# General Circuit questions

Discussion in 'Homework Help' started by SallySmith, May 17, 2012.

1. ### SallySmith Thread Starter New Member

May 17, 2012
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Can anyone answer these questions on the attached schematic? Much appreciated!!

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Q1. This part of a circuit is a DC regulate power supply.
T1 is a main transformer which convert the voltage from the mains to a different, usually lower, voltage. The AC input voltage 220V or 120V is convert to a smaller 12.6V AC voltage. Next this voltage is rectifier (BR1) and smooth by C22 and C16 capacitors.
After this we get a DC voltage (17.7V).
U11 is a voltage regulator. A voltage regulator provides constant DC +12V output voltage and contains circuitry that continuously holds the output voltage at the design value regardless of changes in load current or input voltage.

Q2. U2 - It' a PLD circuit a programmable logic device. Se we can not tell how this circuit work. The only think that we can tell is that this part of a circuit is some kind of a control circuit. Based on the input signal states U2 controls the motor, heater and other part of a circuit.

Q3. This part of a circuit is a motor driver.
U9 provides opto-isolation between low control voltage and and high voltage needed to start the motor.
Based on the input signal states U9 turns on (trigger) the triacs and triac turns the motor to run in reverse direction.

Q4 Well this part of a circuit drivers the external counter.
Q3 is a BJT http://en.wikipedia.org/wiki/2N2222

Last edited: May 17, 2012
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3. ### SallySmith Thread Starter New Member

May 17, 2012
4
0
Thank you for explanations! How did you calculate the stepped down AC voltage after the transformer and how do you calculate the DC voltage?

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,991
1,115
Simply I find the transformer data sheet in the internet
http://parts.digikey.com/1/parts/998361-xfrmr-pwr-115-230v-6-3v-1-6a-14a-10r-12.html
So I know the secondary voltage was 12.6V AC.
From the circuit theory you should know that AC voltage after passing rectifier bridge and smoothed by C1 is equal to

Vdc = Vac *√2 - 2Vd = (12.6V * 1.41) - 1.2V = 17.7 - 1.2V = 16.5V

Where Vd is a bridge diodes voltage drop.

U10 is 7805 ---> 5V DC voltage regulator
U11 is 7812----> 12V DC voltage regulator