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Logic gate output won't go low enough to turn off NPN transistor

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I'm trying to get a 74HC00 (NAND) logic gate's output to turn an NPN transistor (BC337) on and off. But I'm finding the logic gate's output, although an oscilloscope shows it to be going as low as gnd (and below due to ripple) and as high as the 5V power rail (plus some ripple), doesn't seem to ever go "low" enough to switch the transistor off.

Measuring the transistor's base (separated from the NAND's output by a 1K resistor to prevent the NAND gate driving too much current) shows it constantly sitting at the 1.2V or so above ground one would expect to see when it is on, and the NPN's collector, at the the lower end of a high side load, stays at the sort of low voltage one would expect when a current is flowing through the load.

I tried pulling the NAND gate's output to GnD with another 1K resistor, transistor still won't turn off when connected.

I then resorted to having a 10nF cap in series after the 1K resistor, the other end of this cap is held just above ground by a potential divider between the power rail and gnd, so that the cap lets the signal vary about this reference voltage, decoupled from any DC offset in the NAND's output. This way I can get the transistor to go off and on, but it decoupling its base from DC through the capacitor seems an excessive number of components to need just to get a signal from a logic gate output to a transistor. Is there a way which needs less components and doesn't give a situation where if the logic signal were further attenuated by more than 1K  it would no longer be able to get high enough to turn the transistor on.

This is a fast signal, square wave with a frequency of several MHz, rise time and fall time not important so long as it can rise and fall the full voltage in the time available for a period.


The pinout of the BC337 is different to, eg, BC547.
Check that first.

Could you draw out a schematic showing all the details, including the load that the transistor is driving? 

Are you saying you have an output of 0V at the output of the 74HC00, that output is connected to the base of the transistor by a 1K resistor, and the voltage at the base of the transistor is 1.2V?  Is this all true in the steady-state (DC) condition?  Is there any other component connected to the base of the transistor?

Benta beat me to it, but I was also going to ask--are the pins of the transistor connected correctly?

Andy Watson:
If Benta's solution is not correct:
The transistor is probably saturated during the "on" time. It takes extra time to recover from saturation and remove the stored base charge. A quick fix might be to reduce the impedance of the base drive circuit - make the base drive from a potential divider between the output of the logic gate and ground. Aim to get the impedance seen by the base to 100\$\omega\$ or less. Or use a schottkey diode between base and collector to prevent saturation.
Using a low base drive impedance would also mitigate the effects of miller capacitance too.

If you want to use the speed-up capacitor trick - put it in parallel with the 1k resistor.

Try 10K or 100K with 100pf in parallel.


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