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sorasit46
Joined: 07 May 2007
Posts: 69
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 Posted: Tue Mar 25, 2008 6:11 pm |
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| baltazar_aquino wrote: | | you can use the complementary pair NA51W/NA52W . Or the equivalent 2N6121/2N6124 pair. But I dont' know exactly your load current. You just mentioned the relay rating (8A/24V). With these transistors, you can safely switch 3A (4Amax) up to 40V(max) VCE. If the rating is not enough for you, you can use the whooping 2N3055/MJ2955 pair (TO-3 package). In both cases you have to buffer first your PIC pins with smaller signal general purpose transistors (complementary pair also). |
DC Motor specification is 24 Volts 5Amps(typ) 8Amps(max).My area is not sufficiency for TO-3 package.
regard |
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sorasit46
Joined: 07 May 2007
Posts: 69
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| Posted: Tue Mar 25, 2008 7:33 pm |
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My schematic.
Please suggest me too.
regard
[img]http://www.uppic.net/showpic.php?picid=0e5de8d17b354ef356c3f5d1b1b1ea5a[/img] |
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sorasit46
Joined: 07 May 2007
Posts: 69
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| Posted: Tue Mar 25, 2008 7:35 pm |
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| sorasit46 wrote: | My schematic.
Please suggest me too.
regard
[img]http://www.uppic.net/showpic.php?picid=0e5de8d17b354ef356c3f5d1b1b1ea5a[/img] |
Sorry,I can not up my schematic direct to this board. |
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Ttelmah
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| Posted: Wed Mar 26, 2008 3:52 am |
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First, the base drive resistors need to be smaller than 4.7K. The BD140, has a worst case current gain of just 40. With 4.7K, into the base, you will only potentially deliver 40mA to drive the relays!....
You don't show the power connections to the relays. Don't know what the coil ratings are, but the IN4148, is quite a nice 'quick' diode for the transient suppression on these. However is it's current rating high enough?. I doubt it. Same comment applies even worse, to the one across the PWM fet. Take the diode here to the ground pin, otherwise you risk driving the AN0 pin -ve. However see below, for comment on trap diodes.
What supply suppression have you got _close_ to the PIC?.
Add a driver to feed the FET. The STP80N55, has a worst case threshold voltage of 4v. With it's transconductance at 150S, using a PIC pin to drive it at 4.3v (worst case from a 5v supply), it should switch reliably, but it will be dissipating more heat than it should.
Really, I'd add four trap diodes. Two each side of the motor, to the supply rails. The problem you have is that the only thing guaranteed to be there, to snub the motor, if the relay contacts are both open, is the small capacitor. Look at how the diode networks normally present on H-bridges, to see how they should run. With these, you then don't need D3. Add small capacitors across the relay contacts as well.
You will need to add an integrator, between the current sense point, and the PIC input, to get much meaningful data from this. Otherwise you will randonly see high currents as the PWM is on, and low currents as it is off, depending on the sampling times...
Best Wishes |
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baltazar_aquino
Joined: 16 Mar 2008
Posts: 27
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| Posted: Thu Mar 27, 2008 2:15 am |
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| Ttelmah wrote: | | First, the base drive resistors need to be smaller than 4.7K. The BD140, has a worst case current gain of just 40. |
First, when you use a transistor for switching , you should not think of the current gain once the transistor reached saturation which happens at the "ON" condition. It does not apply anymore. Your collector current at saturation is determined only by the following: Rail voltage, VCEsat (~0.2V) & the load resistance. Current gain is suitable only in the active region (ie when you use the device for amplification). Generally, 4.7K is actually more than enough to drive a general purpose relay drive transistor. That's why for switching alone, you can liberally replace transistor with another model of the same type. You are right, the gain of 40 is in the worst case scenario but the worst case is not always the case 
Second, I think we are missing something here. You guys are definitely discussing a schematic diagram which others have not seen. We cannot contribute unless we see what you are talking about. |
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Ttelmah
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| Posted: Thu Mar 27, 2008 3:17 am |
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What you say, is generally correct, once the transistor has _switched_, but not as it switches, when it functions as a normal amplifier. 4.7K, is not good in a whole series of ways. The first is that really a resistor ought to exist from the base to collector, to speed of the switch off. Second, while 4.7K, wll be adequate to get it through the switching state in the on direction, if the gain is 'typical', it won't on these transistors if they have the lowest gain, if the relay resistance is reasonably low (don't know what these relays draw, but generally, a relay able to switch 8A reliably, will have quite a low coil resistance - though because of their inductance, the initial current will be low - which may 'save' the situation I would expect this to be an unreliable solution in some cases...).
Best Wishes |
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baltazar_aquino
Joined: 16 Mar 2008
Posts: 27
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| Posted: Thu Mar 27, 2008 5:33 am |
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| Ttelmah wrote: | ....a relay able to switch 8A reliably, will have quite a low coil resistance - though because of their inductance....
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There is no relationship between the contact & coil ratings. I have here right now two PCB mount type relays , a 10A/28Vdc contact rating , the coil resistance is 380ohms and rated at 12VDC. It just requires around 32mA to energize the coil. Another is a 5VDC coil relay and contact rating is 24VDC/10A , coil resistance is 70 Ohms. This requires around 72mA to energize the coil . Absolutely no relationship.
| Quote: | | Generally, 4.7K is actually more than enough to drive a general purpose relay drive transistor. |
My question is:
Did he use a BD 140 to switch a single 24VDC/8A relay ? if he did- this is a wrong application of this transistor. Instead of advising him to consume more current by lowering the base bias R, i would rather advise him to change to a small signal type transistor like A1015/C1815 which are more than enough to drive these type of relays under any circumstances - as I have done in my designs. BD140 approaches the intermediate power application region (hence you have the thermo tab and the low hFE) and ought not to be used for small signal switching only - using it for this purpose is an overkill and doesn't show sound engineering design principle, unless he just gussed about the value. Anyways, again, the point is, I have not seen the schematic diagram. it will benefit everybody if what you are discussing is visible also to others otherwise, just do it by PM ( no offense ) |
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Ttelmah
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| Posted: Thu Mar 27, 2008 6:09 am |
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Beware of mixing contact ratings, with switching ratings. If you use a 10A relay, to switch a motor drawing 8A, you are asking for rapid failure.... There is a general (but not exact) relationship between coil sizes, and contact sizes/speed of closure and the ability to suppress arcs. For a given _voltage_ relay, this then relates to the current drawn. You can get tiny relays with high contact ratings, but they have slow closure times, compared to versions with larger coils.
Your two relays, are operating at different voltages, which (of course) hides any relationship. Both are basically 0.4W designs.
Try to find a 5V unit able to handle twice the current, and you will find that the coil current also rises.
Best Wishes |
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sorasit46
Joined: 07 May 2007
Posts: 69
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| thank you Telmah & baltazar & webboard |
| Posted: Thu Mar 27, 2008 7:51 am |
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Now I have many knowledge about BD140. I choose BD140 because BD140 can supply 1.5A .Not supply only relay but in the future I can add some equipment such as horn or spotlight.
Now my important question is ....
Please see new schematic.
[img]http://upload.siamha.com/v.php?id=82210ask2jpg[/img]
Please to suggest me again.
Regards |
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baltazar_aquino
Joined: 16 Mar 2008
Posts: 27
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| Posted: Thu Mar 27, 2008 12:02 pm |
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| I sent you a PM. check your folder. |
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