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I found this image of a transistor-based and gate:

enter image description here

and I can't understand what the purpose of the 4.7K resistor is. If the component controlled by the gate is placed between "Out" and ground, then that resistor is just a waste of energy since the circuit should work with or without it.

What am I not understanding?

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    \$\begingroup\$ If you removed the resistor, what would be the lowest voltage that the output node could reach? \$\endgroup\$ – uint128_t Aug 31 '17 at 2:29
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    \$\begingroup\$ More to the point, what would the highest possible output be? \$\endgroup\$ – WhatRoughBeast Aug 31 '17 at 2:50
  • \$\begingroup\$ Without that 4k7 resistor, what would be the output at no-input state (i.e. A and B are floating, as shown in the schematic)? \$\endgroup\$ – Rohat Kılıç Sep 1 '17 at 9:42
  • \$\begingroup\$ @a9556: This is a bad AND gate. The H level at the output will be about 0.6V lower than the H levels at the inputs, i.e. after a few such gates the H level is unuseable. \$\endgroup\$ – Curd Sep 1 '17 at 11:02
  • \$\begingroup\$ @a9556: In digital logic usually a L is not represented by floating (not connected to anything) but by 0V (connected to GND). That's why you need the resistor. \$\endgroup\$ – Curd Sep 1 '17 at 11:06
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First, let me tell you that this AND gate is one of the worst AND gate implementations. It has the several problems. Two of them:

1) It lacks regeneration (ok, this is a common issue on other discrete-component implementations too).

2) Under certain circumstances, it could not even work. If B is high (6V in your circuit), then the output will be: ( Vb - Vbe ) * R2 / (R1 + R2), where R2 is 4.7k and R1 = 10k. In other words, 1.69V. In some cases that value could be too high to be considered a "0".

Similar approach (just with a higher input current) could be achieved with a diode-resistor AND. Still, this would not have the issue of point 2.

That said, the 4.7 kOhm serves as pull down resistor, i.e. to create a "0" output signal, when A or B is "0". Furthermore, a "low" resistor value (few kOhms) is mandatory for the case of B = "1" and A = "0" to work correctly (see point 2). In fact, if it were, let's say, 470 kOhm, with B="1" and A = "0" (Vb = 6V, Va = 0V) you would get Vout= 5.2 V, as the base-emitter junction would act like a diode...

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  • \$\begingroup\$ Why would the output voltage depend on the resistor, which is in parallel to the component? I thought the voltage in each loop is 6V no matter what other loops there are... \$\endgroup\$ – a9556 Sep 2 '17 at 13:22
  • \$\begingroup\$ Because, if A="0", the top BJT will be off, leaving the collector of the bottom BJT in high impedance (i.e. as if there were nothing connected to it). In other words, all that remains is a series of the 10kOhm resistor, the base-emitter junction of the bottom BJT (i.e. a diode) and the pull down resistor. The equation for that loop is: Vb = I * R1 + I * R2 + Vbe (where R1 = 10k, R2=4.7k). Therefore, solving for I * Re (i.e. Vout), you get the expression written on the issue 2, shown in the answer. \$\endgroup\$ – next-hack Sep 2 '17 at 13:50
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If both transistors are on, and there is no resistor, you are shorting out the power supply

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  • \$\begingroup\$ No. If both transistors are on, and there is no resistor to ground (this is NOT equivalent to say "no RESISTANCE to ground") you are NOT shorting the power supply. Unless you say: "and instead of a resistor there is a short". But then the circuit would have no use at all, as OUT = 0 permanently... \$\endgroup\$ – next-hack Sep 1 '17 at 16:23
  • \$\begingroup\$ (I took too much to edit the previous comment) And even with a 0 Ohm, you would not actually short the supply: at least one BJT would operate in its linear region, supplying a Ic which is hfe (100-300) times the Ib (less than 600 uA). This would lead to an Ic between 60 and 180 mA. If you manage to keep Tc=25°C both BJTs survive (see datasheet) the large dissipated power (as much as 1W). \$\endgroup\$ – next-hack Sep 1 '17 at 16:34
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The "primitive" "AND" gate needs an emitter pull down with inputs A or B at 0 V then Vout = High if both inputs are high (5V) and then R reduces the turn off time for a given xx pF load when goes low.

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If the output is connected to a modern 'normal' gate (TTL or CMOS) it will not work without the resistor. It will work with RTL if you have access to such museum pieces from the 1960s.

So it really depends on what you imagine might be the characteristics of whatever you will connect it to.

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This Resistor causes "emitter-degeneration". There are several pros and cons for using such a resistor. I would recommend to use google for further reading

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  • \$\begingroup\$ no it's not the resistor that causes degeneration of the signal; it's how the transistor is used. \$\endgroup\$ – Curd Sep 1 '17 at 11:04

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