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Sheen140
modified 5 years ago

5 Volt Linear Regulator

6
11
253
03:56:22
19v-7.5v input, 4.95-5.45v output
published 5 years ago
PrathikP
5 years ago
I see that there is not voltage reference. Without a voltage reference, you can't get a reliable output voltage. Infact, any changes in the line or load will cause the output voltage to change. In order to make a regulator, you must use a voltage reference. Without that, you can't make a voltage regulator.
Sheen140
5 years ago
@PrathikP If I'm not mistaken this circuit does have a voltage reference. The two diode connected PNP transistors have a constant 0.6 V drop (emitter to base voltage drop) that acts as the voltage reference.
PrathikP
5 years ago
That's only for the current source.
PrathikP
5 years ago
See this http://everycircuit.com/circuit/4745379776823296 this small change makes the regulator massively better.
Sheen140
5 years ago
@PrathikP I suppose it depends what you mean by "better"! Yes using a zener as a voltage reference before the common collector buffer will certainly dramatically reduce voltage variations at the output. However, real world zener diodes require a minimum bias current of at least 5ma in order to provide a stable voltage drop, which means using one in such a circuit will add up to 0.1 Watts of continuous power dissipation. Additionally, using a zener like you have in your circuit limits the output voltage of the circuit to that of the zener, whereas in the original circuit you can set the output voltage to a wide range of voltages (say 3.3 volts, or 2.4, or 5.8, etc). It all depends on what your looking for. If you need a ultra high precision and well regulated output voltage and you aren't bothered by the minor inefficiency a zener reference like the circuit you posted is the way to go. However if you can deal with up to say 0.4 volts of variation over the range of input voltages and you're more concerned with low power draw and fewer components, I'd still say the original circuit could be feasible! But hey I could just be biased! ;)
PrathikP
5 years ago
Punny, but 5mA * 5V is 25mW and not 0.1W! You are using a 20k resistor to set the output voltage, which is fine if the drop across the 20k is sensed by an op amp, but since it's sensed by a BJT that'll create huge problems, like the poor load and line regulations. A zener diode is cheap and easy to work with and you get huge improvements for a small increase in cost. As for varying the voltage in the original circuit, you would need to make modifications.
PrathikP
5 years ago
But if your load is going to be no heavier that 20kohms and you don't mind a couple 100mV or voltage swing, I guess youre good to go with what you have
PrathikP
5 years ago
If you're interested, I've improved the ripple rejection with just an extra resistor and 2 capacitors http://everycircuit.com/circuit/5836464242556928
Sheen140
5 years ago
@PrathikP you would have to calculate the power dissipated by the input voltage (in this case a maximum of 20V) times the 5mA bias current of the zener, giving you a power dissipation of ~20V × 5 mA = ~100mW. Still you are right this is not a major problem at all for 99% of voltage regulators (I mean for most applications who cares about an increase of 0.1 W in power drop if we get such a dramatic increase in line regulation?)
Sheen140
5 years ago
@PrathikP Actually after rereading your comments I think I mostly agree now. I was trying to make a version of a band gap voltage reference but it seems like the simplicity comes at the expense of very poor line regulation (wikipedia seems to say even the worst linear voltage references should only have about 0.1% line variation, whereas mine is like what almost 10%?) I think maybe I was just being too biased at first! Thank you for the analysis PrathikP, I appreciate the wisdom!
PrathikP
5 years ago
A bandgap voltage reference is quite complex. Take this for example http://everycircuit.com/circuit/4976871836745728

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