# Diode relay series circuit

Discussion in 'The Projects Forum' started by xpyd3r, Mar 2, 2008.

1. ### xpyd3r Thread Starter New Member

Jan 14, 2008
9
0
Okay so I'm not really sure what I'm doing, this is the circuit
+ o------|>|-----/\/\/\/\/-------O))))))------o -
Essentially a 9 v power supply, a diode resistor and relay.
So the coil operates between .08w and .17w so to figure out the current needed at 9v, I say P/E=I or about .12w/9v = .016A. Then since the circuit is series, the amperage stays at .016 for everything. So to find the voltage drop of the 9v, I use resistance and current to find the voltage drop. The relay's resistance is 580 and the diodes resistance is 514. 580x.016= 9.28v and 514x.016 = 8.224v. So that can't be right. I'm pretty new but I'm really trying to understand. So could someone help me understand what I did wrong and what I should be doing to make this work? Any and all help is greaty appreciated, so Thanks so much in advance...

2. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
Ok, the diode DOES drop some voltage across itself, but it is a semiconductor, not a fixed resistance.

A standard silicon PN junction diode will drop between 0.6 and 0.7 volts across itself when forward biased. As the current increases from nothing to the diode's maximum current capacity, that voltage will increase slightly.

Let's assume that your battery is putting out 9v (a fresh battery can put out nearly 10v).
Let's also assume that the diode has a Vf (forward voltage drop) of 0.65V
9V - 0.65V = 8.35V remaining.
The problem now is reduced to how much resistance is required to limit current through your coil, which has a resistance of 580 Ohms, with 8.35V applied across it.
Ohm's Law:
Power = Voltage Squared / Resistance in Ohms
P = 8.35^2 / 580
P = 0.12 Watts (approximate)
You do not need a resistor to limit the current, as you already have 0.12W available to power your relay's coil.

Note that it is good practice to place a diode across the coil of a relay. This surpresses a large voltage "spike" of opposite polarity when the flow of current through the coil is interrupted. The inductor attempts to keep the flow of current constant, which causes the voltage polarity across the coil to invert. This momentary "spike" can reach hundreds of volts, and burn out semiconductors.

See the attached schematic.

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3. ### John Luciani Active Member

Apr 3, 2007
477
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You cannot model a diode as a resistor. The voltage across the diode when it is forward biased (as in your diagram) will be constant. The value of the forward voltage is specified
as Vf(max) in the datasheet.

The voltage across the series combination of the resistor and coil will be (9V - Vf).

(* jcl *)

4. ### xpyd3r Thread Starter New Member

Jan 14, 2008
9
0
Thanks you so much, guys. It's a bit confusing for me. So you used the power and the resistance because the resistance is fixed, while if I were to use the current, it would require a different resistance than the one provided by the relay right? And if it was dropping more voltage than it should, then I would have to calculate the value of the resistance for the resistor to keep the relay from burning out right? Sorry if these are stupid questions but just getting a grasp on the theory of everything is a little confusing right now.

5. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
If you select a relay whose coil is rated at some voltage and arrange to energize it with a voltage source close to the value desired, then there is absolutely no danger of burning up the coil. Even though the voltage source may be capable of supplying that voltage at 100 amps, the relay will only use the few milliamps necessary to pull in the armature.

The diode (across the coil, not in series) protects the voltage source and has no effect on the coil.

6. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
Well, the resistance of your relay is fixed at 580 Ohms, and requires a certain amount of power in order for it to pull the contacts closed (or open); according to what you've stated is between .08W and .17W.
Power in Watts = Voltage times Amperes, or as normally expressed, P = EI
So, you can get the operating current limits by taking the square root of (Power / Resistance)
Ihigh=Sqrt(0.17/580) = 17.1 mA (approx)
Ilow=Sqrt(0.08/580) = 11.7mA (approx)
So, there are your current limits. What voltages will cause that much current to flow?
Voltage = Current times Resistance, or E=IR
Elow = 0.0117*580 = 6.786
Ehigh = 0.0171*580 = 9.918

Get very familiar with Ohm's Law.

7. ### rherber1 Member

Jan 6, 2008
15
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Connecting a rectifier across the coil is NOT to protect the voltage source, it is to prevent damage to switching contacts or semiconductors (transistor) due to back emf. Any self respecting power source for powering relays will have extremely low output impedance and back emf won't do any damage to it. However, a high voltage spike of perhaps 500V or more will zap any transistor in series with the coil and will cause pitting of switch or relay contacts where these are used in series with the coil.

It is well to remember that a rectifier across the coil is not a universal panacea for back emf suppression since it increases the release delay quite considerably, and this may not be desirable in many cases. Other techniques are often recommended.

There are many good references on relay technology around. For instance Panasonic http://pewa.panasonic.com/pcsd/tech_info/pdf/rti.pdf