Adapting a circuit for a higher voltage-of alarm clock with dc motor.?

Thread Starter

Harkonnenback

Joined Sep 21, 2021
6
Hi,

I was looking to connect a motor to an alarm clock so that it would start and stop at a certain time.
I found this thread, which was very helpful, on this site: https://forum.allaboutcircuits.com/threads/replace-speaker-of-alarm-clock-with-dc-motor.101453/. It discusses how to do just that, replace the speaker of an alarm clock with a dc motor.

I didn't want to go into an alarm clock and try to work with the internal components, since I am pretty inexperienced, so instead I bought an alarm clock that has an external vibration disk. The disk is meant to go beneath your bed or pillow and shake to wake you up. It connects to the alarm clock via a detachable cord. Anyways, I thought that instead of going into the clock I could just use this external wire to pretty much do the same thing.
Here is a link to the product for reference:
https://www.amazon.com/dp/B000OOWZUK/ref=cm_sw_em_r_mt_dp_F0CJBA2E2CHGQTVT6GFP?_encoding=UTF8&psc=1

I cut off the vibration disk and stripped the wire. I measured the current with a multimeter, and found that the voltage running through the wires jumped around from 0 to 5 to 8 or 9 volts as the alarm went off.

This differed significantly from the voltage discussed in the previous thread, which was a 1.5 volt alarm clock, with the current running to the speakers only going up to 0.8 max.
Currently, I am trying to create the circuit design created in the thread mentioned previously, which to my knowledge was thought up by @Rich2 and drawn up by @absf . However, I wanted to hold off on ordering components until I found out if the design would need to be modified in any way as a result of the higher voltage.

Here is the design. As I understand it, the general idea behind the design is to sense if a current is being sent, and if it is, to activate a second circuit, which has its own battery power source, thus solving the issue of fluctuating or incompatible power coming from the alarm.1632278933825.png

Thanks in advance for your help!
 
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Thread Starter

Harkonnenback

Joined Sep 21, 2021
6
I'm guessing the diodes can remain the same, since they are directing the current. But my main question is should I change the resistor and the BC337?
 

Audioguru again

Joined Oct 21, 2019
3,677
But my main question is should I change the resistor and the BC337?
You never mentioned the extremely important current used by the motor. The datasheet of the BC337 says its max output current can be 500mA when its base current is 50mA. When the motor current is known then the transistor, its base resistor and the capacitor value can be calculated.
 

Thread Starter

Harkonnenback

Joined Sep 21, 2021
6
You never mentioned the extremely important current used by the motor. The datasheet of the BC337 says its max output current can be 500mA when its base current is 50mA. When the motor current is known then the transistor, its base resistor and the capacitor value can be calculated.
Thanks for the reply! How about this? https://www.jameco.com/shop/ProductDisplay?catalogId=10001&langId=-1&storeId=10001&productId=2310215
Data sheet says rated load current is 220 +/- 44 mA, so it shouldn't exceed the BC337 right?
 

Thread Starter

Harkonnenback

Joined Sep 21, 2021
6
Forgot to mention this before, but this was the last post in the previous thread and no one ever responded to it, so I'm uncertain how accurate it is. Do you think it would be better to use a voltage doubler, or was that only because the voltage of the original alarm clock was so low?
On Rich's circuit, it would be much better to use a voltage doubler than a bridge. Place the resistor between the filter cap and the transistor base. A capacitor on the base never develops much voltage and it is drained away almost immediately.
 

Audioguru again

Joined Oct 21, 2019
3,677
The motor is rated at 24V and draws a "locked rotor" current of 905mA each time it starts running.
You do not say what kind of work the motor will be doing so it might not work with only 12V powering it and a driver (Mosfet maybe?) must be able to feed it the high current.
The rectified and filtered speaker signal voltage might be too low to drive a Mosfet.
 

Thread Starter

Harkonnenback

Joined Sep 21, 2021
6
The motor is rated at 24V and draws a "locked rotor" current of 905mA each time it starts running.
You do not say what kind of work the motor will be doing so it might not work with only 12V powering it and a driver (Mosfet maybe?) must be able to feed it the high current.
The rectified and filtered speaker signal voltage might be too low to drive a Mosfet.
So, in order to adapt the circuit for 24V instead of 12V, can I just double the batteries there? (For the work the motor will be doing, I'd like it to wind a string connected to a latch so that it will open the latch)
 

MisterBill2

Joined Jan 23, 2018
9,542
Hi,

I was looking to connect a motor to an alarm clock so that it would start and stop at a certain time.
I found this thread, which was very helpful, on this site: https://forum.allaboutcircuits.com/threads/replace-speaker-of-alarm-clock-with-dc-motor.101453/. It discusses how to do just that, replace the speaker of an alarm clock with a dc motor.

I didn't want to go into an alarm clock and try to work with the internal components, since I am pretty inexperienced, so instead I bought an alarm clock that has an external vibration disk. The disk is meant to go beneath your bed or pillow and shake to wake you up. It connects to the alarm clock via a detachable cord. Anyways, I thought that instead of going into the clock I could just use this external wire to pretty much do the same thing.
Here is a link to the product for reference:
https://www.amazon.com/dp/B000OOWZUK/ref=cm_sw_em_r_mt_dp_F0CJBA2E2CHGQTVT6GFP?_encoding=UTF8&psc=1

I cut off the vibration disk and stripped the wire. I measured the current with a multimeter, and found that the voltage running through the wires jumped around from 0 to 5 to 8 or 9 volts as the alarm went off.

This differed significantly from the voltage discussed in the previous thread, which was a 1.5 volt alarm clock, with the current running to the speakers only going up to 0.8 max.
Currently, I am trying to create the circuit design created in the thread mentioned previously, which to my knowledge was thought up by @Rich2 and drawn up by @absf . However, I wanted to hold off on ordering components until I found out if the design would need to be modified in any way as a result of the higher voltage.

Here is the design. As I understand it, the general idea behind the design is to sense if a current is being sent, and if it is, to activate a second circuit, which has its own battery power source, thus solving the issue of fluctuating or incompatible power coming from the alarm.View attachment 248554

Thanks in advance for your help!
The "Rich2" circuit will certainly work in this application, however, several component values may need to be changed, AND a different switch transistor may be required.
 

Thread Starter

Harkonnenback

Joined Sep 21, 2021
6
The "Rich2" circuit will certainly work in this application, however, several component values may need to be changed, AND a different switch transistor may be required.
How should I figure out what switch transistor to use? Is there a table or equation that would show what would work?
 

MisterBill2

Joined Jan 23, 2018
9,542
To determine the correct transistor choice you need to know the current that your motor will require, and the voltage that it uses for power. Then look at a list of power switching transistors for one able to handle more than the motor current by at least 20%. At this point you will want to decide what sort of heat sink your transistor selected will require. There will be a lot of choices available so cost and availability to you will help make the decision about which one to pick.
After that you will need to consider the required base current to switch on the transistor adequately. That will affect the value of R1 in the "Rich 2" circuit. The value shown is probably good for most applications.
 
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