DTL NAND - understanding

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Hi 'freaks,

here is a NAND made in DTL from a german wiki page

german names:
we use "U" for voltage and "V" for valve.

I understand most of the circuit. R1 keeps the first node at 5V until V1 or V2 conducts. Which happens when U1 or U2 go to ground. V5 and Rc work as an inverter.

I don't understand what V3 and V4 do. Are they supposed to decrease the BE voltage of V5 by two diode-drops? And why goes R2 to ground? Shouldn't it be in series with V4 to limit the base current? Well, R1 already limits it. Then ... does it work as a voltage divider? But if yes, why V3 and V4?

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Quote:
I don't understand what V3 and V4 do
They make sure that V5 if OFF. When either V1 or V2 are grounded there is a voltage drop of about 0.7V at the junction with R1. If that was going directly to the base of the transistor it would never turn off.

R2 makes sure that the base is not affetted by possible leakage.

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

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And if the gate follows another gate the output
of the first gate will not go perfectly down
to 0 Volt, but will be somewhere below 0.5V.
The two diodes switch of V5 even in this case.

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Quote:
They make sure that V5 if OFF. When either V1 or V2 are grounded there is a voltage drop of about 0.7V at the junction with R1. If that was going directly to the base of the transistor it would never turn off.

In other words: If U1 is grounded, then V1 conducts. If there was no diode-drop, then the Node1 (R1-V1-V3-V2) would be at ground. Actually it is at ~0.7V. This makes V5 conduct, which isn't wanted. R2 doesn't change this, because it's parallel. Is this correct?

How do V3 and V4 solve this problem? I guess they are closed because Node1 is at 0,7 V (one diode-drop) but at least 1,4 V (two diode-drops) is needed to conduct two diodes. Now R2 pulls the base down to ground.

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Look at it this way: with V1 and V2 floating, there are two series resistors and two series diodes between the rail and ground - so assuming 0.6v voltage drop for each diode, the voltage at the base wants to be

(5 - (2*0.6))/2 = 1.9v

However, the diode between the base and emitter in the transistor will want to keep 0.6v across itself. Since that is somewhat lower than the 1.9v, there will be current flow into the base and the transistor will turn hard on.

Now if one of the V1 or V2 (or both!) are grounded, the voltage upstream of V3 must drop to 0.6v - and V3 and V4 want to drop a further 1.2v. That would take us into negative volts, and we don't have any of those, so the voltage at the base is now tied down to ground by R2 and the transistor is turned off.

If the V3 and V4 were not present, the circuit could only work if the voltage drop across the input diodes was guaranteed to be greater than that across the Vbe of the transistor - tricky. It would also only work if the signal was actually grounded - which is not guaranteed if it's being fed from a previous gate; it's more likely to be 0.1 to 0.2 volts, depending on the transistor's saturation voltage.

Putting the diodes there (and assuming the same Vak drop in all of them) means that the input level need only drop below Vak+Vbe - approximately 1.2v - to be a guaranteed zero logic input.

So in brief, the diodes are there to improve the noise tolerance of the inputs.