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On low-power radios, the current consumption for reception is similar to the consumption for transmission. For example, the Texas Instruments CC2652 System-on-Chip datasheet claims these values:

  • Active-Mode RX: 6.9 mA
  • Active-Mode TX 0 dBm: 7.3 mA

I've read an explanation that the most energy hungry component is the local oscillator, which generates the high-frequency carrier wave and needs to do that both for reception and transmission. However, it's not clear to me why would the generated sine wave need to be with similarly high amplitude in the case of reception, compared with the case of transmission. An alternative hypothesis is that running all of the (other) analog and digital RF components is what consumes the energy. Can you clear up the confusion?

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    \$\begingroup\$ A local oscillator (LO) is required in both a transmitter and receiver but it's power output would not be at a level suitable for direct transmission. A transmitter will also have a power amplification stage downstream of the LO to drive the antenna. This extra stage may account for the discrepancy with your particular devices although the transmit power (and therefore the power amplification stage) is a small part of the overall power budget as mentioned in the answers below. In a more powerful setup where the transmitter is operating at several watts this would absolutely be the case. \$\endgroup\$ – mhaselup Aug 7 at 1:09
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    \$\begingroup\$ I have difficulty believing that a crystal radio receiver consumes more power than an RF transmitter. \$\endgroup\$ – Hot Licks Aug 7 at 18:07
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    \$\begingroup\$ Though this doesn't answer the OP: many RF communication systems keep their receiver on when they're not transmitting. Questions of power aside, such systems consume a lot more energy for receiving than for transmitting. \$\endgroup\$ – fearless_fool Aug 8 at 12:49
  • \$\begingroup\$ ham.stackexchange.com \$\endgroup\$ – Rodrigo de Azevedo Aug 9 at 18:19
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In short: receiving is much more complicated than transmission.

You'll notice that whatever you measure in the real world is overlaid with noise.

The problem "seeing all this noise with a bit of signal in it, how do I know what the transmitter meant to transmit" is the central problem that communications engineering tries to solve.

So, to receive a couple of bits correctly, your receiver needs to:

  • receive, even if there's no signal on the air, to notice when there's signal. That means the whole receive chain, and a couple-of-megasamples-a-second ADC runs.
  • Detect something like a preamble. That usually involves a correlation. That means, for every new sample (couple of millions per second), take the most recent e.g. 2000 samples and compare them to a known sequence
  • When there actually is detection of signal, correct all influences of the channel that are bad for your type of transmission. Depending on the system, this involves:
    • Frequency correction (no two oscillators in this universe are identical. Your receiver has a different frequency than your transmitter, and that breaks basically everything that isn't very basic. You need to estimate the frequency error, which typically involves tracking phase errors, or doing statistics, and then multiplying with a synthesized sinosoid or adjusting a power-hungry oscillator)
    • Timing estimation (your sampling is not synchronous to when the transmitter transmitted a symbol. fix that. Typically involves complex multiplications, time-shifting filters or adjustable and power-hungry oscillators.)
    • Channel equalization (your signal doesn't only take the shortest path. Multiple reflections reach the receiver. If the time difference between the shortest and longest path are not negligibly small compared to a symbol duration, you need to remove the echoes. Typically, involves solving an equation with a lot of unknowns or something similar, and application of a filter, which is quadratically in complexity to channel length, at best)
    • Phase correction (your channel still might rotate the phase of your received symbol. Calls for a phase-locked loop or some other control mechanism)
  • Symbol decision (great! After all these corrections, you, if everything goes right (it almost certainly doesn't do 100%), you only got the symbol that was sent, plus noise. So, which symbol was sent? Do a guess based on a defined decision algorithm, or do a guess and say "I'm 89% percent certain")
  • Channel decoding (The transmitter didn't just transmit the data bits – it added forward error correction redundancy, which allows you to correct errors that you still make. These algorithms can be very computationally intense.)
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    \$\begingroup\$ Don't forget the Low Noise Amplifier at the front end -- it's the one bit of analog circuitry that usually draws considerable current. \$\endgroup\$ – fearless_fool Aug 8 at 12:41
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    \$\begingroup\$ ah, indeed, but that often is overestimated. Just because a good LNA operates at an inefficient operational point doesn't mean it uses much power in absolute numbers. In case of low power integrated receivers, you're right, might be significant. \$\endgroup\$ – Marcus Müller Aug 8 at 13:50
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    \$\begingroup\$ FWIW, I served on the IEEE 802.15.4 (aka ZigBee) standards body, and my company co-designed the CC2420. In low-power LAN systems, the LNA is a significant contributor to power consumption. \$\endgroup\$ – fearless_fool Aug 9 at 14:06
  • \$\begingroup\$ @fearless_fool nice! \$\endgroup\$ – Marcus Müller Aug 9 at 15:12
  • \$\begingroup\$ By the way, all of your points are spot on: it's no coincidence that our brains contain more machinery for hearing than for speaking. \$\endgroup\$ – fearless_fool Aug 9 at 19:45
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Well, first off, you're looking at a microcontroller. The datasheet also says it will draw 3.4mA without any radios on, so you can only attribute 3.5mA to radio reception.

And then, if you look at the block diagram, you find this:

RF block diagram

That's right, a software-defined radio with its own ARM core. This allows, as TI says, an awful lot of future-proofing; they can add support for new protocols on the 2.4GHz band just by releasing some new firmware for that core. But can you really blame a CPU that does sample-by-sample digital processing of radio signals if it uses a few milliwatts of power? I'd say it's pretty efficient for what it is.

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  • \$\begingroup\$ The high Rx current consumption is a very common trend present in low-power radios. I don't think its worth focusing on the particulars of the example too much. \$\endgroup\$ – kfx Aug 7 at 9:53
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    \$\begingroup\$ @kfx I think it is, because it is also a good illustration of that trend: you get integrated radios, and those are complex devices. Also, your question is about a specific device, as the power consumption numbers are not applicable to anything else. \$\endgroup\$ – Marcus Müller Aug 7 at 12:23
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However, it's not clear to me why would the generated sine wave need to be with similarly high amplitude in the case of reception, compared with the case of transmission.

The thing is, it's not a high amplitude. 0 dBm of transmit power is 1 milliwatt and a power of 1 milliwatt into 50 ohms is 224 mV RMS i.e. a small voltage but, a voltage that is also about right for an IF stage and a mixer in a receiver.

Just look at the current consumption of the transmitter - 7.3 mA. If its power supply is 5 volts then that's a power consumption of 37 mW to emit 1 milliwatt on the antenna. Not efficient.

I might be more interesting to ask why the transmitter consumes 37 mW in order to deliver 1 mW to its antenna.

Both transmitter and receiver need oscillator hence, low power systems will consume about the same current consumption if using the same sort of power supply voltage.

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    \$\begingroup\$ I don't think all of the current consumed during transmission is used for the actual transmit power. If you take the increase in current from Rx toTx, 0.4 ma, and multiply by 5 volts, that is an additional 2 milliwatts which is twice the transmit power. That implies an efficiency of 50% which is not unreasonable. \$\endgroup\$ – Barry Aug 6 at 20:29
  • \$\begingroup\$ Not that it changes much, but I want to clarify that the voltage range of that particular chip is 1.8V - 3.6V, not 5V. \$\endgroup\$ – kfx Aug 6 at 20:31
  • \$\begingroup\$ @Barry well, I'm certainly not saying all of the current is used for power to the antenna. \$\endgroup\$ – Andy aka Aug 6 at 21:36
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Signal reliability may be improved either by putting more power into the transmission, or putting more effort into extracting a signal from the "noise" (unwanted nearby signals). The reason the receivers for these little radios need more power than the transmitters is that the transmitter power is restricted by emissions regulations. One could design lower-power receivers rather easily if they only had to receive signals that were much stronger than any other nearby signals.

By comparison, a crystal radio set consumes almost no power, but will have very poor adjacent signal rejection. I have actually used a crystal set to receive broadcasts, but I was in a city with one local AM radio station. I would not expect to get good results trying to use a crystal set in the Chicagoland area, which has multiple 50,000 watt AM stations at 670, 720, 780, 890, among others (unfortunately, a wire broke on my cats-whisker set and I would have to disassemble it rather thorougly to repair it).

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