Push/pull from different rails query

[SparksFly]

Hi,

I have a query regarding transistors in push/pull configuration for switching between +ve/-ve rails.

Referencing the example at the top of this page http://www.ecircuitcenter.com/Circuits/pushpull/pushpull.htm

In my application, VS1 will be a 0 to 3V3 digital signal, which I want to convert to +/-5V (eg 0V becomes -5V, 3V3 becomes +5V).

I understand 3V3 to saturate the NPN b-e, resulting in output +5. However I'd need -0.7V instead of 0V input to saturate the PNP side because the output is referenced to 0V. Correct?

How would I go about resolving this? I thought I could perhaps add a potential divider to the output to replace the 0V with ~1.5V. I haven't seen any diagrams online that do this, so I would guess that's because it's a bad idea.

I'd prefer to use basic components if possible (no op-amp, comparator or ICs etc).

Can anyone point me in the right direction?

Thank you!

 

[Bluejets]

What do you intend to use it for..??

You do realise the circuit you have shown is used just to explain the operation.
Actual will be more to it than shown above.

 

[AnalogKid]

That circuit (and the others on the linked page) will not do what you want to do. A 3.3 V signal will not saturate an NPN emitter follower whose collector is at 5 V. It can be done with 4 transistors, 2-NPN and 2-PNP.

ak

 

[Audioguru]

You must learn about how transistors work so see the output voltages of this circuit.

 

[Harald Kapp]

This is one (out of many other) possibility how to construct such a circuit:

It is by no means optimized for component values, speed, power consumption etc.
It is only meant to visualize the complexity of such a circuit as you wish to build.

 

[SparksFly]

Thanks. I do know that the circuit wont work, I'm hoping to get some advice to design it so it is working.

A 3.3 V signal will not saturate an NPN emitter follower whose collector is at 5 V. It can be done with 4 transistors, 2-NPN and 2-PNP.

Ok, do you mean configured as two darlington or Sziklai pairs? One for the push and one pair for the pull? If so, I still cant see how I'd achieve the -V switching.

You must learn about how transistors work so see the output voltages of this circuit.

I can understand that if Vin=0 then Vout = 0. How do you get 2.6V though?

This is one (out of many other) possibility how to construct such a circuit:

Thank you, I will study it now.

 

[Harald Kapp]

I can understand that if Vin=0 then Vout = 0. How do you get 2.6V though?

The voltage at the emitter is the voltage at the base (3.3 V) minus the base-emitter voltage V_BE which is ~ 0.7 V.

 

[SparksFly]

Aha, thank you! I've downloaded LTSpice and found the source of my misunderstanding. I was following the "transistors are like switches" analogy too closely. I was treating it a bit like a relay switch.

Resolved, thank you!

 

[Audioguru]

An emitter-follower has a voltage gain of almost 1 but also has a voltage difference of -0.7V. Therefore with its base at +3.3V its emitter is at (3.3V - 0.7V=) +2.6V.
A transistor with its emitter at 0V and its output at its collector resistor has a voltage gain of up to 200 times (!) so an input of 3.3V to its series base resistance will saturate it and its output will be close to 0V.
A darlington-follower also has a voltage gain of almost 1 but has a voltage difference of 1.4V or more.