Looking at a Grid Doctor 300 to run some radios when the power is out , up on a mountain just testing or field day etc. This "generator" has a 320 wh battery . how long will it run a 10 watt radio' assume 25 percent of the time transmitting and the 75 percent receiving so 15 mins transmitting and 45 receving out of 1 hr. I would think it could run it a few hours ? 3 or 4?? tks
Solar Generator
- Thread starter MackinNC
- Start date
How much current does the radio draw when transmitting at 10 watts output?
My guess is around 10-13 hours figuring a two Ah draw. Much depends on the power draw during TX and RX. With a 100W solar panel connected, you’re going to have energy to spare and will re-charge the battery while you have sunlight. Figure 4 Ah from the solar panel in angled and plenty of sunlight. That gives you two Ah to power and two Ah towards charging. With five hours of direct sunlight you’ll push 10 Ah back into the battery the next day. Also, you noted “radios”, so there’s more amps being drawn.
One of my smaller solar generator systems is:
- Ecoflow River 2 Max 512 Wh LiFePO4 battery ($400)
- 100W flexible solar panel ($80) a bit more if you need folding
- Cable to connect solar to battery ($25)
The River 2 Max comes with a 12v DC power cord, so you can charge it while driving to top it up as well. I recommend a LiFePO4 (LFP) battery over a Lithium Ion in whatever solar generator you buy.
One of my smaller solar generator systems is:
- Ecoflow River 2 Max 512 Wh LiFePO4 battery ($400)
- 100W flexible solar panel ($80) a bit more if you need folding
- Cable to connect solar to battery ($25)
The River 2 Max comes with a 12v DC power cord, so you can charge it while driving to top it up as well. I recommend a LiFePO4 (LFP) battery over a Lithium Ion in whatever solar generator you buy.
Thanks, a little more hours than I thought but good to know. I fiqured it as drawing 2- 2.5 watts but didnt want to say as to see what others thought.
I would only run 1 radio at a time.
so now I wonder if I get 10 hrs out of a 10 watt radio then a 50 watt will cut run time to just a couple hours??
You have answered your question. The battery has a capacity of 320 watt-hours. How many watts does the radio consume when it is keyed up? If a radio operates at 50% efficiency, it will consume 20 watts of power when transmitting at 10 watts.
The calculation is straightforward: 320 Wh divided by 20 watts = 16 hours of transmitting. This means you can key the mic for 16 hours. With a 25% duty cycle of 15 minutes of transmission (TX) and 45 minutes of reception (RX), the operation time would be less than 60 hours, or around 2 days.
To get more accurate, measure the power the radio uses during receiving. For example, if it uses 4 watts and runs 45 minutes = 4 watts x .75 hours = 3 wh hours. 20 watts for 15 minutes = 20w x .25h = 5wh. You would consume 8 wh per hour. 320 wh / 8wh/h = 40 hours.
Remember that:
Watt Hours = Watts x Hours
Amp Hours = Amp x Hours.
Now go getter done.
The calculation is straightforward: 320 Wh divided by 20 watts = 16 hours of transmitting. This means you can key the mic for 16 hours. With a 25% duty cycle of 15 minutes of transmission (TX) and 45 minutes of reception (RX), the operation time would be less than 60 hours, or around 2 days.
To get more accurate, measure the power the radio uses during receiving. For example, if it uses 4 watts and runs 45 minutes = 4 watts x .75 hours = 3 wh hours. 20 watts for 15 minutes = 20w x .25h = 5wh. You would consume 8 wh per hour. 320 wh / 8wh/h = 40 hours.
Remember that:
Watt Hours = Watts x Hours
Amp Hours = Amp x Hours.
Now go getter done.
Last edited:
JDKelley
Just call me "Sparks." Or "Lucky."
"Watt-hours" is exactly that - figure it to be the number of watts the battery will deliver for one hour. So, 320W-h means 320W for one hour.
Now, you can play with that - any combination that multiplies out to 320 will work. So, 80W x 4 hours, for instance. Or 20W for 16 hours.
Bear in mind that listed power on a radio is transmit power, not power draw. This means that your 10W radio will put out a 10W PEP signal, but it will consume slightly more 'leccy generating it and pushing it out - I'd not be surprised if your consumption figure weren't closer to 13-14W. So, figure 16W (to make the maths come out even,) for 20 hours. 16W x 20h = 320 W-h. YMMV, TANSTAAFL. Bear in mind that there are inversion and conversion losses, for instance, so you're not going to get the full 320 W-h. But, since you're not drawing the full 16W you're using to figure your battery longevity, that's essentially a non-issue, you can probably still count on 20h runtime. Longer, with the solar panel - unless it gives you a good 20-30W charge current, it will at least slow down the draw loss by partially replacing what you're taking (keeping the maths simple, say you've got a 10W panel. You're drawing 16W from your radio. 16 W - 10 W = 6 W, so that's what you're drawing from the battery. Or, since batteries themselves are rated in Ampere-hours, let us apply Watt's law, and 6 W / 12 VDC = 0.5 ADC. If you can somehow make up that half-Ampere of charge, you can keep your battery static. Perhaps a wind turbine? A water wheel, if you've got a stream handy? Be creative!)
Ohm's Law: E = I x R. E = Volts [V] (Electromotive force.) I = current [A] (I don't know why.) R = Resistance [ohms]. "AC" or "DC" is usually specified, as in "VAC" or "ADC." Resistance doesn't care.
Watt's Law: P = E x I. P = Power [W] (Watts. Doesn't care what sort of current.) E = Volts [V] (Electromotive Force.) I = Current [A] (Again, I don't know why they use "I" for this.)
These two are often combined as: P = [I x R] x I, or P = I^2 x R. Engineers use this combined form a lot when figuring transmission lines (when they're figuring "I-squared-R" losses.)
If you're going to get into any sort of electronics, or bone for your ham ticket, tattoo those two formulae on the inside of your eyelids. As they told us in training: "Gentlemen, you will see this material again." (This was your cue to write. It. Down. If I say that, find a notebook.)
Now, you can play with that - any combination that multiplies out to 320 will work. So, 80W x 4 hours, for instance. Or 20W for 16 hours.
Bear in mind that listed power on a radio is transmit power, not power draw. This means that your 10W radio will put out a 10W PEP signal, but it will consume slightly more 'leccy generating it and pushing it out - I'd not be surprised if your consumption figure weren't closer to 13-14W. So, figure 16W (to make the maths come out even,) for 20 hours. 16W x 20h = 320 W-h. YMMV, TANSTAAFL. Bear in mind that there are inversion and conversion losses, for instance, so you're not going to get the full 320 W-h. But, since you're not drawing the full 16W you're using to figure your battery longevity, that's essentially a non-issue, you can probably still count on 20h runtime. Longer, with the solar panel - unless it gives you a good 20-30W charge current, it will at least slow down the draw loss by partially replacing what you're taking (keeping the maths simple, say you've got a 10W panel. You're drawing 16W from your radio. 16 W - 10 W = 6 W, so that's what you're drawing from the battery. Or, since batteries themselves are rated in Ampere-hours, let us apply Watt's law, and 6 W / 12 VDC = 0.5 ADC. If you can somehow make up that half-Ampere of charge, you can keep your battery static. Perhaps a wind turbine? A water wheel, if you've got a stream handy? Be creative!)
Ohm's Law: E = I x R. E = Volts [V] (Electromotive force.) I = current [A] (I don't know why.) R = Resistance [ohms]. "AC" or "DC" is usually specified, as in "VAC" or "ADC." Resistance doesn't care.
Watt's Law: P = E x I. P = Power [W] (Watts. Doesn't care what sort of current.) E = Volts [V] (Electromotive Force.) I = Current [A] (Again, I don't know why they use "I" for this.)
These two are often combined as: P = [I x R] x I, or P = I^2 x R. Engineers use this combined form a lot when figuring transmission lines (when they're figuring "I-squared-R" losses.)
If you're going to get into any sort of electronics, or bone for your ham ticket, tattoo those two formulae on the inside of your eyelids. As they told us in training: "Gentlemen, you will see this material again." (This was your cue to write. It. Down. If I say that, find a notebook.)
As AK9R initially questioned, what is the current draw at 10W? Knowing how much your radio draws during TX as well as during RX will provide more accurate numbers. I have a Yaesu 991A that is a power hog, compared to purpose built QRP radios. While it will draw no more than 8 amps at 5W using FM mode, it draws 1.25A while just receiving. She's a thirsty one. Contrasting to that, a Yaesu 817ND draws no more than 1.5A during 5W transmit on FM and no more than 0.35A during RX. That is a big difference between the two radios.
What radio model are you using? Perhaps we can figure out the current draw knowing that if you don't have something to measure it.
What radio model are you using? Perhaps we can figure out the current draw knowing that if you don't have something to measure it.
