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I am building the following OR gate:

schematic of discrete OR gate

What is the purpose of the 4.7k ohm resistor at the bottom? Will the circuit work without it?

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  • \$\begingroup\$ There are two buffers connected in parallel (If you delete one transistor then you will get an emitter follower). So, the purpose of the emitter resistor is the same as in an emitter follower. About the second question, well, it depends on where the output is connected to. \$\endgroup\$ – Rohat Kılıç Apr 15 '17 at 21:26
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Either transistor can drive the output 'high', though this kind of rudimentary gate loses voltage from input to output, so a chain of many 'or' gates will, at some number, fail to function.

The resistor pulls the output low when both transistors are off.

Whether it will 'fail to work' without it depends on what it is connected to and how logic levels etc. are defined, so that's not an answerable question.

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With Emitter Followers, the Re must always be just enough smaller than your load, so DC current must exceed trace capacitance AC current.

It affects the slew rate from emitter current.

When driving capacitance on cable or tracks with x pF/cm, the same rule applies and Ie=CdV/dt and Re*C=T defines the rise & fall time.

Also for best speed Rb/Re ratios are well defined with compromise for efficiency. Typ 10:1

Even Re=1k can be a problem >10MHz.

enter image description here

So old CMOS is Zout= 300~1k for 15~3V (CD4xxx) and 50 ohms for 74HCxx and 25 ohms for 74ALVCxx for output impedance.

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