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I have an IC which outputs pulses over an open collector pin. I need both these pulses and their inverses in my circuit. I want to do the inversion by using as few as possible transistors and without using any ICs.

enter image description here

First, I tried simply inverting the signal by a single BJT as seen below. But this won't work. Because, when Qin is on, SIG will be low, and when Qin is off, again SIG will be low since this time base of Q1 will ground the signal. The signal will be lost in any case. Connecting a resistor to the base of Q1 may seem to be a solution, but this way, the signal will be attenuated by this resistor network (it will be less than 12V).

enter image description here

Next, I tried first generating a copy of the signal by and emitter follower circuit (by Q1 in the image below), then do the inversion by this copy of this signal. But again the same thing happened; now the signal is lost through the bases of Q1 and Q2.

enter image description here

In my final attempt, I connected R3 to prevent the loss of signal. By doing so, the signal is not lost, but it is attenuated by the voltage divider network built up with R1 and R3. I added Q3 to pull it up to 12V level as seen below.

enter image description here

I think this final circuit will do what I want, but this, time the circuit looks too complex to me. Is there a simpler way to do this with at most two external transistors?

(Note: Only 12V is available in the circuit. The frequency of the signal (clock pulse) is 20kHz. The signal and its inverse will drive separate totem-pole transistor pairs. They should be pulled up by 10k\$\Omega\$ resistor at most when high, and should be directly shorted to the ground when low.)


EDIT:

I rebuilt up my circuit according to both Brian's and Vladimir's suggestions. I implemented two methods in one circuit. Then I simulated them. Here is the schematic:

enter image description here

Results of Brian's circuit:
enter image description here
(Results are almost similar without the totem pole stage.)

Results of Viladimir's circuit:
enter image description here
(Results are much more terrible without the totem pole stage.)

I'm loosing my hopes on obtaining crystal clear signals. It is only 20kHz and I'm having this much of troubles. Maybe I should give up my stubbornness and use a logic gate IC.


EDIT 2:

After Viladimir's comment about IRF530 being a wrong choice of MOSFET, I started to try out different transistor types. I got a very good result when I changed Q1 with 2N2222 and M1 with BSS138. The new and better signal timings are in the image below. And these timings are without a totem pole drive; I removed it; it doesn't make a difference now.

enter image description here

However, this was just a simulation. Will I get the same (or similar at least) results in the real life? If yes, what was the problem with Q1 and M1? What features of 2N2222 and BSS138 made this circuit run better? I will be glad if someone makes a short comment on this.

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In your second circuit, simply connect R2 from Q1 emitter to Q2 base instead of ground. –  Brian Drummond Jun 3 at 16:45
    
I think that's because the IRF530 is not suited for this application... –  Vladimir Cravero Jun 3 at 19:32
    
Regarding your updated question: I haven't looked at it in any level of detail, but my first guess looking at the data sheets is that the high input capacitance of the IRF530 was your main problem (670pF versus 27pF for a BSS138). –  Jules Jun 4 at 0:48
    
@Jule If 670pF was the problem, then how do they drive power mosfets with 28nF gate capacitance at 500kHz with a similar totem pole (link)? –  hkBattousai Jun 4 at 4:44

3 Answers 3

up vote 7 down vote accepted

Your first circuit is perfect... If you use a mosfet instead of a bjt. A mosfet does not pull anything low when it is on, it indeed is a capacitive load but you are going at 20kHz so that should not be an issue at all. Just be careful: your mosfet should support a quite high \$V_{GS}\$, i.e. 12V, that's not something all MOS are happy to do.

Your circuit should look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

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What you're looking for is a single-input, differential output buffer. There are lots of examples for this online, for example the last on on this page. It uses 2 transistors.

http://www.interfacebus.com/Transistor-Differential-Amplifier-Circuit-Description.html

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Problem with the differential amplifier is the choice of R3, and how it influences the logic threshold of the circuit. I think it needs a little more to be a useful logic buffer. –  Floris Jun 3 at 17:33

Your first circuit was almost there. The problem was your analysis.

But this won't work. Because, when Qin is on, SIG will be low, and when Qin is off, again SIG will be low since this time base of Q1 will ground the signal. The signal will be lost in any case. Connecting a resistor to the base of Q1 may seem to be a solution, but this way, the signal will be attenuated by this resistor network (it will be less than 12V).

When Qin is on, yes, SIG is low. This means INV is high. When Qin is off, the base emitter junction of Q1 is held low, but INV is low.

What you need is a resistor between SIG and the base of Q1. As you point out, doing this will attenuate the value of SIG, but this is not a problem. Replace R1 with 1 k, and put a 100k resistor from SIG to the base of Q1 (and, just to be safe, another 100 k from the base of Q1 to ground. SIG will be reduced by 1%. Is that really a problem? Otherwise, the circuit will work just fine.

By comparison, your last circuit will drive SIG appropriately, except that the output voltage will be no more than about 11 volts. Your currents are low enough that the emitter-base voltages will actually be less than the standard assumption of 0.7 volts, or the result would be even worse. As it stands, assuming 2N3904 transistors, the voltage drop across R1 is about 0.1 volts, and the drop across Q1 and Q1 about 0.45 volts each. Final SIG output is about 11 volts. If a 1% drop (.12 volts) is unacceptable, why is a drop of 1 volt OK?

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