even if your transistor was wired up correctly it has nothing controlling the base-emitter that would cause switching to occur.

   I know you say BAT2 is a temporary stand in for a differential voltage to feed to the transistor. However, I see nowhere from which you can derive that in your circuit nor do I see how you plan to apply a differential voltage to a single transistor.

You might want to go back to the drawing board for this one. Start simple with one function (like just switching on the lamp manually) and and build the circuit up to the functionality you want. It looks like you skipped a few steps and fumbled along the way.