Electric motor relay and high current question

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Thread Starter

paulsheer

Joined Sep 30, 2015
15
Update: I am pleased to report that the circuit below works perfectly. In my test it draws about 0.5 Amp on the base and 20Amp on the emitter and hardly gets warm. The real-life motor will draw double, but still miles from the limits of this transistor. (I am told this is all deducible from the spec sheet by any sophomore electrical engineer.)

So this is exactly what I wanted and solves my problem for the trivial sum of $40.

I am thoroughly astonished that the posters below could be so disparaging and foolish. It is all so unnecessary.

I would really suggest that the moderators modify the forum to allow a thread-creator to kick people off his thread. As it stands, sabotage gets a free pass.

(I have been blocked from making new posts to this thread.)

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Hello,

I have a motor running off a relay control system as shown in figure (A). I don't want to change the relay control system. It produces a 12V output and can push 18 Amps max. This all works fine.

I am changing to a new motor which can spike to 100Amps on startup, but normally draws about 40Amps. The relay cannot handle this.

I want to put NPN transistor "BUT30V" as shown in figure (B). BUT30V can handle 100 Amps.

See the specs for BUT30V of here. (I've ordered it and it's $40 shipped.)

www.st.com/content/ccc/resource/technical/document/datasheet/ae/8e/30/ce/bd/88/4a/17/CD00000772.pdf/files/CD00000772.pdf/jcr:content/translations/en.CD00000772.pdf

Will this work? I'm trying for something simple that's dead easy to wire up.

What current leakage can I expect when the relay is off?

Thanks!
 
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MaxHeadRoom

Joined Jul 18, 2013
28,688
You would be better off with a Mosfet or BJT.
Why not just use a motor contactor instead of a S.S. device to replace the relay?
Max.
 

Thread Starter

paulsheer

Joined Sep 30, 2015
15
every time I ask on these forums I get messages telling me to do something different and always strategically leaving out information so that this different course is closed

and then also not answering the question at hand

in other words, not helping at all

<snip>
 
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MaxHeadRoom

Joined Jul 18, 2013
28,688
I am just telling you what I would consider from a 50yr background in Electrical and Industrial Electronics, you DID say you wanted something easy to wire up.
I am sorry if you found it offensive.:rolleyes:
Max.
 

EM Fields

Joined Jun 8, 2016
583


Hello,

I have a motor running off a relay control system as shown in figure (A). I don't want to change the relay control system. It produces a 12V output and can push 18 Amps max. This all works fine.

I am changing to a new motor which can spike to 100Amps on startup, but normally draws about 40Amps. The relay cannot handle this.

I want to put NPN transistor "BUT30V" as shown in figure (B). BUT30V can handle 100 Amps.

See the specs for BUT30V of here. (I've ordered it and it's $40 shipped.)

www.st.com/content/ccc/resource/technical/document/datasheet/ae/8e/30/ce/bd/88/4a/17/CD00000772.pdf/files/CD00000772.pdf/jcr:content/translations/en.CD00000772.pdf

Will this work? I'm trying for something simple that's dead easy to wire up.

What current leakage can I expect when the relay is off?

Thanks!
With the relay OFF, the leakage current will be that given by the data sheet, but you'll have to do some interpolating. Best bet for something definitive is to ask ST.

If you have to use a solid state solution why have you chosen to use an emitter follower instead of a common emitter configuration?

If you want something dead simple, why can't you use a contactor, as MaxHeadRoom suggested?
 

EM Fields

Joined Jun 8, 2016
583
every time I ask on these forums I get messages telling me to do something different and always strategically leaving out information so that this different course is closed

Can you cite an example, please?

and then also not answering the question at hand
in other words, not helping at all

Sometimes the question at hand is badly formulated and needs to clarified before a definitive answer can be given, so what one might take as an effrontery is just an attempt at trying to get better data.

is it more important to you to show how smart you are? why bother logging in then?
If you have to ask for help, then it should be a foregone conclusion that you're looking for someone who is smarter than you are who can answer your question, and whose answer shouldn't be denigrated with rancor.

A kinder, gentler, way would be to ask for clarification if the answer isn't clear.
 
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MaxHeadRoom

Joined Jul 18, 2013
28,688
An age old problem :
When someone ask a question with little history in the forum, there is no indication as to their level of expertise, you may be talking down to them, or way over their head.
I posed an alternative as usual, as there is no point at this time in getting into technical explanation that may, or may not be understood.:(
If anyone seriously is looking for help, they usually come back with questions if not completely clear to them.
Max.
 

EM Fields

Joined Jun 8, 2016
583
An age old problem :
When someone ask a question with little history in the forum, there is no indication as to their level of expertise, you may be talking down to them, or way over their head.
True enough, but if they initially write something like "How many watts are there in an ohm?" and they seem to be sincere, their expertise needs to brought up to snuff, kick and scream as they may.

I posed an alternative as usual, as there is no point at this time in getting into technical explanation that may, or may not be understood.
I disagree. Nip the hard problem in the bud and throw a stick out there which, when retrieved, will get get closer to home.

If anyone seriously is looking for help, they usually come back with questions if not completely clear to them.
Max.
OK...
 

EM Fields

Joined Jun 8, 2016
583
It seems that instead of actually asking for help, beforehand, you've ordered the part, clueless, and what you're looking for, now, is for someone to dig you out of the hole you've dug for yourself or for someone to take the blame for your incompetence when the transistor's magic smoke is let loose.

MaxHeadRoom and I have both made suggestions which can only help your case, and yet you choose to discount them as being "not straight answers"?

Good luck on your own from now on, from my end.

Mod edit: cleaned up a bit.
 
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profbuxton

Joined Feb 21, 2014
421
Ok , heres my $100 worth.Sketch as shown.
(1) Will it work. No. Why. You have not shown any base bias resistor. Have you allowed for a substantial heat sink for the transistor. that may take up space you dont have I don't know.
(2) Leakage current. Since you have already ordered same you would have perused the data sheet and can read the leakage current from same.
(3) Dead easy to wire. If thats what you want then, as suggested earlier, use a contactor to handle the current. Much simpler.

PS. If thats not to your liking feel free to ignore and not respond in any fashion.
 

Thread Starter

paulsheer

Joined Sep 30, 2015
15
Keep upping the ante and someone might actually be enticed into designing something for you which might keep your clients from discerning - if only momentarily - your own incompetence.
I don't have clients

Mod edit: snipped argumentative comment
 
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EM Fields

Joined Jun 8, 2016
583
<snipped quote>
You didn't provide enough information about the motor to determine whether your emitter follower circuit would work, and the transistor's leakage current data is given by the data sheet, although not with the Vce(o) you mentioned, which is why I referred you to ST instead of contacting them myself and posting what I found.

And, "<snip>" ?
 
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bertus

Joined Apr 5, 2008
22,277
Hello,

From your drawing it is not clear how the relais is switching.
With the information you are giving, we can not give you a decent advice.

As for the leakage current, this is written on the wiki page:
---
In bipolar junction transistors, the emitter current is the sum of the collector and base currents. Ie = Ic + Ib. The collector current has two components: minority carriers and majority carriers. The minority current is called the leakage current[clarification needed].

Leakage current is generally measured in microamperes. For a reverse-biased diode it is temperature sensitive. Leakage current must be carefully examined for applications that work in wide temperature ranges.
---

https://en.wikipedia.org/wiki/Leakage_(electronics)

Bertus
 

JohnInTX

Joined Jun 26, 2012
4,787
@paulsheer I have deleted your disrespectful comments. That is not the way we work around here. If you don't like the answer a member has given, you should explain why it doesn't work for you and continue from there in a constructive manner. As long as that happens, the thread will continue and hopefully, you can get the help you need. If, however, you persist in your conduct, your privileges on AAC will be revoked.

For the other members: I have edited some of the replies as well.
 
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Thread Starter

paulsheer

Joined Sep 30, 2015
15
I am asking for the favor of answering two simple questions.

"Will it work?"
"What is the leakage current?"

I have gotten the answer to the first question. Which is "no" for which I now owe $100. Now I need an answer to the second question.

It is a very simple question to answer.

I have even provided a link to the datasheet in the first post. All one would need to do is click on the link and read a few lines. It should take such experts as yourselves mere seconds to do so and reply with a figure or range of values in milli-amps.

However it would seem that some people have alternative "agendas" that prevent them from answering this straight question.

We are now into over 12 hours and 16 posts and the simple answers which consume mere seconds of people's time are not answered.

Yet I was very polite and specific in my first post.

I would suggest to you that it is not myself that is at fault and that you rather reprimand other forum members.

In their own way, they are impugning the reputation of this web site by obfuscating and misdirecting.
 

JohnInTX

Joined Jun 26, 2012
4,787
I would suggest to you that it is not myself that is at fault and that you rather reprimand other forum members.
In their own way, they are impugning the reputation of this web site by obfuscating and misdirecting.
I disagree. Your second post jumped on Max for not giving you the answer you expected. He apologized but you escalated and it got worse from there. I don't condone all of the member's responses to your conduct and some of their replies have been snipped, too.

FWIW, in all of the caterwauling, your question was actually answered:
Put the load in the collector. Use a base resistor. (@EM Fields @profbuxton )
Leakage current is not as simple (@EM Fields and @bertus) as you may like. The datasheet specifies ICER as 1-5ma with specific conditions applied. Anything more detailed than that and you need to be talking to ST - not railing at members who are trying to help you.

The thread will stay open as long as the discussion pertains to the original question. Personal taunts, attacks, accusations of hidden agendas, disrespectful conduct towards ANY member, new or veteran, will not be allowed.
 
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Thread Starter

paulsheer

Joined Sep 30, 2015
15
"The datasheet specifies ICER as 1-5ma with specific conditions applied"

Thank you sooo much!!!

That is exactly what I wanted to know. 1 to 5 milliamps. Excellent.

You see how simple that was?? John, you are a credit to electrical engineers everywhere.

I have a followup question now, but I think I need to take a break first and recuperate!!! It's very stressful trying to draw blood from a stone :) I think I'll rest for the remainder of the weekend. Ha ha.
 

JohnInTX

Joined Jun 26, 2012
4,787
OK, I guess, but its not a complete answer. I think you would be well served to consider the suggestions of the others, most of whom have much more practical experience in this area than do I.
 

Thread Starter

paulsheer

Joined Sep 30, 2015
15
if i went into a discussion about each suggestion i would still not have an answer even after 100 posts

and i have an answer now
 
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