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I have the Tecsun PL-310ET and PL-660 SW portables which are spec’d for 5V and 6V power sources respectively. Would it damage them if I tried to power them with a 9V Bioenno battery (actual steady voltage ~9.6V)?

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  • $\begingroup$ Welcome to ham.stackexchange.com! $\endgroup$ – rclocher3 Dec 11 '19 at 0:28
  • $\begingroup$ If the manufacturer doesn't explicitly say that the input voltage range can go up to 9V, then I would go on the assumption that catastrophic damage would occur. $\endgroup$ – Duston Dec 11 '19 at 20:41
  • $\begingroup$ To be safe use 4 or 5 rectifiing diodes in series with 9V battery and you will be in ballpark of safe voltage (there is 0.7V drop on each diode) $\endgroup$ – Pedja YT9TP Dec 21 '19 at 8:56
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I wouldn't risk directly connecting a 9.6v battery to a radio that expects 5-6v.

The simplest solution would probably be the diodes mentioned in the comment to your question, but as the battery voltage dropped the voltage reaching the radio would drop as well, and as the battery got low the output voltage after the diodes might be too low to power the radio, wasting the last bit of the battery that could otherwise be used.

I would recommend a linear regulator such as a 7805 (for 5v) or 7806 (for 6v). Here's how it's wired, and if you don't have them you can leave out the capacitors for this sort of application:

circuit diagram for 7805 regulator from https://www.electronicshub.org/understanding-7805-ic-voltage-regulator/

With a 78xx regulator the voltage to the radio won't drop until the battery has reached the regulator's drop-out voltage, and the battery will likely be dead before that happens, at least for the 7805. For the 6v radio, you might consider an LDO (low drop-out) version of the 7806 regulator to squeeze every last bit out of the battery, or just use a 7805, which is much more common and easier to find than a 7806 anyway, as the 6v radio can probably run fine on 5v.

It's worth keeping in mind that either the diodes or the linear regulator will waste a significant amount of battery power as heat. Their main advantages are that they are simple to build and won't introduce noise into the radio. More efficient regulators are commonly available but tend to be noisy, especially for sensitive radio receivers, but if you need to conserve battery you should do some further research into switching regulators and buck converters, or just use a battery the radio can handle directly.

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    $\begingroup$ The diode drop makes the last little bit of the battery capacity inaccessible, but a linear regulator wastes a significant fraction of all of the battery capacity. If the battery is putting out 9.6V then the 7806 is 62.5% efficient and the 7805 is 52% efficient. $\endgroup$ – hobbs - KC2G Feb 18 at 20:32
  • $\begingroup$ @hobbs-KC2G is correct. A better solution would be a decent 5 VDC power supply, such as those ubiquitous "wall-warts". $\endgroup$ – Mike Waters Feb 18 at 22:37
  • $\begingroup$ @hobbs-KC2G I thought of that, and the diodes will waste just as much as a linear regulator if they're dropping as much voltage, so I didn't mention it. I considered mentioning a switchmode buck converter for efficiency but wanted to keep my answer simple. Also, I've had bad luck with cheap buck switchers and radio equipment, those things always seem to be noisy as all getout; linear regulators or diodes at least output clean power, if wastefully. $\endgroup$ – MoTLD Feb 19 at 1:39
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    $\begingroup$ @MikeWaters You're probably correct, if the wall wart put out clean enough power not to make the radio noisy, but the question specifically asked about powering the radio from a 9.6v battery. $\endgroup$ – MoTLD Feb 19 at 1:43
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    $\begingroup$ Edited to note the inefficiency of diode droppers and linear regulators, thanks @hobbs-KC2G. $\endgroup$ – MoTLD Feb 19 at 1:53

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