maychang

Sirigu1992 posted on 2018-9-3 18:58 Teacher, I want to simulate the entire circuit in multisim to see how changing the capacitance and load resistance affects the output voltage, but the circuit...

"I want to simulate the entire circuit in multisim to see how changing the capacitance and load resistance affects the output voltage" The component parameters have little effect on the simulation results.

西里古1992

maychang posted on 2018-9-3 20:46 "I want to simulate the entire circuit in multisim to see the effect of changing the capacitance and load resistance on the output voltage" Component parameters...

Can K7812-500R3 be replaced by LM7812 in the simulation component? What do the letters in the prefix mean? I looked at the input voltage range and the maximum output current are similar.

西里古1992

maychang posted on 2018-9-3 10:04 The working principle of capacitor voltage reduction is that the capacitor limits the current. The current output by the rectifier bridge is basically fixed. The load (from the 24V voltage regulator to the right)...

The input voltage range of K7812 manual is (15V~36V). If the voltage range is not reached when the AC input is 220V, does it mean that there is a problem of low output voltage of the rectifier bridge?

maychang

Sirigu1992 posted on 2018-9-4 13:10 Can K7812-500R3 be replaced by LM7812 in simulation components? What do the letters in the prefix mean? I checked the input voltage range and it is about the same...

"Can K7812-500R3 be replaced by LM7812 in simulation components?" It should be possible. "What do the letters in the prefix mean?" That depends on the datasheet released by the manufacturer.

maychang

Sirigu1992 posted on 2018-9-4 13:16 The input voltage range of K7812 manual is (15V~36V). If the voltage does not reach this range when the AC input is 220V, does it mean...

I have said it more than once before, for example on the 27th floor: "The working principle of capacitor voltage reduction is that the capacitor limits the current, and the current output by the rectifier bridge is basically fixed." The voltage at both ends of the rectifier bridge output DC changes with the load. I also said on the 27th floor: What you said "at 220V, the input end is only about 11V~12V", which is obviously because the load current is too large, resulting in a low output voltage of the rectifier bridge. The reasons may be: 1. The capacity of the voltage-reducing capacitor is insufficient. 2. The load current is too large.

西里古1992

maychang posted on 2018-9-4 14:38 I have said it more than once before, for example on the 27th floor: "The working principle of capacitor voltage reduction is that the capacitor limits the current, and the current output by the rectifier bridge is basically fixed...

Teacher, if you have time, please take a look at how I determine the capacitance range of C101 capacitor. Is it correct?

maychang

Sirigu1992 posted on 2018-9-4 16:05 Teacher, if you have time, please take a look at how I determine the capacitance range of C101 to see if it is correct

K7812 efficiency is not calculated in this way. In fact, there is no need to calculate K7812 efficiency. We only need to know the load current. The load current includes the current through the two relay windings (according to you, the relay winding is 160 ohms, so the current of the two relay windings is 0.15A), plus other circuits (those transistors that control the action of the relays, etc.), plus the 8-pin M54123 at the top of the 26th floor picture (I don't know what it is, but this thing is also powered by K7812), plus the quiescent current of K7812 (estimated to be about 4mA, please check the datasheet for details). Leave a certain margin (if so much current is not needed, the excess current will flow away from the voltage regulator DZ101). The above calculation is the DC current, multiplied by 1.1 to get the AC current effective value. Calculate the capacitance of C101 based on the AC current effective value.

maychang

Sirigu1992 posted on 2018-9-4 16:05 Teacher, if you have time, please take a look at how I determine the capacitance range of C101 to see if it is correct

When calculating C101, be sure to calculate it based on the lowest grid voltage. The grid voltage regulations allow negative 10%, that is, 198V. But the actual situation may be lower.

maychang

Sirigu1992 posted on 2018-9-4 16:05 Teacher, if you have time, please take a look at how I determine the capacitance range of C101 to see if it is correct

It is not appropriate to use a 24V voltage regulator for DZ101. The minimum input voltage of K7812 is about 15V, so a 16~18V voltage regulator is enough.

西里古1992

maychang posted on 2018-9-4 17:07 K7812 efficiency is not calculated in this way. In fact, there is no need to calculate K7812 efficiency. We only need to know the load current. The load current includes the two relays...

The M54123 is a leakage detection chip. The power supply working current is 8mA, which is an order of magnitude lower than the load current of the two resistors of 0.15A. The static current of K7812 is also the same. Then the working current of the transistors Q101 and Q102 that control the relay to be attracted should also be very small when they are turned on (I am not very good at calculating the working current of the transistor). So I think the DC load current is at most 0.16A. The main load is the two resistors. Then the effective value of the AC current is 0.176A, the capacitive reactance Xc=100πC, U=198V, C is about 2.8uF... Just add the load currents together, which means that the currents at both ends of the input and output of K7812 are the same.

西里古1992

maychang posted on 2018-9-4 17:07 K7812 efficiency is not calculated in this way. In fact, there is no need to calculate K7812 efficiency. We only need to know the load current. The load current includes two relays...

K7812 static current datasheet is not found, only the no-load input current is 0.2mA

maychang

Sirigu1992 posted on 2018-9-4 18:41 The M54123 is a leakage detection chip. The power supply working current is 8mA, which is an order of magnitude lower than the load current of the two resistors of 0.15A. The K7...

"That means the current at the input and output of K7812 is the same." The input current is slightly larger than the output current, and the extra current is the current consumed by K7812 itself.

maychang

Sirigu1992 posted on 2018-9-4 18:45 I didn’t find the K7812 static current datasheet, only the no-load input current is 0.2mA

“I didn’t find the K7812 static current datasheet, only the no-load input current is 0.2mA” The no-load input current is the current consumed by the K7812 itself when the output is open. But this is not the static current, because the current consumed by the K7812 itself changes with the output current. However, the current consumed by the K7812 itself will not change much, and it is estimated that 4mA has a margin.

西里古1992

maychang posted on 2018-9-4 18:58 "I can't find the K7812 static current datasheet, only the no-load input current is 0.2mA" The no-load input current is the current consumed by the K7812 itself when the output is open...

In that complete circuit diagram, the 12V power supply is basically used to power M54123. The datasheet says the power supply current is 8mA, and there is a 0.15A resistive load. The green light is basically very small, 0.003A, and the static current is 0.4A. The collector current Ic=1.5A written in the datasheet for the control relays Q101 and Q102 should be the maximum current carrying value of the tube. When the tube is turned on, 12V is also used to power it. So what is the load current of this part?

maychang

Sirigu1992 posted on 2018-9-4 19:42 In that complete circuit diagram, the 12V power supply is basically used to power M54123. The datasheet says the power supply current is 8mA, and there is also a 0.15A resistive load, green...

"Then the collector current Ic=1.5A written on the datasheet for controlling relays Q101 and Q102 should be the maximum current carrying value of the tube." Yes.

maychang

Sirigu1992 posted on 2018-9-4 19:42 In the complete circuit diagram, the 12V power supply basically powers the M54123. The datasheet says the power supply current is 8mA, and there is also a 0.15A resistor load, green...

"When the tube is turned on, 12V also powers it, so what is the load current?" When Q101 and Q102 are turned on, they are in saturation state, and the collector current is mainly determined by the power supply voltage and the load connected to the collector (relay winding and 10 ohm resistor). When Q103 and Q104 are saturated and turned on, the collector current is determined by the power supply voltage and R119 and R120.

西里古1992

maychang published on 2018-9-4 20:28 "When the tube is turned on, 12V also supplies power to it, so what is the load current?" When Q101 and Q102 tubes are turned on, they are in saturation state, and the collector current...

Teacher, I have a few questions about the datasheet. Let me talk about them one by one. In this selection table, in the efficiency column, the minimum (VIN)/maximum (VIN)@full load, what do those numbers mean, 94/91, 84/85, etc. . . . . .

西里古1992

maychang published on 2018-9-4 20:28 "When the tube is turned on, 12V is also supplied to it. What is the load current?" When Q101 and Q102 are turned on, they are in saturation state, and the collector current...

There are many nominal input voltages in this output characteristic. I checked and found that the nominal voltage means the output voltage of the power supply when there is no load and no current output. How to understand the nominal input voltage? What is the input voltage of the power supply in the above situation?

西里古1992

maychang published on 2018-9-4 20:25 "Then the collector current Ic=1.5A written in the datasheet of Q101 and Q102, which control the relay, should be the maximum current carrying value of the tube." Yes.

The two figures give the positive output efficiency/input voltage (full load) and the positive output efficiency/output load (nominal input). Is the output current percentage the output current/maximum output current? Then, can we determine the corresponding efficiency by determining the output current based on the maximum output current? And can we estimate the corresponding input voltage based on the efficiency in the left figure? Is this how you read the graph?

maychang

Sirigu1992 posted on 2018-9-5 09:52 Teacher, I have a few questions about the datasheet. Let me talk about them one by one. In this selection table, in the efficiency column, the minimum (VIN)/maximum (VIN)@full...

It turned out to be a domestic chip with a Chinese manual. It seems that I made a mistake. This thing is not a linear regulator. Please post the complete datasheet. It is impossible to read it page by page.