I would like to build a power supply for a Vintage analog synthesizer kit that I picked up. There are no requirements provided besides voltage. It requires +/-15vdc and 5vdc. Has a microcontroller, 4 gate ic, comparator, a 555 timer and about 5 op amps. I do not have the specific consumption of the ICs yet so ball park would be fine. Would would be the best transformer ratings to use for this. I have heard 15-0-15 and I have heard 18-0-18. I would like some help with this and a refresher on the mathematics so that I can help myself in the future. I intend to build a few +/-15vdc, 5vdc power supplies over the next while. They will probably vary in physical size and current capability. I intend to use 79xx and 79xx regulators. How would I accomodate a system that exceeds the current max for 78xx and 79xx regulators?
Thanks
Thanks
A handy guide from Hammond is located here. It includes useful thumb rule equations.
A more detailed analysis can be found in this post.
Does the synthesizer have a built-in speaker? If so, I'd estimate a couple hundred mA for its amplifier. Add a couple hundred mA for the 555 (that's its max output), and a couple hundred for the op amps. Those are the parts I'm guessing use the ±15V (or any part of it), which totals much less than the 1A rated current for 78xx and 79xx regulators.
A 7805 should provide plenty of current for the MCU and logic.
So given an estimated total of 1.75 amps, a 15-0-15 @ 2A or 3A transformer will provide the needed power (with some to spare), using a half-bridge rectifier with capacitor filter. The half-bridge uses a regular bridge rectifier, but the transformer center-tap is grounded with the load(s). This in effect makes two full-wave supplies. Bridge + supplies the positive voltage and bridge - supplies the negative voltage.
The 78xx and 79xx can likely get by with small heat sinks; a 5V drop at 750mA is only 3.75 watts. The 7805 running off a 15V winding may have to dissipate about 15 watts. A finned heat sink like you see with the smaller chip amps will work for that.
To scale up for a bigger system, the logical thing to do is use a bigger regulator, like the LM350 or LM338. Higher currents probably require discrete series pass transistors. These can be used with any of the regulators already mentioned, or they can be designed using the old LM723 or even just a zener diode. These circuits are usually shown in the respective datasheets.
A more detailed analysis can be found in this post.
Does the synthesizer have a built-in speaker? If so, I'd estimate a couple hundred mA for its amplifier. Add a couple hundred mA for the 555 (that's its max output), and a couple hundred for the op amps. Those are the parts I'm guessing use the ±15V (or any part of it), which totals much less than the 1A rated current for 78xx and 79xx regulators.
A 7805 should provide plenty of current for the MCU and logic.
So given an estimated total of 1.75 amps, a 15-0-15 @ 2A or 3A transformer will provide the needed power (with some to spare), using a half-bridge rectifier with capacitor filter. The half-bridge uses a regular bridge rectifier, but the transformer center-tap is grounded with the load(s). This in effect makes two full-wave supplies. Bridge + supplies the positive voltage and bridge - supplies the negative voltage.
The 78xx and 79xx can likely get by with small heat sinks; a 5V drop at 750mA is only 3.75 watts. The 7805 running off a 15V winding may have to dissipate about 15 watts. A finned heat sink like you see with the smaller chip amps will work for that.
To scale up for a bigger system, the logical thing to do is use a bigger regulator, like the LM350 or LM338. Higher currents probably require discrete series pass transistors. These can be used with any of the regulators already mentioned, or they can be designed using the old LM723 or even just a zener diode. These circuits are usually shown in the respective datasheets.
VA is apparent power, obtained by multiplying the secondary volts by the secondary amperes. It's the same formula as for dc watts because it ignores the ac phase angle (theta).
.707 and 1.414 are reciprocals -- 1/1.414 = .707. For a sine wave, the peak voltage is 1.414 times the rms voltage. 1.414 is the square root of 2. These numbers are derived from the sine wave equation, Vp sin theta.
.707 and 1.414 are reciprocals -- 1/1.414 = .707. For a sine wave, the peak voltage is 1.414 times the rms voltage. 1.414 is the square root of 2. These numbers are derived from the sine wave equation, Vp sin theta.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.