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I've been studying and building JFET circuits to serve as a high-impedance input to a DIY guitar amp. In my research on voltage divider biasing, I've come across a circuit in Nuts & Volts Magazine that has me perplexed. It's shown on the left in the diagram below.

What I've seen in textbooks and other examples on the internet has always looked more like the circuit on the right.

enter image description here

I'm trying to figure out what the advantage of a gate resistor (R3) is. Resistor values were not shown in the original circuit, but I got the feeling R3 would be on the order of 1M or more with R1 and R2 being lower. This would keep the input impedance high, while allowing more significant current to flow through R1 & R2.

My question is this: What is the advantage of having R3 as opposed to using large value biasing resistors for R5 & R6 in the textbook circuit? Couldn't I get a 1M input impedance using 2.2M resistors for R5 & R6 just as easily as using a 1M R3 and lower value R1 & R2? It seems the only difference is more current flowing through R1 & R2. How is this an advantage?

The only thing I can think of is that such an arrangement would allow a person to replace R1 and R2 with a potentiometer and adjust the bias without adversely affecting input impedance. But beyond that, I'm struggling to see a reason.

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    \$\begingroup\$ What do you see when you replace the DC biasing circuits with their respective Thevenin equivalents? \$\endgroup\$ Commented Jun 8, 2021 at 16:25
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    \$\begingroup\$ @RohatKılıç, That's just it. If I use R5 & R6 = 2.2M, the input impedance is the parallel combination of R5 & R6, or 1.1M. If I use R3 = 1M and 10k for R1 & R2, input impedance is R3 in series with R1|| R2, or 1M + 5k. Not a significant difference. No current should flow through the gate (or R3), so the voltage should whatever the divider network is set for. So why would I use three resistors to accomplish what I can do with two? That's what I'm not seeing. \$\endgroup\$
    – Dave H.
    Commented Jun 8, 2021 at 16:45
  • \$\begingroup\$ @G36, can you elaborate on the capacitor across R2? I can see that it would bypass R2 for AC signals, but if R2 were on the order of 10k, would the difference be significant? I can definitely see an RC filter with R3 and that capacitor to filter AM radio interference, so I'll try it. \$\endgroup\$
    – Dave H.
    Commented Jun 8, 2021 at 17:20

7 Answers 7

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If you add a large capacitor across R2 it will make a difference. In that case, the power supply noise won't be amplified by the amplifier.

schematic

simulate this circuit – Schematic created using CircuitLab

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Historically, high value resistors (especially carbon film types) have tended to have a lot more drift (both with temperature and with time) than lower values. If one of the resistors in the divider shifts proportionally more than the other the bias point drifts.

If the series high value resistor shifts in value there is no great change. For example, here is a snippet from a Yageo datasheet for their carbon film resistors:

enter image description here

By using the lower values with a series resistor, the bias drift with temperature is reduced by a factor of 5:1 or so. A considerable improvement for just adding a cheap resistor.

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  • \$\begingroup\$ It's worth noting that any JFET bias here will have to be a quite small voltage in order not to forward bias the gate, so you have R1»R2. That makes the circuit on the right have much larger resistors. \$\endgroup\$
    – user107063
    Commented Aug 7, 2023 at 21:23
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The usual reason for using the left hand circuit instead of the right hand circuit for biasing is to enable the placement of a fairly large electrolytic capacitor across R2 to stop noise & ripple from the supply getting into the signal path. With the right hand circuit the equivalent filtering technique would be to split R5 into two resistors, a small resistor (top one) and a larger resistor (lower one) and insert a fairly large electrolytic capacitor from the junction between them and ground. The drawback of this second filtering technique is that it reduces the input bias point below mid-supply level when used to bias an op amp.

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Actually, it looks like you already have the answer and your assumption is correct.

Both circuits are identical.

The only advantage of the left-most circuit is that it allows you to use a single very large resistor (R3) instead of two, in case it's hard or impossible to find higher values. For example, imagine you only have 1M resistors in your inventory and you can't find or buy 2.2M resistors, so you can use low-value resistors (e.g. 10k) along with only one 1M resistor to keep the input impedance around 1M and bias the amplifier.

It also may not be easy to find very large resistors having the exact ratio when you need a specific gate voltage. In that situation, the 3-resistor biasing may help. Because it would be easier with lower resistors to match an exact ratio (in other words, exact gate voltage).

That's it. Nothing special.

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  • \$\begingroup\$ Pehaps noise could be a factor? Lower value gate resistor means less thermal noise? \$\endgroup\$
    – Dave H.
    Commented Jun 8, 2021 at 17:16
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    \$\begingroup\$ @DaveH. could be. Also please see my edited answer. \$\endgroup\$ Commented Jun 8, 2021 at 17:18
  • \$\begingroup\$ @DaveH.Thermal Johnson noise is identical for Thevenin equivalents. After all, you could split the conducting layers of a resistor into lots of imaginary small parallel and serial resistors. There may be a difference regarding other noise types. \$\endgroup\$
    – user107063
    Commented Aug 7, 2023 at 21:02
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Another advantage of the left hand circuit, when decoupled with a large capacitor, is that the bias voltage can be shared between amplifiers, such as both channels in a stereo circuit, with the channels isolated from each other by R3.

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  • \$\begingroup\$ A clever trick to clone voltage in several places... \$\endgroup\$ Commented Jun 9, 2021 at 18:49
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Well, let me get involved in unraveling the role of the humble resistor...

Another clever trick is to connect the lower end of the capacitor not to ground but to the JFET source. Then the R3 left end will follow its right end and the R3 resistance will be enormously (virtually) increased.

The name of this clever trick is "bootstrapping".

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R3 sets the biasing current to be in some relation to the current induced by the input signal. aka. the biasing current is set to have less changes than the input then what the gate sees is the actual input signal and not some multiplication of both. in sum with the proposed capacitor the biasing current is in a bandwidth relation to the input signal and its noise is shunted through the capacitor hopefully.
Impedance is set by R3 plus R2. R2 and cap form a high pass shut to ground. thus R3 sets which -analog- bandwidth the bias has. the trio R3+R2/cap sets the whole of the input bandwidth.

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