Voltage Substractor simple circuit

[Helena Wells]

I have created a voltage adder using this circuit where Va and Vb are the input voltages.

And Vout is between the emitter and the resistor. How can I modify it to a voltage substractor?

 

[berkeman]

You show an Emitter Follower configuration there, and yes the voltage gain is positive (and close to 1) through it.

Have you looked the "Common Emitter" configuration with an NPN transistor yet? The gain through that circuit is negative. Have a look at CE circuits, and learn how to set the voltage gain close to negative 1.

And a step better is to start learning opamp circuits. There you can get much more accurate gain control and biasing for things like adders and subtractors.

 

[Helena Wells]

You show an Emitter Follower configuration there, and yes the voltage gain is positive (and close to 1) through it.

Have you looked the "Common Emitter" configuration with an NPN transistor yet? The gain through that circuit is negative. Have a look at CE circuits, and learn how to set the voltage gain close to negative 1.

And a step better is to start learning opamp circuits. There you can get much more accurate gain control and biasing for things like adders and subtractors.

This is just a common emitter amplifier with 2 inputs instead of 1.

 

[Svein]

What you have drawn is a primitive "OR" gate (if one input is "high", the output is "high". Here is a schematic of a "NOR" gate (common emitter instead of common collector).

 

[Helena Wells]

What you have drawn is a primitive "OR" gate (if one input is "high", the output is "high". Here is a schematic of a "NOR" gate (common emitter instead of common collector).
View attachment 267043

Actually 2 inputs with 2 voltages are supposed to increase base current even more than 1 voltage source - > more collector current and less voltage drop on the transistor so Vout will be close to Vcc.

 

[Svein]

Googling "voltage subtractor circuit" return a lot of answers. Her is one circuit:

(https://circuitdigest.com/electronic-circuits/differential-amplifier-or-voltage-subtractor-circuit)

 

[Helena Wells]

Googling "voltage subtractor circuit" return a lot of answers. Her is one circuit:
View attachment 267044
(https://circuitdigest.com/electronic-circuits/differential-amplifier-or-voltage-subtractor-circuit)

Yes I know but I don't want op amps involved.Only common emitter amplifiers.

 

[berkeman]

This is just a common emitter amplifier with 2 inputs instead of 1.

No, you've drawn a "Common Collector" or Emitter Follower configuration. The Common Emitter configuration is different:

https://41j.com/blog/wp-content/uploads/2014/12/npn_configs1.png

 

[berkeman]

Yes I know but I don't want op amps involved.Only common emitter amplifiers.

CE amps are inverting (voltage gain is negative), and Emitter Follower amps are non-inverting. If you really want to build a subtractor, you need at least one part of that circuit to be inverting, no? And the other part to be ________

 

[Helena Wells]

No, you've drawn a "Common Collector" or Emitter Follower configuration. The Common Emitter configuration is different:

https://41j.com/blog/wp-content/uploads/2014/12/npn_configs1.png

View attachment 267046

I don't agree see the Common Collector amplifier has a slightly different construction.

 

[berkeman]

I don't agree see the Common Collector amplifier has a slightly different construction.

How so?

 

[Helena Wells]

Will this work?

 

[Helena Wells]

How so?

What you have drawn is an inverter not a common collector amplifier.

 

[Baluncore]

Yes I know but I don't want op amps involved.Only common emitter amplifiers.

Voltages can be converted to currents by resistors.
Current Mirrors use pairs of common emitter transistors to add and subtract currents.

Would you consider using current mirrors to make the voltage arithmetic circuit.
https://en.wikipedia.org/wiki/Current_mirror#Circuit_realizations_of_current_mirrors

If you can define the arithmetic problem, I can design a mirror.

 

[Helena Wells]

Hmm ok.

 

[Baluncore]

 

[Helena Wells]

View attachment 267049

?

 

[Baluncore]

The current that flows through Q1 will mirror through Q2, because their base and emitter voltages are the same.
The current that flows through Q3 will mirror through Q4.
The current that flows through Q5 will mirror through Q6.

Q1 is the total summing currents. Q1, Q2, Q5, Q6 are all equal.
Q3 is the total subtracting currents. Q3 and Q4 are equal.

The Q4 current is removed from the collector of Q6 before the remainder flows through R5.
The output voltage is developed across R5.

Do you have a specific question?

 

[Helena Wells]

The current that flows through Q1 will mirror through Q2, because their base and emitter voltages are the same.
The current that flows through Q3 will mirror through Q4.
The current that flows through Q5 will mirror through Q6.

Q1 is the total summing currents. Q1, Q2, Q5, Q6 are all equal.
Q3 is the total subtracting currents. Q3 and Q4 are equal.

The Q4 current is removed from the collector of Q6 before the remainder flows through R5.
The output voltage is developed across R5.

Do you have a specific question?

I want to substract voltages.

 

[Baluncore]

If you put voltage X on a sum input and voltage Y on a sub input,
then the output voltage will be X - Y.

 

[Svein]

Actually 2 inputs with 2 voltages are supposed to increase base current even more than 1 voltage source - > more collector current and less voltage drop on the transistor so Vout will be close to Vcc.

1. If you want an addition circuit, remove the diodes (they turn the analog circuit into a digital)
2. If you want to subtract, invert the signal you want to subtract and then add (a standard common-emitter circuit will invert the signal). Figure out:
   1. How to ensure that the gain of the inverter is 1.0
   2. How to ensure that the DC level of the inverted signal is the same as the non-inverted.

 

[Helena Wells]

If you put voltage X on a sum input and voltage Y on a sub input,
then the output voltage will be X - Y.

OK.

 

[Helena Wells]

1. If you want an addition circuit, remove the diodes (they turn the analog circuit into a digital)
2. If you want to subtract, invert the signal you want to subtract and then add (a standard common-emitter circuit will invert the signal). Figure out:
   1. How to ensure that the gain of the inverter is 1.0
   2. How to ensure that the DC level of the inverted signal is the same as the non-inverted.

If I remove the diodes current may flow from 1 input to a second input and I don't want that.And I have posted another circuit which makes the subtraction.

 

[Svein]

If I remove the diodes current may flow from 1 input to a second input and I don't want that.

Then isolate the input voltages using emitter followers.

 

[Helena Wells]

Then isolate the input voltages using emitter followers.

And make it even more complex? No thank you! Simplicity is king.

 

[hutchphd]

The circuit you originally proposed will not work as you hope.
The base current will be provided almost entirely by the higher input voltage source and the output voltage will be that voltage (minus a diode drop). As @Svein has pointed out it will act as a weird OR gate.
You need to understand this. Can you play with these circuits somehow (either on the bench or simulation)?

 

[Svein]

And make it even more complex? No thank you! Simplicity is king.

Sigh. Do you want accuracy or simplicity? Your adder will not work as an adder (as several others have mentioned). I have not seen your subtract circuit.

I think I will have to paraphrase what a university prof once said: "If there is no requirement that the circuit should work, I can make it really simple".

 

[Borek]

Actually 2 inputs with 2 voltages are supposed to increase base current even more than 1 voltage source

Unless I am horribly wrong - no. The output of your circuit will be that defined by the higher of the two voltages V_A and V_B.

 

[Averagesupernova]

Unless I am horribly wrong - no. The output of your circuit will be that defined by the higher of the two voltages V_A and V_B.

You are NOT horribly wrong.

 

[Helena Wells]

Unless I am horribly wrong - no. The output of you

You are NOT horribly wrong.

Ok but what if the voltages are the same?

 

[DaveE]

And make it even more complex? No thank you! Simplicity is king.

The problem with this statement is that while the schematic may be simple, the basic BJT device is anything but simple. Look at the schematic of a simple op-amp, like LM741. The reason they have so many transistors is to compensate for non-linear and non-ideal effects. In practice, for analog circuits you will most often see transistors used in pairs to help compensate for the imperfections of the basic devices. Analog transistor circuits just aren't simple in the real world.

It's really great that you are exploring this because it is the best way to learn, at a fundamental level, about analog circuitry. However, in a real EE lab a simple op-amp is the same size, the same cost, and works better than a single discrete transistor. Discrete transistors, in practice, are use for high frequencies, high power, and the simplest digital/switching applications.

Also, as I think has already been said, in your original circuit, if one diode is on the other will be (mostly) off. It is a classic logic circuit, like voting. Not a voltage input adder at all. You can see this if you consider a diode model that acts as a perfect switch, either 100% on or 0% on.

 

[berkeman]

After a bit of thread cleanup, this thread is done. Thanks everybody for trying to help Helena with her growing understanding of electronics.