# Voltage Substractor simple circuit

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
Mentor
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. • • DaveE and etotheipi
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). 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
berkeman
Mentor
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: • etotheipi
berkeman
Mentor
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 ________ berkeman
Mentor
I don't agree see the Common Collector amplifier has a slightly different construction.
How so?

Will this work? 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.

• DaveE and hutchphd
Hmm ok.

Baluncore • berkeman
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?

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.

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.

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