Flyback Diode Needed?

[ThomsCircuit]

Hope my schematic is legible enough.
I have been shown to always place a diode across the relay but what about the FAN in this project?
This circuit is 5V. The switch is a TTP223. Relay is a 5v cube style which the ECAD model was incorrect. NO & NC swapped. I find that happens often.
The Fan is a 40mm cpu type fan. Well probably used on video cards nowadays. Brushless if that matters. Ive add a few diodes in series to drop the voltage to the fan down to 3.7v and added a pot to slow it down more if needed. so my Q is do i need a diode across the fans plug? (center screen nearest the relay.) I have some knowledge that the diode diverts current eliminating spikes but do I need one for this small fan?

 

[Externet]

The fan is not driven by a transistor/mosfet but by relay contacts; not affected by spikes. No need for the diode.

 

[ThomsCircuit]

The fan is not driven by a transistor/mosfet but by relay contacts;

thank you.
Your reply to my previous post got my wheels spinning. I like using relays. I understand them but i can see that the transistor can acomplish the same with a smaller footprint. Difficult for me to figure out which one to use.

 

[ChrisP58]

thank you.
Your reply to my previous post got my wheels spinning. I like using relays. I understand them but i can see that the transistor can acomplish the same with a smaller footprint. Difficult for me to figure out which one to use.

Do you understand the transistor that drives the relay? It is no different.

I don't know the details of your fan and relay, but you could probably replace the relay coil with the fan and it would work just fine. The fan might need more current, but that's just a matter of changing to a different transistor.

 

[Diver300]

You appear to have an LED in parallel with the fan, which will limit the voltage to the fan, and may burn out if you turn up the potentiometer too far.

I agree with the others that a transistor could replace the relay. You could also use at transistor with a heatsink to regulate the voltage instead of using the series diodes and the potentiometer.

In this case the relay is not switching much current. If you do have relays switching a lot of current or inductive loads, a flyback diode on the coil can be a problem. They slow down how fast the relay's magnetic field reduces, so it can slow the opening of the NO contact, leading to more arcing on the contacts. The solution is a resistor in series with the flyback diode. The value should be 5 - 10 times the relay resistance, which will lead to a voltage spike that is 5 - 10 times the supply voltage. Obviously the transistor has to be rated to that, but the voltage is still being limited, so the voltage needed for the transistor is known. Without a flyback path, the voltage can be much larger, and will depend on lead capacitance and how fast the transistor is switched off.

 

[ThomsCircuit]

he solution is a resistor in series with the flyback diode. The value should be 5 - 10 times the relay resistance, which will lead to a voltage spike that is 5 - 10 times the supply voltage.

Songle 5V relay

• **Coil resistance: 70 ohms**
• Coil rated voltage: 3-48VDC
• Coil pick-up voltage: 5 VDC
• Coil Current: 71-90mA

The resistor here would be 350 - 700 ohms
So if I were to use a transistor how would i calculate the requirements?

you could also use at transistor with a heatsink to regulate the voltage instead of using the series diodes

I have on hand BD243C. Could you show how i would limit the fan to say 3.8 volts?

 

[ThomsCircuit]

You appear to have an LED in parallel with the fan, which will limit the voltage to the fan

i was wanting to use the led (white Fv=3.0) to provide visual feedback if i adjust the pot.

 

[ChrisP58]

What are the spec's of your fan? Can you post a link for it?

 

[ThomsCircuit]

Vf1 = fan voltage, (5)
Va = fan amps, (0.14)
Rf = Vf1/Va (36)

What are the spec's of your fan? Can you post a link for it?

i had looked this up a few days ago. this ok?

 

[ThomsCircuit]

voltage from pin 1 of the ttp-223 is about 4.9v
My power supply is a 12V 10A Transformer stepped down to 5.5v using a buck converter

 

[Diver300]

Songle 5V relay

• **Coil resistance: 70 ohms**
• Coil rated voltage: 3-48VDC
• Coil pick-up voltage: 5 VDC
• Coil Current: 71-90mA

The resistor here would be 350 - 700 ohms
So if I were to use a transistor how would i calculate the requirements?

The Coil rated voltage can't be 3 - 48 V when the pick-up voltage is 5 V. I think that 3 - 48 V is the range of voltages available for relays of that type. I would guess that voltages of 3, 5, 12, 24 and 48 V, and maybe some others, are available.

If your coil resistance is 70 Ohms, and you have a 350 Ohm resistor in series with the flyback diode, and the supply is 5 V, the voltage spike when the transistor turns off will be around 25 V, so the transistor rating needs to exceed that.

What happens is the the coil current will be 5/70 = 71 mA. When the transistor turns off, the inductance of the coil means that the current in the coil cannot change instantaneously, so initially the 71 mA will flow in the resistor. The voltage across the resistor is therefore 0.071*350 = 25V.

You probably don't need to have the resistor with the fan current being that small.

 

[Diver300]

i was wanting to use the led (white Fv=3.0) to provide visual feedback if i adjust the pot.

You need a resistor in series with the LED. A red one would have a lower voltage. Then the brightness would give some indication of the voltage.

 

[Diver300]

I have on hand BD243C. Could you show how i would limit the fan to say 3.8 volts?

That circuit should work. You need a potentiometer that is 1 kOhm or more to avoid overloading the TTP223. You need a transistor with enough gain so that it can drive the fan without taking too much current from the potentiometer, so a gain of 200 or more is needed with a 1 kOhm potentiometer. The transistor will produce some heat, so if the fan is running at 3.5 V and taking 140 mA, the voltage across the transistor is (5 - 3.5) = 1.5 V and the power is 1.5 * 0.14 = 0.21 W.

The BD547 is more suitable than the BD243C, because the BD547 has enough gain, while the BD243C does not. The BD547 will get hot, but the 0.21 W is much less than it is rated at, and it will only be about 40°C hotter when running. It may be too hot to touch, but well under its maximum temperature of 150 °C.

 

[ThomsCircuit]

Coil rated voltage can't be 3 - 48 V

I wondered about that myself. Thank you. Ive been reading about bjt and mosfet. Their differences, current vs voltage. Its still pretty confusing but your insight here is greatly appreciated.

 

[Externet]

View attachment 133423
That circuit should work...

Hello Diver300.
Can you tell why the fan load is placed on the emitter instead of the collector ?
When is it convenient to configure the load in series to emitter or to collector ?
How does the base current changes in one or the other layout ?
What is the emitter voltage when off and when conducting ?

 

[Diver300]

Can you tell why the fan load is placed on the emitter instead of the collector ?

It is much easier to make the transistor operate in a linear mode when the load is on the emitter.
(edited for typo)

When is it convenient to configure the load in series to emitter or to collector ?

When the load is connected to the emitter, the voltage gain is 1, so any change in input voltage will result in the same change of output voltage. When the load is connected to the collector, as small change in input voltage will result in a large change in output voltage, but it's not possible to know exactly what input voltage is needed for a particular output voltage, so it is very difficult to make the transistor do anything other than turn fully off or fully on.

How does the base current changes in one or the other layout ?

The gain of a transistor will vary with temperature, and between otherwise identical transistors.

When the load is connected to the emitter, the base current is equal to the load current divided by the transistor gain, so the exact base current will vary, so a circuit needs to allow for a variable base current.

When the load is connected to the collector, the circuit design controls the base current. If too little current is used, the voltage to the load will be reduced, but it's not possible to predict the current needed exactly, so usually the base current is calculated with a safety factor.

What is the emitter voltage when off and when conducting ?

When off, the base and emitter voltages will be zero.

When the TTP223 output is on, the base voltage is set by the potentiometer, and can be anywhere in the range 0 - 5 V. The emitter voltage will be about 0.6 V lower than the base voltage.

 

[Nigel Goodwin]

It is much easier to make the transistor operate in a linear mode when the load is on the collector,

I presume you meant 'emitter' there?

 

[Diver300]

I presume you meant 'emitter' there?

Yes, I did. I've corrected the typo.

 

[ThomsCircuit]

Driver300, Mr. Goodwin, and Externet.
You have been so helpful. I may not be a seasoned expert on bjt & mosfets but if i were invited to a transistor party im sure i could hold my own for a while.

 

[ThomsCircuit]

That circuit should work.

Just be certain i get this right I am to eliminate the relay and re-configure my pot (replacing R2 on my sch with the Pot) and my transistor (547)
EDIT. My current Pot is 500r but i do have 1K as well.