I’m using the 12v squirrel cage blower in an automatic temperature controller setup for a smoker but need to slow it down. Being an electronics novice, can anyone let me know what size pot I’d need to use? Cheers
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Hi Lance, Welcome to the forum 
What are you driving the blower with?
The product page for the Squirrel Cage Blower lists an operating voltage of 6-12 Volts.
If you power it with a lower voltage it should run a bit slower.
You could just strap a potentiometer in series with it, but it’s a messy way of doing things as you are effectively just burning up power in the potentiometer instead of the motor, so would need one with a power rating similar to the motor.
Alternatively, you could PWM control your blower power supply to make it adjustable that way.
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Hi all
Hold up. I think you might need a wire wound rheostat. Pots to handle that sort of current would be a bit scarce. There is one little point often forgotten. When a pot has a power rating, could be 0.5W, 1W or whatever, that rating applies only over the whole of the resistive element. When using a part of this element as in a rheostat connection only a part is used and the power handling rating has to be reduced accordingly.
PWM control is the better way to go.
Cheers Bob
PS. Isn’t a squirrel cage motor AC???.
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Hi Bob,
Good point, I figured there was a reason I don’t see 10 Watt potentiometers floating around. Definitely one of those problems that seems like there may be a quick easy shortcut, but has other consequences.
I thought so too! I think in this case they are referring to the shape of the blower fan blades looking like a squirrel cage, in much the same way a squirrel cage rotor does. A bit of a misleading name and until I read the label was not quite sure.
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Hi Trent
Aint that the truth. I have always known them as centrifugal blowers and yes they do pump quite a bit of air for their size. I have come across them in higher power transmitters and I am not sure but they may even had squirrel cage motors in them. They always had to start the right direction.
Don’t look much like what I would imagine a squirrel cage looks like. But in all fairness although I have seen squirrels running around the trees in the UK I have never seen a cage for one.
Cheers Bob
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Hi Trent. It’s powered by a 12v transformer/PID/SSR setup. I found the plans on a UK BBQ site. He was using a 14CFM fan and said he powered his down with a pot in series. This is what he said:
“A rotary style potentiometer wired in series with the fan provides a bit of manual speed control. All this does is allows you to slow down the top speed of the fan. The controller does not vary fan speed, it just turns it on and off. If your fan is over powered (like mine is) then when the controller switches the fan on, it blasts air in too quickly to the fire, it heats up very quick and then the temp over shoots. Throttling the fan down using the potentiometer allows a more subtle and gentle control, which helps the PID controller.”
Cheers.
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Hi Lance
Actually AD-DC Adaptor. Bit of a difference, “Transformer” implies AC output.
I hope that bit of mains wire from the PID controller to the SSR is not carrying mains voltage. I think most SSRs are 3-30VDC operating voltage. You must have been lucky to be able to find a DC SSR. Most I have seen have been zero crossing switches and as DC does not get to zero will never switch
You should ask where this guy found a pot with enough current capability to use as a speed control. Also if he is still using it. Might have fried by now. If you only want to slow it down a bit measure the current then calculate a resistor to drop the voltage to 9 or 10 volts or whatever you need A bit wasteful on power heating up a resistor but might be the way to go if you don’t want to have continuous control over speed. Don’t forget to calculate the resistor power requirement too, I will probably get quite hot. If you don’t know how to do this refer to Michael’s recent post and video here
Cheers Bob
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Hi Bob. Thanks for the reply. As I said in my initial post, I’m a novice at this stuff and I’d never heard of PID’s and SSR’s until I started this project. Unfortunately, the bloke who posted the plans stopped replying to comments about 18 months ago. If I understand correctly, using the fan that Trent posted a pic of above where the current is 0.9 amps, to drop it down to 9 volts, I’d need a 10 ohm resister. I hope that’s right. I’ve attached the link to the plans if you’d like to have a look. Thanks for your help, I appreciate it.
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NO WRONG.10Ω at 0.9A would drop 9V. If you need 10V you will be dropping 2V (12-10). To do this at 0.9A the resistor would be 2.2Ω (2/0.9). At the reduced voltage the current will be a bit reduced also so it will not quite drop the full voltage. Bit of a circle really but there will be an equilibrium somewhere. 2.2Ω is a preferred value but if this is not slow enough 2.7Ω is the next preferred value then 3.3Ω. The resistor power rating with these values ranges from 2 to 2Watt so I would go for 5W. Be a bit careful as it will probably get hot.
Cheers Bob
PS I don’t think you looked too closely at the video I linked
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I thought I looked closely Bob, but maths isn’t my strong suit and I got a bit confused. I think I’ve got it now, the voltage I use in the ohms law calculation is the voltage I want to lose, not the voltage I want to end up with.
Thanks again,
Lance
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Hi lance
That is correct. Don’t worry, you are not the first and I don’t think you will be the last to fall for that one. Having learned through a mistake you are unlikely to make the same one again.
The power you lose is dissipated as heat (got to go somewhere) so be careful touching, HOT.
Cheers Bob
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Hi Lance,
I can’t stress this enough, but you should avoid dealing with mains voltage wherever you can. While some projects may be made slightly simpler by switching AC, yours isn’t one of them, and the risk of death outweighs the mild convenience of using someone else’s design by many orders of magnitude. Your current design exposes 240V in a few places, one small slip up and you could be touching it. I suggest dismantling your project as it is and discussing something simpler and safer with us.
You can use an enclosed DC plug-pack to supply your DC voltage for your fan, and use a MOSFET to switch your fan with PWM.
What sort of outputs does your PID controller have? How much current does your fan want?
Keen to get you back on track.
-James
Hi James
Yes. I am sure 12VDC PID units would be available. Don’t know about higher output 12V plug packs though. To be conservative would probably need about 3A +. An adaptor as shown may be the only way.
Looking at the pic Lance posted he is using a 12V output AC-DC adaptor to power the fan. It looks like the PID module is 240VAC powered. I think he has used a bit of mains cable to connect the PID to the SSR and I think this a low DC voltage to switch the SSR.
I agree with your point re handling mains voltages. I also note the mains has no earth wire. It could be that the PID and the adaptor are double insulated and make no provision for an earth connection.
I Think the use of a bit of mains cable for the PID to SSR connection is a mistake. If someone walked up to this scenario and did not know they could well be misled.
Cheers Bob
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Hey Lance,
Michael has just recently made a new tutorial that may help out if you’re getting started with it:
It seems James and Bob are on the money here. If you go with PWM for control, you don’t have to worry at all about the resistance on your power supply, essentially all you’re doing is only pulsing the power on long enough to get the motor accelerating in pulses so it maintains a lower speed. On a scope this would look like this, with the variations depending on what’s known as the duty cycle:
When you say it’s an automatic temperature controller setup, are you using a microcontroller for it? If so, most are quite easy to script to increase or decrease the duty cycle on one of their GPIO which with an appropriate relay or MOSFET can be used to control the 12V DC headed to your squirrel cage blower (I’m with Bob and Trent here, that name is certainly a misnomer. Squirrel cage motors are usually used for applications where you need to have the motor last for a really long time without wearing or having to worry about replacing brushes such as in a ceiling fan, this is basically just a fan that blows outwards sitting on a brushed DC motor. I made this a link to the wiki for it if anyone reading is curious about it 
)
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Hi James.
I appreciate your concerns WRT dealing with mains voltage and my intention is to box everything up. The output of the PID controller is 12VDC and the fan draws .9A.
Cheers
Lance
Hi Bryce
I’m using a PID controller as in the pic above. The only reason I called it a squirrel cage blower is that’s how it’s described on the Core Electronics website.
Cheers
Lance
Hi Guys
Obviously I want to do this as safely as possible. I’ve just re read the instructions that came with the PID and it says that it can use three different power supplies including 12-24VDC, I’ve attached a copy. If I used 12VDC as the power supply, am I safe to assume that I could run the blower directly from the SSR?
I think that the reason the bloke who designed this used a pot is because, until you have the fan running into the smoker, you don’t know how fast you want it to run to prevent temp. overshoots. I had another read of his post and there was a question from someone else re the pot and he posted this link https://uk.rs-online.com/web/p/potentiometers/5028621/?searchTerm=5028621&relevancy-data=636F3D3126696E3D4931384E525353746F636B4E756D6265724D504E266C753D656E266D6D3D6D61746368616C6C26706D3D5E5C647B367D247C5E5C647B377D247C5E5C647B31307D2426706F3D313426736E3D592673743D52535F53544F434B5F4E554D4245522677633D4E4F4E45267573743D35303238363231267374613D3530323836323126
Thanks for you help, much appreciated.
LancePID Controller.pdf (371.4 KB)
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I can see no reason not to run the PID controller from 12v.
There is an inconsistency in the information you have provided. This is part of the quote from the site: “The controller does not vary fan speed, it just turns it on and off.”. If that’s what’s happening then he has it wired wrong. The Relay Contact Output will just turn the fan on and off. To get fan speed control you need to use the Voltage Pulses Output. That should output either variable timed pulses of fixed width or fixed time pulses of variable width, which in either case will control the fan speed directly.
Can you confirm from that site that he is using the Relay Contact Output because if he is then from my reading of the PID controller spec sheet you already have everything you need for continuous control of fan speed through the Voltage Pulses Output with everything operating at 12v.
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Hi Jeff
I can’t see anywhere where he refers to any of the outputs he used. As I’ve said to the others, I’m a bit of a novice at this stuff. Does using the Voltage Pulses Output mean that I connect the fan directly to the PID and if so, which terminals would I use? The manual that came with the PID has connection setup diagrams for the 100V and 100R whereas the one they sent me is a 100VH which doesn’t have terminal 7. I’ve rescanned the whole manual and attached it and a pic of the back of the unit.
Cheers
LancePID Controller.pdf (1.5 MB)
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I had assumed that 100V and 100R referred to two sets of terminals. The specs use the term ‘modes’, not ‘models’ and there is a parameter setting to select between PID and On/Off. But the two wiring diagrams seem to indicates different models, although it could indicate simply two different ways of wiring the same set of terminals. If there are two models that would explain why the web site doesn’t indicate which set of terminals he selected - there wasn’t a choice. But the original comment stands - if he claimed that the controller only turns the fan on and off then that suggests he used either the relay model or the dual-mode model in relay mode.
I would guess that a HV model is a version of the V, but it would be good if you could confirm that. If it is, or if the difference really is a mode and not a model, then the question goes back to your very first comment - how have you determined that it needs to be slower? In other words, does it slow down if you heat the thermocouple up? If it doesn’t, have you checked the parameter for Control Output mode is set to PID and (I presume) the Output Mode is set to Time Duty? If those settings are correct the issue might be with the thermocouple setting or the PID parameters, or both.
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