Jump to content

Got My New MXT500 - Not Impressed


marcspaz

Recommended Posts

6 hours ago, gman1971 said:

To @Mtkester first off, sorry to hear, here are some thoughts:

There are a lot of affordable LMR mobile radios on the used market that will work very well for your GMRS application. If you don't want to venture into the latest Motorola XPR radios, I would look into a used Motorola CDM1250 UHF. Those radios will cost a whole lot less than $400, have receives that will blow your mind how good they are along with top notch transmitter stability; those are used in light repeater duties, so that speaks volumes for those radios: they are indestructible. The CPS programming software can be found online at no cost. The cable can be had from eBay for next to nothing too. I have limited experience with these CDM radios, but I could help you get started if you chose to go with one of those. I know other members here have ample experience with the CDM radios, I am sure they would be willing to help too

Another affordable radio to look into is the Vertex Standard EVX-5400 mobile (LCD display), there are two variants with two power levels each variant: 5400-D6-25, 5400-D6-50, 5400-D7-25, 5400-D7-50. D6 = means 400 to 470mhz coverage, and you guessed right, D7 = 450 to 512 mhz coverage, and the number as you guessed is 25watts, or 50watts.  I would go for a 5400-50(or 45)-D6 The CPS can also be found for free online. Those radios are nice because they are FM/DMR capable if you ever chose to get your amateur license to operate on DMR. However, after using XPR5550e radios for a while I can say the EVX-5400 receivers seem to be one notch below the XPR radios. With that said, I still use an EVX-5300 (NO display) as my primary business site radio, and with a preamp and some filters it is as good as the XPR5550e. This radio has been running in non-stop "light repeater duty" 24/7 for 2 1/2 years now; no issues. Several other members here besides me also own Vertex Standard  EVX radios too, and they seem to like them a lot. The programming cable can be tricky, since most of the knockoff cables will not work (as tested)... not sure if the BlueMaxers cables will work.... but If you decide to go with an EVX-5400 and need help getting started, let me know.

Personally, and given the fact you've paid $400 for the Midland radio, I would be looking at nothing else but an XPR5550e, either one of those two models will work for you: AAM28QPN9WA1AN or AAM28QPN9RA1AN (both models are the 400-470Mhz bandsplit and 25-50W range, again if you get the ticket you can use it for ham too) The CPS 16.0 can be found for dirt cheap on eBay, and the programming cable is pretty cheap too, or you can make your own for <5 bucks if you feel courageous. The XPR5550e is, IMO, the best mobile radio you can buy in the <400 dollar price range today. If you ever decide to go with a 5550e, drop me a line, I'll help you get started with it.

And an honorable mention goes to the Motorola APX radios, which are usually well above the >400 bucks mark, and a word of caution is that if you find something too good to be true in APX radios, suspect it might be stolen. That is another reason why I never took a plunge in the used APX radio market... 

Cheers, and again, I am sorry to hear what happened.

G.

Appreciate the hood info, will be looking at those. I honestly just bought mine for the purposes of reviewing it, was never planning on keeping it, I’m more than happy enough with my KG-1000G to want or need to buy anything else. 

Link to comment
Share on other sites

1 hour ago, Hunter399 said:

Due to not testing power with a dummy load?

No.  Regardless of if you are using a dummy load or an antenna, there are 2 formulas used for calculating power, resistance, current and voltage.  The first is Voltage / Current * Resistance.  If you have any 2 of the 3, you can find the 1 unknown value.  The second is Power / Voltage * Current.  Again, if you have any 2 of the 3, you can find the 1 unknown value.

 

We are going to use the second of the 2 formulas.  We know you have 7.5 amps at 14.6vdc.  We are going to multiply the two for a total of 109.5 watts of total consumed power by the radio.  The transmitter has an efficiency rating of 55% (per the manufacturer, and a very common value).  That means we are going to take 45% away from the 109.5w, being 109.5 * 0.55 = 60.225 watts out to the antenna or dummy load.  If your meter is reading 40w, then either the watt meter is not correct or the measured voltage or current are not correct. 

Link to comment
Share on other sites

5 minutes ago, marcspaz said:

No.  Regardless of if you are using a dummy load or an antenna, there are 2 formulas used for calculating power, resistance, current and voltage.  The first is Voltage / Current * Resistance.  If you have any 2 of the 3, you can find the 1 unknown value.  The second is Power / Voltage * Current.  Again, if you have any 2 of the 3, you can find the 1 unknown value.

 

We are going to use the second of the 2 formulas.  We know you have 7.5 amps at 14.6vdc.  We are going to multiply the two for a total of 109.5 watts of total consumed power by the radio.  The transmitter has an efficiency rating of 55% (per the manufacturer, and a very common value).  That means we are going to take 45% away from the 109.5w, being 109.5 * 0.55 = 60.225 watts out to the antenna or dummy load.  If your meter is reading 40w, then either the watt meter is not correct or the measured voltage or current are not correct. 

Hi Marc,

When calculating the transmitter efficiency rating, is that done based on the power that actually goes into the transmitter stage or does it include all power that is dissipated by the voltage regulator, power used by the receiver, etc? The reason I ask is because in a case like this, where 14.5 Vdc is delivered to the power input of the radio, if the radio has a voltage regulator to reduce that to 13.8 Vdc which is then delivered to the radio, then the actual power delivered to the radio would be 13.8 Vdc x 7.5 A or 103.5 W.  55% of that is still ~57 W, so maybe it's a moot point.

Link to comment
Share on other sites

22 minutes ago, Sshannon said:

Hi Marc,

When calculating the transmitter efficiency rating, is that done based on the power that actually goes into the transmitter stage or does it include all power that is dissipated by the voltage regulator, power used by the receiver, etc? The reason I ask is because in a case like this, where 14.5 Vdc is delivered to the power input of the radio, if the radio has a voltage regulator to reduce that to 13.8 Vdc which is then delivered to the radio, then the actual power delivered to the radio would be 13.8 Vdc x 7.5 A or 103.5 W.  55% of that is still ~57 W, so maybe it's a moot point.

 

I wish I could answer that question accurately.  Very little information about the radio that was given to me.  As you noted, I'm not sure it matters either way, based on the math you shared.

Link to comment
Share on other sites

@Sshannon  If I were to guess, I would say it might be based on the minimum voltage needed for the radio to work.  Given that they said it should be close to 8 amps (46w +/-) at full power and it has a 55% efficiency rating, the math would work out to 11.5vdc.  That kind of makes sense... most 12vdc systems run on as little as 11.0vdc and max out at about 15vdc.

So, you may be on to something there.

Link to comment
Share on other sites

21 hours ago, marcspaz said:

No.  Regardless of if you are using a dummy load or an antenna, there are 2 formulas used for calculating power, resistance, current and voltage.  The first is Voltage / Current * Resistance.  If you have any 2 of the 3, you can find the 1 unknown value.  The second is Power / Voltage * Current.  Again, if you have any 2 of the 3, you can find the 1 unknown value.

 

We are going to use the second of the 2 formulas.  We know you have 7.5 amps at 14.6vdc.  We are going to multiply the two for a total of 109.5 watts of total consumed power by the radio.  The transmitter has an efficiency rating of 55% (per the manufacturer, and a very common value).  That means we are going to take 45% away from the 109.5w, being 109.5 * 0.55 = 60.225 watts out to the antenna or dummy load.  If your meter is reading 40w, then either the watt meter is not correct or the measured voltage or current are not correct. 

So as long as my power going to the radio is correct I shouldn’t be concerned it’s only putting out the 40 watts? 

Link to comment
Share on other sites

32 minutes ago, Hunter399 said:

So as long as my power going to the radio is correct I shouldn’t be concerned it’s only putting out the 40 watts? 

 

 No, I'm sorry.. That is not what I am trying to say. What I mean is, none of your measurements are in agreement with one another.. Therefore I do not trust the 40 W reading.

 

I believe that either you need to calibrate your watt meter, your amp meter, your volt meter, some combination of the three or all three.

Link to comment
Share on other sites

5 hours ago, marcspaz said:

 

 No, I'm sorry.. That is not what I am trying to say. What I mean is, none of your measurements are in agreement with one another.. Therefore I do not trust the 40 W reading.

 

I believe that either you need to calibrate your watt meter, your amp meter, your volt meter, some combination of the three or all three.

I need to measure the volts better, I wasn’t thinking of the loss across how far the wire is i know the amps are spot on. 

Link to comment
Share on other sites

On 1/18/2022 at 10:42 PM, Mtkester said:

I just put my pile of $400 garbage, otherwise known as an MXT500, on the meter as well and she peaked out at 36.4 watts using a 13.8 volt, 30 amp power supply. Already submitted a return request for it. I knew it was going to be disappointing… I didn’t believe Midland would actual add fraud to the long list of this radio’s shortcomings. 

Have they returned it and did the new one read higher? Debating returning mine as well

Link to comment
Share on other sites

@Hunter399... that's cool!  Glad they sent it over to you.  I haven't sent mine back yet because I need it for an offroad event on the 30th.  Hopefully they take care of us.  Looking forward to seeing what they do for you.

 

For what its worth... my MXT400 has more measured output power than my MXT500.  I would like to have that corrected, for sure.  I miss the old days when we just had to spin a POT and make it happen.  Now the FPP alignments lock us out.

Link to comment
Share on other sites

37 minutes ago, PRadio said:

Hi Marc, how is the receive sensitivity for the mxt500? Does it seem to be better than the BTEC you sold me?

Thanks - Phil

 

Hey Phil!  I didn't do any significant technical testing on the MXT500 beyond the peak power.  I can grab a signal generator and test it when it gets back from Midland. 

Link to comment
Share on other sites

1 hour ago, DavidB said:

Marc,  Did you get your radio back yet? Results?  I sent mine back (only getting 30W) and still waiting....

 

 

No, I actually needed it for a trip over the past weekend.  I just shipped it out today.  However, based on feedback from Midland as well as others, I expect it returned by 02/18.  I will share the results as soon as I get the radio back.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

By using this site, you agree to our Terms of Use and Guidelines.