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Help me generate needed frequency with SG3525

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Full Member level 5
Apr 12, 2007
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i have to construct an inverter using SG3525 a pulse width modulation controller.
please how can i generate the frequency that i needed using this device has i ve been having differnet formulae for from different datasheets.

f = 1.18÷Ct*Rt and

f = 1÷Ct (0.7Rt*3Rd)

f= frequency
Ct = timing capacitor in µf
Rt = timing resistor in kΩ
Rd = discharge resistor (this resistor might be ommited sometimes)

please i am confused becaue the two formulea are not giving the the same frequncy. which one is right.

good circuit or refrence will be appreciated

Re: help with SG3525

I don't have a paper, but I went to the ON Semi webiste and in their datasheet you can find the actual oscillator schematic:
See Fig. 7.
So let's try to derive our formula and see which one is correct.

From that figure, we can see that the cap is charged by a constant current, given by the current mirror built with Q1, Q3, Q5. The charging current is then:

Since Vbe≈0.6V, it follows that Ic≈3.8V/Rt

So the capacitor will charge linearly from a voltage V1 to a voltage V2. The comparator built with Q6, Q9 determines when the cap charges up and when it dischanrges, based on these two voltage levels, V1 and V2. To determine the charge time, we must know V1 and V2.
From the schematic, you see that V2 is simply given by the resistive divider 7.4K and 14K. So

V1 is the lower threshold and is obtained by adding the 2kohm resistor in parallel with the 14k one, by turning on Q11 when we reach V2. So the 14k now becomes:

With that,

So now we have the charging current and the voltage levels.
Then the charging time is:

But Ic=3.8/Rt, so

For the discharge time, we will neglect the case of low Rd and we will neglect the voltage drop across Q4, Q2 and assume they are perfect switches, just like we assumed above that Q11 was also perfect.

The cap discharges from V2 through Rd, following an exponential law:

Note that the cap tends to discharge towards V3, which is simply V3=Rd*Ic, that ism the voltage drop across Rd due to the charging current, which never gets turned off in this chip. Incidentally, that is why Rd is limited to 500 ohms, so that with even the maximum charging current of 1.9mA, which occurs for Rt=minimum, the voltage across the cap still crosses the V1 threshold.

Anyway, we will assume that Rd is sufficiently low for the cap's voltage to cross V1.

Since the cap discharges exponentially, we can calculate the discharge time by setting the voltage on the cap equal to V1, since once it reaches V1 the comparator will turn off Q2, Q4, allowing the cap to charge up again.


But V3=Rd*Ic=Rd*3.8/Rt=3.8*Rd/Rt

So you can now calculate td for a ratio of Rd/Rt. For Rd/Rt=10, we get:
td=-RdCt*ln((0.956-0.38 ) / (3.27-0.38 ))=1.613*RdCt
But we set Rd=0.1Rt, so we can say:

So the frequency is:

I have to stress that this formula is only valid for Rd=0.1Rt. For other ratios, you will get different values. But you should always get the right frequency if you just calculate tc and td separately and from that the frequency, without imposing any conditions on the ratio of the two resistors.

It seems the first formula you have was derived in a similar manner, but for a different ratio of resistors. I haven't tried to see what happens if the ratio is even larger, perhaps the constant really approaches 1.18. You can try that.

I don't know how the second formula was arrived at, but it could be that other things have been taken into consideration, or other assupmtions have been made. In fact, I am not sure it is correct, since dimensionally it is wrong: in the denominator you have Ct multiplied by Rt and Rd. That is not a time constant.

I hope this helps.
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