Auto accessory wiring

[TiredGun]

A wiring question regarding the installation of two 50-watt mobile radios. I have two 50-watt radios, pulling 15 amps each when transmitting on their respective highest power setting. I do not intend to operate both radios at the same time, but am installing them for insurance for communications when in the backcountry where my CB or cell phone cannot do the job. The radios are a GMRS 50-watt and a VHF/UHF 50-watt ham radio. Now, finally, is here my question with, hopefully, sufficient detail and specifics. I want to install an auxiliary fuse block that services the radios. However, I am trying to determine where to put a relay or relays as part of the system. I am running 10 gauge fused wire (both ground and power are fused) directly from the battery. The fuses are 20 amps each. I am presently considering two options: (1) First option is wiring a switched 4-pin, 40/30 amp relay directly from the wires running from the battery, controlled by an ignition switch so that the system does not work until the ignition is activated. The relay then would be wired to an auxiliary fuse block connecting the 87-pin to the fuse block's positive pole and grounding the relay (85-pin) to the fuse block's negative pole which is also receiving the negative 10 gauge wire directly from the battery. The relay, of course, is receiving its power via the 30-pin directly from the 10 gauge wire running from the battery. Now, each radio will be wired to the fuse block, receiving power from a positive pole that has been fused with a 20-amp fuse per positive pole and wiring the radios' grounds to the respective poles on the fuse block. (2) The other option is to directly power the fuse block from the battery, then power one or two switched 20 amp relays from the fuse block controlling each radio.

Does any of this make sense? Is there something I am missing here? Thank you for your help.

 

[BillK]

Not sure why you need the relay. Just over complicates it. I have installed more than a few Ham Radio rigs in cars and have always wired them directly to the battery with fused cables.

 

[Firebrick43]

relays engaged with ignition do keep from running down the battery if you forget to turn off the radios. I would do it but that is the main/only reason I would.

Why would you fuse the ground? There is no good reason to do so.

 

[infinkc]

You just need one 10ga positive wire from the battery for both. Use the car chassis for ground.

I assume the radios don’t have an accessory wire, that is why you want to use the relay.

 

[fitter30]

Wouldn't want to connect to ignition switch in case something goes wrong. They cost to much to repair. Fuel pumps are relay operated and fused. Pilot relay controlling a two pole power relay. Every off the battery.

 

[kbeefy]

Why would you fuse the ground? There is no good reason to do so.

In a worst case scenario the wiring for the radio could be used by another high amp load as a ground circuit, possibly burning up the radio.

Like welding, lightning strike, faulty engine ground and starter finding that as the path of least resistance etc...

I typically only fuse the positive side.

 

[Firebrick43]

In a worst case scenario the wiring for the radio could be used by another high amp load as a ground circuit, possibly burning up the radio.

Like welding, lightning strike, faulty engine ground and starter finding that as the path of least resistance etc...

I typically only fuse the positive side.

That would be a peculiar series of events for everything to happen that way. In all of those cases the amps will still go thru the radio before blowing the ground fuse still doing the damage.

 

[kbeefy]

um, nope. If the fuse can only carry 20a then thats all the rest of the circuit sees. Thats the point of a fuse, or circuit breaker.

 

[Jeff Ivers]

I am confused. I always thought that watts divided by voltage equals amps, which would make each device about a 4 amp draw. I also thought devices were rated at max power, so how could these devices draw 15 amps?

 

[wyliesdiesels]

I am confused. I always thought that watts divided by voltage equals amps, which would make each device about a 4 amp draw. I also thought devices were rated at max power, so how could these devices draw 15 amps?

The 50watts is the RF transmitter power of the radio. That doesnt mean it only draws 4amps on the 12v power supply side

 

[Fasthotrod]

I am confused. I always thought that watts divided by voltage equals amps, which would make each device about a 4 amp draw. I also thought devices were rated at max power, so how could these devices draw 15 amps?

Technically, you are not wrong... Watts (power) is equal to Voltage * Amperes.

So if you were to run the numbers at the DC side, 15A at 12 VDC = 180 watts. So what gives?

When we start talking about radios (and many other loads) there are some losses in the system. The radio may be able to transmit a 50 watt signal, but the radio circuits will also require power to operate, and there may be components within the radio that have a momentary, high ampere draw to power up, so we need to account for that as well.

So if the radio was transmitting at 50 watts, and the radio circuits needed 40 watts to operate, we're at 90 watts... but if the transmitter or power supply had an initial inrush current, it might spike at say 140 watts.

140 watts = 12 VDC * 11.667A

So the wiring to the radio is likely sized for 15A to meet the needs of the radio without blowing a fuse unnecessarily, and is fused to protect the conductor, not necessarily the radio. (I would be surprised if the radio did not have it's own internal fuse/circuit protection to protect the radio.)

Hope this helps.

Mark

 

[TobeyA]

I'd power the fuse block via the relay. But yes, it makes sense. I have an aux fuse block on my race car powered exactly the same way.

 

[Jeff Ivers]

The 50watts is the RF transmitter power of the radio. That doesnt mean it only draws 4amps on the 12v power supply side

Technically, you are not wrong... Watts (power) is equal to Voltage * Amperes.

So if you were to run the numbers at the DC side, 15A at 12 VDC = 180 watts. So what gives?

When we start talking about radios (and many other loads) there are some losses in the system. The radio may be able to transmit a 50 watt signal, but the radio circuits will also require power to operate, and there may be components within the radio that have a momentary, high ampere draw to power up, so we need to account for that as well.

So if the radio was transmitting at 50 watts, and the radio circuits needed 40 watts to operate, we're at 90 watts... but if the transmitter or power supply had an initial inrush current, it might spike at say 140 watts.

140 watts = 12 VDC * 11.667A

So the wiring to the radio is likely sized for 15A to meet the needs of the radio without blowing a fuse unnecessarily, and is fused to protect the conductor, not necessarily the radio. (I would be surprised if the radio did not have it's own internal fuse/circuit protection to protect the radio.)

Hope this helps.

Mark

Thank you both for the clarifications. I think I have learned something new today, but still not sure I completely understand. I guess I am used to dealing with things like light bulbs and electric heaters where the power output is the figure used in the equation to determine amperage load. It seems that you are telling me that when it comes to a radio transmitter, the rated power output is irrelevant to the load of the device. In such circumstances is a separate load rating supplied with the device that must be used when making wiring decisions?

 

[75gmck25]

This is something similar to your option 1.
- 10 gauge power wire from battery positive directly to a relay (30 terminal), with a fusible link (30 amp rating, IIRC) near the battery to protect the wire (you could also use a fuse).
- The other side (87 terminal) of the relay is connected to a junction block that has one 10 gauge wire in from the relay, and multiple terminals for output connections. Each power output has a fusible ink or fuse. In your case this would be your fuse block.
- All grounds (including 85 relay terminal) are connected to the chassis, frame, engine or body.
For my newer, more electrically sensitive items (wideband AFR gauge, etc.) I have run ground wires all the way back to a common ground point. There are no fuses, switches or relays in the ground wires.
- The relay has an activation wire on terminal 86 running to an ignition-switched source (active in both accessory and run position) on the main fuse block. When the truck is running or in the accessory position it will provide power from the battery to the auxiliary fuse block.
4 pin Relay pinouts are:
30 - power feed from battery
85 - ground
86 - switched hot from ignition-on power
87 - to fuse block

The wiring in my old truck is actually a little more complicated than above because I have a 2nd/auxiliary battery and the relay is designed to isolate the batteries if the truck is not running and only allow a power draw off the auxiliary battery.

 

[Chukster]

relays engaged with ignition do keep from running down the battery if you forget to turn off the radios. I would do it but that is the main/only reason I would.

Why would you fuse the ground? There is no good reason to do so.

There are one or 2 brands of radios that call for fusing both sides of the power wires; I've seen the rationale, but never been swayed by it. (short version: I think the rationale is dumb, but some engineer somewhere has decided, so...)

FWIW, TiredGun, there is a product called ChargeGuard which you can use; it will power up when the engine starts, and stay on, feeding power downstream for a selectable time. ( https://www.havis.com/product/chrggrd_unv_cntrlmdl-38782-0/ ).
Also, I put this in my car for my ham radios (a 2M, a 2M/440 dualbander, and a 60W. 6meter rig) : https://www.eaton.com/us/en-us/catalog/emobility/series-15600-atc.html, with 6 positions. So, Battery->BIG Fuse->ChargeGuard-> Fuseholder & distribution point.

Works great.

 

[slimpickins]

Thank you both for the clarifications. I think I have learned something new today, but still not sure I completely understand. I guess I am used to dealing with things like light bulbs and electric heaters where the power output is the figure used in the equation to determine amperage load. It seems that you are telling me that when it comes to a radio transmitter, the rated power output is irrelevant to the load of the device. In such circumstances is a separate load rating supplied with the device that must be used when making wiring decisions?

This is exactly why all the LED manufacturers now include the "LUMENS" output along with the watts. The light from a 60W incandescnet bulb can now be produced with about 7 or 8 watts by an LED. Everyone was used to talking about a 60W or a 100W light bulb and we all knew what that meant. Nobody except nerds ever talked about lumens. Now, the Watts is the input power and the Lumens is the output power. The 50 W rating is the tx power or output power (think of it a "lumens" of radio power or how bright is the signal from a given radio.) Then you need to know how much power it takes to produce a radio signal that provides 50 watts of radio "lumens".

To put it another way:
Incandescent bulb: Input power 60W; output light: 800 lumens.
LED bulb: Input power: 9W; output power: 800 lumens.

Radio #1: Input power 15A@12V = 180W; Output (tx) power: 50 watts
Radio #2: Input power: 30A@12V = 360W; Output power 40 watts.

It is a matter of different efficiencies.

 

[Chukster]

...

Radio #1: Input power 15A@12V = 180W; Output (tx) power: 50 watts
Radio #2: Input power: 30A@12V = 360W; Output power 40 watts.

It is a matter of different efficiencies.

Also, the fuse rating for the radios is set above the level the radios will draw in transmit; ferinstance, I looked up the current draw in transmit for Kenwood TK-880, a 25 watt UHF commercial dash-mount radio, and also the TK-880H, a 40 watt radio

25 watt radio actually draws maximum 8A at 13.6Volts; that means about 100 watts, split between RF radiated power and power in the radio turned into heat, running the circuits, lighting the backlighting on the display, yadda, yadda. Service manual calls for a fuse of 10 amps.

40 watt radio draws 12A, that means total power consumed of about 163 watts; service manual calls for a fuse of 15 amps.

VHF radios will be similar currents, tho not completely equal.


You'd have to step way back in time to find radios (like Motorola Motrac) which had tube finals, they would draw comparatively huuuuge amounts of power, because they would have to draw current to get the filaments hot enough to work!! Cars with weak electrical systems were a big problem; I don't know if the stories of cars slowing down or stalling when the radio transmitted are true, or just apochryphal!