I would like to send an audio signal from a Presonus HP4 headphone amplifier to a Zoom H4n "handy" recorder, but I am having difficulty understanding the role of impedance in this setting.

My thought was to simply use a 0.25-inch male stereo TRS (connected to one of the HP4 headphone outputs) to dual 0.25-inch male TS (connected to the two H4n "phone" jacks) "breakout" cable. However, the "Headphone Impedance Working Range" for the HP4 is listed as 32 Ohms to 600 Ohms; whereas the "Input impedance" for the (unbalanced) 0.25-inch "phone" inputs on the H4n is listed as 480 kOhms. So, if I am understanding this correctly, it seems to me that these impedances are "mismatched" by a factor of 1000 (which seems strange.)

Here are the relevant HP4 specs:

HP4 specs

Here are the relevant H4n specs:

H4n specs

I'll note that the manual for the H4n also states that the 0.25-inch "phone" inputs can be used for "instruments (e.g., electric guitar, keyboards) or line level sources". Given the stated input impedance of 480 kOhms, this sounds to me like a "Hi-Z" input; but these are also described as "phone" jacks in the manual. This leads me to one of the following three (not necessarily mutually exclusive) conclusions:

  1. This "discrepancy" of impedance is not actually all that important.
  2. I don't understand impedance well enough. (This is quite probable.)
  3. The H4n manual contains typos and/or "imprecise" language.

Perhaps someone could tell me which of the three conclusions listed above best describes what is going on here, or, perhaps, there are more conclusions that I——certainly a novice in this area——am overlooking.

(I'll note that the more mathematical details any response includes, the better.)

  • \$\begingroup\$ Seems backward. Why not send the INPUT of the headphone amp to the Zoom instead, and then the output of the zoom to the input of the headphone amp? \$\endgroup\$ Aug 1, 2021 at 17:59
  • \$\begingroup\$ The 0.25" connectors are often called "phone plugs" and "phone jacks" regardless of the actual application. \$\endgroup\$ Aug 1, 2021 at 18:02
  • \$\begingroup\$ No need to impedance match here. As long as voltage levels are compatible, simple connection will do. \$\endgroup\$
    – user16324
    Aug 1, 2021 at 18:14
  • 1
    \$\begingroup\$ Golden rule in audio: you don't impedance match, you impedance bridge; since 99% of the equipment is active and has some kind of input buffer you don't care about power transfer but only about voltage transfer. For pickups and similar the issue is on the frequency response since you are creating RC networks with the impedances and the cable capacitance (which is more than you would think). Also some pickups have complex output impedances which worsen the issue \$\endgroup\$ Aug 2, 2021 at 6:09
  • 1
    \$\begingroup\$ Ambiguous, but I suppose a better choice than "confessional" ;) \$\endgroup\$
    – LShaver
    Aug 2, 2021 at 17:51

4 Answers 4


It's actually number 1: It doesn't matter. An audio amplifier can (almost) always drive a lighter load than it is designed for.

The "maximum 600 Ohms" claim is most likely related to the output voltage that the headphone amplifier can provide; higher-impedance headphones need more voltage. This is irrelevant for your recorder.

  • 2
    \$\begingroup\$ I disagree with the "always" in your statement. There are some audio amplifiers that are unstable when driving loads that are higher impedance than they are rated for. \$\endgroup\$
    – jwh20
    Aug 1, 2021 at 17:49
  • \$\begingroup\$ Impedance matching at audio frequencies is about power transfer, so it's relevant after the power amplifier. Signal transmission is through voltage, and the power doesn't matter, so impedance matching doesn't matter either (within bounds, e.g a short would be too low impedance). \$\endgroup\$ Aug 2, 2021 at 10:44

That may come as a surprise, but all those different interfaces, such as headphone connector, line level connectors, and mic connectors need to have their own parameters such as voltage ranges and impedances to best work with the thing it is connected to.

So an headphone output connected to line or mic input may not sound so good because they don't have compatible impedances and voltage levels.

Frankly, connecting a headphone output to line input makes no sense, since the headphone amp has a line level monitor output for that purpose.

Now, as per your questions:

  1. The discrepancy is important and you need it. Modern audio equipment and interfaces use voltage as the means of audio signal transfer.

Therefore you generally want as low output impedance as you can get, and as high as input impedance you can get, so that an input sees the maximum voltage the output can provide.

If you match the impedances, you maximize received power at the input, but you lose half of the voltage. That's why impedance matching is not used in modern analog audio interfaces. Few decades ago, things were different, and equipment might have had say 600 ohm input and output impedances.

But at present, you don't need and don't even want to match the impedances.

  1. It may not about understanding impedance itself. It's just how it is used in audio gear and why.

  2. Not really, it defines voltage ranges and impedances for each interface.

  • \$\begingroup\$ Yes, I considered using the monitor outputs on the headphone amp, but I noticed that these are balanced inputs (and the 1/4" inputs on the recorder are unbalanced). Of course, I am now realizing that this likely doesn't matter... (I just don't currently have enough 1/4" cables lying around to test this.) \$\endgroup\$ Aug 1, 2021 at 20:30

The H4n is a great device, if pricey (but not overly priced.)

Balanced and unbalanced/single-ended

I guess the first thing to discuss is the meaning of balanced and unbalanced, since it's more a technical issue between experts and has far less real meaning to consumers. Between experts, as I've interpreted what I have read (as I'm no expert here, just a hobbyist trying to follow along at times), it appears that balanced means differentially amplified.

Many phantom-powered, XLR microphones (not all by any means) are balanced in exactly this sense. Figure 1(c) in an answer by Transistor provides an example of what I mean here.

Unbalanced is often taken as synonymous with single-ended. That would be like Figure 1(a) in the above link.

Microphones are transducers. Speakers are transducers. These are devices that convert between dynamic electronic signals and some dynamic physical property of interest. Since humans don't directly perceive (well) electronic signals directly, transducers are what make an electronic device useful to us.

Transducers can be operated in balanced mode and they can be operated single-ended, or unbalanced. In addition to the balanced XLR microphone design just mentioned, there are bridged amplifier outputs to drive speakers and these may also be considered as balanced outputs. Amplifier outputs can also be unbalanced -- that's quite commonly the case, in fact.

In between the input transducers and the output transducers, amplifier systems may have intermediate stages. And once again, these stages can also be balanced (differential) or unbalanced (single-ended.)

That's my hobbyist perceptions of what I've read from experts (Burkhard Vogel, for example.)

[Since I speak German as a 2nd language I can't help add that Der Vogel is "the bird" while Die Vögel is the plural form, as in "the birds."]

The point to the above is not to get too "hung-up" on the idea of balanced inputs and outputs. It's a specialist term used between specialists to discuss differential vs single-ended propagation of signalling. (There are noise and other technical reasons for caring, and/or paying for balanced signalling.)

The term does not necessarily also mean that the output impedance equals the input impedance, though that may be the case at times for technical reasons.

When should the impedance be matched?

When the specifications say they should be.

Or when you are wanting to deliver the maximum power into an output transducer or get the maximum power from an input transducer. There is a maximum power theorem that says that the maximum power is transferred when the output impedance matches the input impedance.

A low impedance output to a high impedance input works to get most of the signal voltage across, one to the other.That's the goal. This is neither a balanced/unbalanced question nor an impedance-matching question. It's just a matter that most amplifier systems operate with voltage-signalling and you want to get the maximum signal across, one to the other. Not the maximum power (except again in those rare cases where that's important.)

You still have to meet specifications, though. There are reasons not to have extremely high impedances involved, for example. (Noise is one such reason.) There are reasons where you want extremely high impedance. (A voltmeter is one such reason.)

So, read the specs and make sure that you are within range of what they can handle. If your input source tells you that it cannot drive a \$480\:\text{k}\Omega\$ input, then your H4n unbalanced input doesn't meet the specs and probably won't work as well as it should. But it's unlikely that you will have a microphone source that works poorly with such a high impedance input. Still, it's possible. I think the H4n designers figured that anyone using an unbalanced input isn't focused explicitly on noise issues as the most important factor, or else they wouldn't be using unbalanced inputs. So they focused instead on presenting as light a load as possible in order to have the least possible impact on the input source, despite noise considerations of doing so.

Just use the H4n and be happy. ;)

Maximum power theorem

Well, you said you wanted mathematics. :) Tip me over with a feather, will you?

Suppose you have a source of power which has an intrinsic source voltage and source impedance and you want to extract the maximum possible power. You have the freedom to choose the load impedance. So you want to know, "What's the optimal impedance to use, in this case?"


simulate this circuit – Schematic created using CircuitLab

$$\begin{align*} I_{_\text{TOTAL}} &= \frac{V_{_\text{SRC}}}{R_{_\text{SRC}}+R_{_\text{LOAD}}} \\\\ P_{_\text{LOAD}} &= I_{_\text{TOTAL}}^2\cdot R_{_\text{LOAD}} \\\\ &=\left[\frac{V_{_\text{SRC}}}{R_{_\text{SRC}}+R_{_\text{LOAD}}}\right]^2\cdot R_{_\text{LOAD}}\tag{1}\label{1} \end{align*}$$

You want to maximize Eq. \$\ref{1}\$. This is done by taking its derivative (slope equation) with respect to \$R_{_\text{LOAD}}\$ and solving for the case where that slope becomes zero. (Where the slope is zero is either a maximum or a minimum.)

The derivative is:

$$\begin{align*} \text{D}\bigg[P_{_\text{LOAD}}\bigg] &=\text{D}\left[\left(\frac{V_{_\text{SRC}}}{R_{_\text{SRC}}+R_{_\text{LOAD}}}\right)^2\cdot R_{_\text{LOAD}}\right] \\\\ \text{d}\,P_{_\text{LOAD}} &= \left(\frac{V_{_\text{SRC}}}{R_{_\text{SRC}}+R_{_\text{LOAD}}}\right)^2 \left(\frac{R_{_\text{SRC}}-R_{_\text{LOAD}}}{R_{_\text{SRC}}+R_{_\text{LOAD}}}\right)\text{d}\,R_{_\text{LOAD}} \\\\ &\therefore \\\\ \frac{\text{d}\,P_{_\text{LOAD}}}{\text{d} R_{_\text{LOAD}}} &= \left(\frac{V_{_\text{SRC}}}{R_{_\text{SRC}}+R_{_\text{LOAD}}}\right)^2 \left(\frac{R_{_\text{SRC}}-R_{_\text{LOAD}}}{R_{_\text{SRC}}+R_{_\text{LOAD}}}\right) \end{align*}$$

Setting that to zero, we know that the first factor can't be zero so we can focus on making the second factor equal to zero, instead. The numerator there makes this obvious:

$$\begin{align*} R_{_\text{LOAD}}&= R_{_\text{SRC}} \end{align*}$$

So, there's the maximum power theorem.

When does maximum power transfer matter?

One of the few times where any consideration of maximum power may be wanted is with headphones where their impedance can vary widely. These typically today vary from \$8\:\Omega\$ to \$600\:\Omega\$. (Only one manufacturer actually makes the same model in such a wide range, though.) Some older ones, difficult to acquire today, were designed for crystal radios when they were far more popular. The better quality ones may present typically somewhere between \$2\:\text{k}\Omega\$ and \$4\:\text{k}\Omega\$.

Higher-impedance headphones use very thin wire, on the order of 20 microns in diameter, offering a lower movement mass. Because the wires also leave less air between windings, the voice-coil's magnetic field is stronger. When a manufacturer takes good advantage of these benefits, the result is a more sensitive and better sounding product with lower distortion.

Suppose a cellphone designed to accept a \$32\:\Omega\$ headphone load that provides, at most, \$2\:\text{mW}\$ (maximum volume setting.) Assuming the cellphone's output impedance is very low by comparison, and it will be quite close to zero given the usual feedback factors involved, this implies \$V_{_\text{RMS}}\approx 250\:\text{mV}\$.

But what about jacking in a professional headphone, rated at \$600\:\Omega\$ impedance? That same cellphone would now only deliver \$100\:\mu\text{W}\$.

A \$32\:\Omega:600\:\Omega\$ audio transformer would match the impedance. The ideal turns-ratio would be about 1:4.33 -- computed as \$1:\sqrt{\frac{600\:\Omega}{32\:\Omega}}\$. Then the same \$600\:\Omega\$ headphone would be reflected over to the cellphone and presented as the expected \$32\:\Omega\$ load. (Add the transformer to the headphone first and then plug the resulting pair into the cellphone.)

Ignoring insertion losses for now, the RMS voltage appearing at the headphone is now \$V_{_\text{RMS}}\approx 1.1\:\text{V}\$ and delivers, you guessed it \$2\:\text{mW}\$, into the headphone load. (A real audio transformer will have insertion losses. But not uncommonly near \$1\:\text{dB}\$.)

More on this topic can be found in "Ch. 15: Audio Transformers" from Glen Ballou's 5th edition of "Handbook for Sound Engineers," 2015. See, in particular, the discussions concerning Figure 15-24:

enter image description here

So that may be reason to match impedance. But this is an uncommon circumstance where the intent is to maximize power delivered into a transducer. For most audio hook-up arrangements, this isn't the goal.

  • 1
    \$\begingroup\$ But you really don't want to match source impedance to headphone impedance either. You want the lowest impedance possible on your amplifier to drive any kind of transducer such as headphone or speaker. \$\endgroup\$
    – Justme
    Aug 1, 2021 at 19:22
  • \$\begingroup\$ @Justme I explained my reasoning. What are your thoughts about why that is wrongly said? It matches my understanding of the physics involved between transducer and human ear well. But I may have a flawed understanding of the physics. (I have used this understanding in the past, with expected results.) \$\endgroup\$
    – jonk
    Aug 1, 2021 at 19:35
  • 1
    \$\begingroup\$ Driving things like speakers and headphones with matching impedance is generally not done due to power lost in the source impedance. You want to minimize the source impedance regardless of load impedance. That's how you get the most power into headphones and speakers with least losses. I mean, would you drive 600 ohm headphones with an amplifier with 600 ohms output impedance? The damping factor would be too high. Half of the power wasted in source impedance. With near zero output impedance, nearly all power is going to the speaker. Maximum power theorem does not apply to audio interfaces. \$\endgroup\$
    – Justme
    Aug 1, 2021 at 19:58
  • \$\begingroup\$ @Justme Take the \$600\:\Omega\$ to \$600\:\Omega\$ case you mention. If the driver source impedance really is \$600\:\Omega\$, and let's say the source RMS is \$10\:\text{V}\$, and I apply an \$8\:\Omega\$ speaker to that, I get \$\approx 2.2\:\text{mW}\$ delivered. If I instead use a \$600\:\Omega\$:\$8\:\Omega\$ audio transformer (fairly efficient) then I get \$42\:\text{mW}\$ into the primary, which appears at the secondary, and is then reduced to \$21\:\text{mW}\$ at the speaker. This is an improvement by a factor of 10, despite losses. Where am I missing an important detail? \$\endgroup\$
    – jonk
    Aug 1, 2021 at 21:14
  • \$\begingroup\$ The problem of matching is having source impedance equal to load impedance, 600 to 600 or 8 to 8. Matching, or rather, trying to convert between impedances made sense with valves as they had kilo-ohms (if not tens) of impedance at few hundred volts with milliamps of current, while speakers were about 8 ohms. In modern audio, impedance bridging is used. With near-zero output impedance, you need only about 0.41V at 51mA to get same 21mW into 8 ohms. Much more suitable for modern devices where headphone impedances are 16-32 ohms. And 1 mW is loud enough to get damage hearing in few hours. \$\endgroup\$
    – Justme
    Aug 1, 2021 at 21:42

Output Impedance MONITOR:
1⁄4”TRS, impedance-balanced 51Ω. Unity gain.

This means the output is the same as your input.

Recorder will be high impedance like 10kΩ so there is no attenuation or effect on the headphone output. (Update 480 kOhm)

The speaker output will be << 1 Ohm so a load from 8 to 600 ohms does not attenuate much but affects power consumption only and is current limited and protected.


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