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Schematic for voltage drop (From 12VDC to 6VDC ) Answered

Hi all ^^
I am fixing a cassette deck ,but i am facing a problem.
The original motor runs on 12 Volts DC ,but it is dead .
I bought a new motor from the same company and with the same specifications (2400RPM).
The problem is that it runs on 6Volts and not on 12Volts,  as the original is .
Since the deck is old ,this is as close as i can get to finding a motor .
The motor itself has a circuit within it ,that keeps the voltage and the RPM constant . So even if the voltage drops, this does not affect performance . If i use 12Volts ,I might destroy the circuit . If i use resistors to lower the voltage only ,then it does not have enough electrical current . Those motors need to draw freely electrical current ,since they draw more current when they are stressed ,so that they can keep spinning constantly at the same speed .This is why i need a schematic to lower the voltage ,but i don't know where to find one .
The 12VDC comes from the main board of the cassette deck .
I have a service manual that i can publish if needed .

Thanks and best regards ,


If the motor just needs a voltage around 6V, you might be able to go without a regulator and just put some diodes into the supply line. Each diode will drop about 0.6..0.7V. So you can start with 10 diodes (1N4001 or similar will be fine) and measure the voltage - preferably when the motor uses the most current as the voltage drop increases with the current. If the voltage is too low, remove a diode and try again.

If you have an 7805, you can get close to 6V by applying a pair of diodes in series with pin 2 back to the voltage source.. It will yield 6.2V from output pin 3 to the V source return. (not to pin 2!!).

Note that using a pair of schottky diodes or one each of a schottky and a normal 1n4001 may get you closer to the target

at a 12V supply voltage, you may want to add a few 1N4001 diodes in series on the input or a power resistor (I'd prefer the diodes since they'll drop a relatively constant voltage) as well, by the way, since the voltage difference between the source and the output will be dissipated by the regulator. Minimizing that value reduces the thermal loading. 1/2 amp should be no problem.

There are 6V linear regulators, like the xx7806 that would be easy to use. They have 3 leads, input, ground, and output. The only other components needed would be capacitors on the input and output, as recommended by the datasheet of the part. The datasheet should have some basic schematic of a typical circuit in it.

These regulators commonly come in the TO-220 package, and the ratings will usually say something like 1 or 1.5 Amp. The current rating itself must be taken with a grain of salt, as the maximum allowable current will really be determined by how much power the part will dissipate. All the voltage above the output voltage will be dropped across the part, and it will carry the same current as your load. If you power it from 12V, it will drop 6V across it times the 60mA, for a power dissipation of about 0.36Watt. It should easily handle that without special heatsinking. The higher currents during startup of the motor will be brief and infrequenc, so it shouldn't cause a problem there either.

Thanks very much for the answer ! It helped allot .
I will make it work tomorrow .

If a lower voltage will work, a 5V regulator is easily available, although you may be able to find a 6V regulator. Conventional linear regulators say they need the supply voltage to be at least 7V above the output, but you should have no problems with just 6V.

Thanks allot for the reply .
I only wonder if it will handle the electrical current.,since i found a 5 v regulator ,but it is rated at 100mA . The motor usually gets between 40 and 60mA ,but sometimes it could eat as much as 200-300mA when it gets harder for it to move . This may also happen in the beginning of the momentum when it starts to rotate , thus drawing more current in order to start moving the capstans .Will this increase in consumption harm the regulator ?It should last just a small amount of time .

You should be able to get a regulator that can deal with up to 0.5A continuous, and they are commonly available for up to 2A.

For a couple of seconds, a little over-current should be fine if you only have a small voltage drop. The regulator has to burn off the excess power, so is designed to dissipate the full 100mA at 25V when connected to a 30V supply.

The best thing to do would be to check the datasheet and see if it has a specification for pulsed current, and if it has any current regulation/cut off featured.